🧠:<think>Alright, so I have this problem about a Christian spiritual warfare movement's growth modeled by a logistic function. There are two parts: the first one is to find the time it takes to reach half the carrying capacity, and the second is to modify the model with a sinusoidal growth rate factor. Let me tackle each part step by step.Starting with part 1. The logistic growth function is given by:[ N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{-rt}} ]We have the parameters: ( N_0 = 1000 ), ( K = 50,000 ), and ( r = 0.1 ). We need to find the time ( t ) when the number of followers ( N(t) ) reaches half of the carrying capacity. Half of 50,000 is 25,000. So, we set up the equation:[ 25,000 = frac{50,000}{1 + frac{50,000 - 1000}{1000} e^{-0.1t}} ]Let me simplify this equation step by step. First, let's compute the denominator:[ frac{50,000 - 1000}{1000} = frac{49,000}{1000} = 49 ]So, the equation becomes:[ 25,000 = frac{50,000}{1 + 49 e^{-0.1t}} ]To solve for ( t ), I can start by multiplying both sides by the denominator:[ 25,000 times (1 + 49 e^{-0.1t}) = 50,000 ]Divide both sides by 25,000:[ 1 + 49 e^{-0.1t} = 2 ]Subtract 1 from both sides:[ 49 e^{-0.1t} = 1 ]Divide both sides by 49:[ e^{-0.1t} = frac{1}{49} ]Now, take the natural logarithm of both sides:[ ln(e^{-0.1t}) = lnleft(frac{1}{49}right) ]Simplify the left side:[ -0.1t = lnleft(frac{1}{49}right) ]I know that ( ln(1/x) = -ln(x) ), so:[ -0.1t = -ln(49) ]Multiply both sides by -1:[ 0.1t = ln(49) ]Now, solve for ( t ):[ t = frac{ln(49)}{0.1} ]Calculating ( ln(49) ). Since 49 is 7 squared, ( ln(49) = 2ln(7) ). I remember that ( ln(7) ) is approximately 1.9459. So,[ ln(49) = 2 times 1.9459 = 3.8918 ]Therefore,[ t = frac{3.8918}{0.1} = 38.918 ]So, approximately 38.92 years. Let me check my steps again to make sure I didn't make a mistake. Starting from the logistic equation, plugging in the values, simplifying, solving for ( t ). It seems correct. Maybe I should verify with another approach.Alternatively, I know that in logistic growth, the time to reach half the carrying capacity is related to the doubling time, but in this case, it's not exactly doubling since it's logistic. However, the calculation seems solid. So, I think 38.92 years is correct. Maybe round it to two decimal places, so 38.92 years.Moving on to part 2. The scholar wants to include a spiritual factor ( S(t) ) which is a sinusoidal function:[ S(t) = A sin(omega t + phi) ]Given ( A = 0.05 ), ( omega = pi/2 ), and ( phi = 0 ). So,[ S(t) = 0.05 sinleft(frac{pi}{2} tright) ]The original logistic growth function is:[ N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{-rt}} ]We need to modify this to include the influence of ( S(t) ) on the growth rate ( r ). So, the growth rate ( r ) is now a function of time, ( r(t) = r + S(t) ). Wait, but is it additive or multiplicative? The problem says "influence of ( S(t) )", so it's probably additive. So, the new growth rate is ( r(t) = r + S(t) ).Therefore, the modified logistic function becomes:[ N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{-int_0^t r(t') dt'}} ]Because in the logistic equation, the exponent is the integral of the growth rate over time. So, since ( r(t) ) is time-dependent, we need to integrate it from 0 to t.So, let's compute the integral:[ int_0^t r(t') dt' = int_0^t (r + S(t')) dt' = int_0^t r dt' + int_0^t S(t') dt' ]Compute each integral separately.First integral:[ int_0^t r dt' = r t ]Second integral:[ int_0^t S(t') dt' = int_0^t 0.05 sinleft(frac{pi}{2} t'right) dt' ]Let me compute this integral. Let me set ( u = frac{pi}{2} t' ), so ( du = frac{pi}{2} dt' ), which means ( dt' = frac{2}{pi} du ). So, the integral becomes:[ 0.05 times frac{2}{pi} int_{0}^{frac{pi}{2} t} sin(u) du ]Compute the integral of ( sin(u) ):[ int sin(u) du = -cos(u) + C ]So,[ 0.05 times frac{2}{pi} left[ -cosleft(frac{pi}{2} tright) + cos(0) right] ]Simplify:[ 0.05 times frac{2}{pi} left[ -cosleft(frac{pi}{2} tright) + 1 right] ]Multiply constants:[ frac{0.1}{pi} left[ 1 - cosleft(frac{pi}{2} tright) right] ]So, putting it all together, the integral of ( r(t') ) from 0 to t is:[ r t + frac{0.1}{pi} left[ 1 - cosleft(frac{pi}{2} tright) right] ]Therefore, the exponent in the logistic function becomes:[ - left( r t + frac{0.1}{pi} left[ 1 - cosleft(frac{pi}{2} tright) right] right) ]So, substituting back into the logistic equation:[ N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{ - r t - frac{0.1}{pi} left[ 1 - cosleft(frac{pi}{2} tright) right] } } ]Simplify the exponent:[ - r t - frac{0.1}{pi} + frac{0.1}{pi} cosleft(frac{pi}{2} tright) ]So, the exponent is:[ - r t - frac{0.1}{pi} + frac{0.1}{pi} cosleft(frac{pi}{2} tright) ]Therefore, the modified logistic function is:[ N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{ - r t - frac{0.1}{pi} + frac{0.1}{pi} cosleft(frac{pi}{2} tright) } } ]Alternatively, we can factor out the constants:[ N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{ - r t } e^{ - frac{0.1}{pi} } e^{ frac{0.1}{pi} cosleft(frac{pi}{2} tright) } } ]But I think the first expression is sufficient. So, that's the new expression for ( N(t) ) incorporating the spiritual factor ( S(t) ).Let me double-check my steps. I considered ( r(t) = r + S(t) ), integrated it over time, computed each integral correctly, substituted back into the logistic equation. Seems good. The integral of the sinusoidal function was handled by substitution, leading to the expression involving cosine. Yes, that seems right.So, summarizing:1. The time to reach half the carrying capacity is approximately 38.92 years.2. The modified logistic function is:[ N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{ - r t - frac{0.1}{pi} + frac{0.1}{pi} cosleft(frac{pi}{2} tright) } } ]I think that's the solution.Final Answer1. The time to reach half the carrying capacity is boxed{38.92} years.2. The modified logistic growth function is boxed{N(t) = frac{K}{1 + frac{K - N_0}{N_0} e^{ - r t - frac{0.1}{pi} + frac{0.1}{pi} cosleft(frac{pi}{2} tright) } }}.</think>

🧠:<think>Alright, so Emma is working on her physics application for Cambridge, and she's diving into quantum mechanics. She came across this equation, which is the time-dependent Schrödinger equation. It looks like she's dealing with a particle in a magnetic field and wave functions. Cool, quantum mechanics is always fascinating!She has two sub-problems to solve. Let me tackle them one by one.Sub-problem 1: Solving the Time-Independent Schrödinger Equation for a Harmonic OscillatorOkay, so the potential is given as ( V(x) = frac{1}{2}momega^2 x^2 ). That's the standard harmonic oscillator potential. I remember that the solutions to the Schrödinger equation for this potential are well-known. The energy eigenvalues are quantized and the eigenfunctions are Hermite polynomials multiplied by a Gaussian.First, let's write down the time-independent Schrödinger equation for this potential:[ -frac{hbar^2}{2m} frac{d^2 psi_n(x)}{dx^2} + frac{1}{2}momega^2 x^2 psi_n(x) = E_n psi_n(x) ]This is a second-order differential equation. The solutions are the eigenfunctions ( psi_n(x) ) with corresponding eigenvalues ( E_n ).I recall that the energy levels for a quantum harmonic oscillator are given by:[ E_n = left(n + frac{1}{2}right)hbaromega ]where ( n = 0, 1, 2, ldots ). So, the ground state energy is ( frac{1}{2}hbaromega ), the first excited state is ( frac{3}{2}hbaromega ), and so on.The eigenfunctions are given by:[ psi_n(x) = left(frac{momega}{pi hbar}right)^{1/4} frac{1}{sqrt{2^n n!}} H_nleft(sqrt{frac{momega}{hbar}} xright) e^{-frac{momega x^2}{2hbar}} ]where ( H_n(x) ) are the Hermite polynomials of degree ( n ).So, for each ( n ), we have a specific eigenfunction and energy level.Sub-problem 2: Time Evolution of a Superposition StateEmma wants to find the time evolution of a wave function that's a superposition of the first two stationary states. The initial wave function is given by:[ psi(x,0) = A(psi_0(x) + psi_1(x)) ]First, we need to determine the normalization constant ( A ). Since the wave function is a superposition of two orthonormal eigenfunctions, the normalization condition is:[ int_{-infty}^{infty} |psi(x,0)|^2 dx = 1 ]Substituting ( psi(x,0) ):[ int_{-infty}^{infty} |A|^2 (|psi_0(x)|^2 + |psi_1(x)|^2 + psi_0(x)psi_1^*(x) + psi_0^*(x)psi_1(x)) dx = 1 ]But since ( psi_0 ) and ( psi_1 ) are orthonormal, the cross terms integrate to zero. Also, the integral of ( |psi_n(x)|^2 ) is 1 for each ( n ). Therefore:[ |A|^2 (1 + 1) = 1 implies |A|^2 = frac{1}{2} implies A = frac{1}{sqrt{2}} ]So, the normalized initial wave function is:[ psi(x,0) = frac{1}{sqrt{2}} (psi_0(x) + psi_1(x)) ]Now, to find ( psi(x,t) ), we need to consider the time evolution of each eigenstate. Each stationary state ( psi_n(x) ) evolves in time as:[ psi_n(x,t) = psi_n(x) e^{-i E_n t / hbar} ]Therefore, the time-evolved wave function is:[ psi(x,t) = frac{1}{sqrt{2}} left( psi_0(x) e^{-i E_0 t / hbar} + psi_1(x) e^{-i E_1 t / hbar} right) ]Substituting the energy eigenvalues ( E_0 = frac{1}{2}hbaromega ) and ( E_1 = frac{3}{2}hbaromega ):[ psi(x,t) = frac{1}{sqrt{2}} left( psi_0(x) e^{-i frac{1}{2}omega t} + psi_1(x) e^{-i frac{3}{2}omega t} right) ]We can factor out ( e^{-i frac{1}{2}omega t} ) to simplify:[ psi(x,t) = frac{1}{sqrt{2}} e^{-i frac{1}{2}omega t} left( psi_0(x) + psi_1(x) e^{-i omega t} right) ]This is the wave function at time ( t ). It's a superposition of the ground state and the first excited state, each evolving with their respective phase factors.Summary of Thoughts:For Sub-problem 1, recognizing the harmonic oscillator potential was key. Knowing the standard solutions for energy levels and eigenfunctions made it straightforward. For Sub-problem 2, understanding that each eigenstate evolves with its own phase factor was essential. Normalizing the initial wave function was also a crucial step to ensure the probabilities sum to 1. The final wave function is a time-dependent superposition, which would exhibit interference effects if measured.I think Emma is on the right track. She just needs to present these solutions clearly, making sure to explain each step, especially the normalization and the time evolution part. It's important to highlight how the superposition leads to time-dependent probabilities, which is a fundamental concept in quantum mechanics.Final AnswerSub-problem 1: The energy eigenvalues are ( E_n = left(n + frac{1}{2}right)hbaromega ) and the corresponding eigenfunctions are ( psi_n(x) = left(frac{momega}{pi hbar}right)^{1/4} frac{1}{sqrt{2^n n!}} H_nleft(sqrt{frac{momega}{hbar}} xright) e^{-frac{momega x^2}{2hbar}} ).Sub-problem 2: The time-evolved wave function is ( psi(x,t) = frac{1}{sqrt{2}} left( psi_0(x) e^{-i frac{1}{2}omega t} + psi_1(x) e^{-i frac{3}{2}omega t} right) ).So, the answers are:For Sub-problem 1:Energy eigenvalues: boxed{E_n = left(n + frac{1}{2}right)hbaromega}Eigenfunctions: boxed{psi_n(x) = left(frac{momega}{pi hbar}right)^{1/4} frac{1}{sqrt{2^n n!}} H_nleft(sqrt{frac{momega}{hbar}} xright) e^{-frac{momega x^2}{2hbar}}}For Sub-problem 2:Wave function: boxed{psi(x,t) = frac{1}{sqrt{2}} left( psi_0(x) e^{-i frac{1}{2}omega t} + psi_1(x) e^{-i frac{3}{2}omega t} right)}</think>

🧠:<think>Alright, so I'm trying to tackle this problem about a journalist analyzing empathy spread in a community using a directed graph model. The problem has two parts, and I need to think through each step carefully.Starting with part 1: We have a directed graph G with n vertices and m edges. Each edge has a weight representing the influence score. The task is to find a Hamiltonian path that maximizes the total influence score. Then, we need to prove that this problem is NP-complete and discuss the best-known algorithmic approach.Okay, so first, what's a Hamiltonian path? It's a path that visits every vertex exactly once. In this context, it would represent a sequence of interviews where each person is interviewed exactly once, and the total influence is the sum of the weights along the path.Now, the problem is to find such a path with the maximum total weight. This sounds a lot like the Longest Path problem in a graph. I remember that finding the longest path is different from the shortest path because, while the shortest path can be found efficiently with algorithms like Dijkstra's, the longest path is more tricky.Wait, isn't the Longest Path problem NP-hard? Yes, I think that's right. For general graphs, it's NP-hard. But here, the graph is directed, so does that change anything? Hmm, no, I don't think so. The directed version is still NP-hard. So, finding a Hamiltonian path that maximizes the total influence is equivalent to finding the longest path in a directed graph, which is NP-hard. Therefore, it's computationally challenging, specifically NP-complete.Wait, hold on. Is it NP-complete or just NP-hard? Because Hamiltonian path is a decision problem, which is NP-complete. The optimization version, which is finding the longest path, is NP-hard. So, in this case, since we're asked to find the path, it's an optimization problem, so it's NP-hard. But the question says to prove it's NP-complete. Maybe they consider the decision version, like whether there's a path with total influence at least K. That would be NP-complete.So, to prove that finding such a Hamiltonian path is NP-complete, I can reduce a known NP-complete problem to it. The classic approach is to reduce the Hamiltonian path problem, which is already NP-complete, to our problem. If we can show that solving our problem can be used to solve the Hamiltonian path problem, then our problem is at least as hard as Hamiltonian path, which is NP-complete.Alternatively, since the problem is about maximizing the total influence, which is essentially the longest path problem, and since the longest path problem is NP-hard, our problem is also NP-hard. But since it's an optimization problem, it's not technically in NP, so it's NP-hard, not NP-complete. Hmm, maybe the question is considering the decision version, so it's NP-complete.I think the key point is that finding a maximum weight Hamiltonian path is equivalent to the longest path problem, which is NP-hard. So, it's computationally challenging.Now, for the best-known algorithmic approach. Since it's NP-hard, exact algorithms are not feasible for large n. So, we need approximation algorithms or heuristic methods. But wait, the longest path problem is not just NP-hard; it's also APX-hard, meaning that it doesn't have a polynomial-time approximation scheme (PTAS) unless P=NP. So, we can't approximate it within any constant factor in polynomial time.Therefore, the best-known approaches are likely to be exponential-time algorithms with some optimizations, like dynamic programming with state compression or branch-and-bound methods. Alternatively, heuristic methods like genetic algorithms or simulated annealing might be used in practice, but they don't guarantee optimal solutions.Wait, but for specific cases, like if the graph has certain properties, such as being a DAG, the longest path can be found in linear time. But in the general case, especially with directed graphs without any restrictions, it's still NP-hard.So, summarizing part 1: The problem is NP-hard (or NP-complete if considering the decision version), and the best-known algorithms are exponential-time, possibly with some optimizations, or heuristic methods for approximation.Moving on to part 2: The journalist adds k new edges with influence scores from a distribution P with mean μ and variance σ². We need to analyze the expected change in the maximum total influence score and discuss the impact of μ and σ².Hmm, so initially, we have a graph G, and after adding k edges, we have a new graph G'. Each new edge has a weight from distribution P. We need to find the expected change in the maximum Hamiltonian path score.First, the maximum Hamiltonian path score in G is some value, say S. After adding k edges, the maximum score might increase because we have more possible edges to choose from. The expected change would be E[S'] - S, where S' is the new maximum score.But how do we model this? The addition of edges can potentially create new paths or improve existing ones. However, since the new edges are added with random weights, their influence depends on μ and σ².If μ is high, the new edges are likely to have higher weights, which could significantly increase the maximum path. Conversely, if μ is low, the new edges might not contribute much. The variance σ² affects how much the weights can vary. Higher variance means some edges could be much higher or lower than μ, potentially offering larger improvements but also more uncertainty.But quantifying the expected change is tricky. Maybe we can model it probabilistically. Each new edge has a weight X ~ P with E[X] = μ and Var(X) = σ². When we add k such edges, the expected total weight added is kμ. However, this doesn't directly translate to the maximum path because the edges are added to specific pairs of vertices, and whether they are used in the maximum path depends on their position in the graph.Wait, but in the maximum Hamiltonian path, only n-1 edges are used. So, adding k edges might provide alternative paths, but the maximum path will choose the best possible edges. So, the expected improvement would depend on how likely the new edges can be part of a better path.Alternatively, perhaps we can think of the maximum score as the sum of the top n-1 edges in the graph. But no, because the edges have to form a path, so it's not just the sum of the top edges, but a connected sequence.This seems complicated. Maybe we can consider that adding edges with higher μ increases the potential maximum, and higher σ² allows for more variability, which could lead to higher maximums if some edges are much larger.But in expectation, how does it change? Maybe the expected maximum increases by some function of μ, σ², and k. But I'm not sure about the exact form.Alternatively, perhaps we can model the problem as a random graph where edges are added with certain weights, and then the maximum Hamiltonian path is analyzed. But I don't recall specific results about this.Wait, maybe we can use linearity of expectation. The expected value of the maximum path after adding edges is the expectation over all possible additions of k edges. But the maximum is a non-linear function, so linearity doesn't directly apply.Alternatively, perhaps we can consider that each new edge has a certain probability of being included in the optimal path. If the new edge has a high weight, it might replace an existing edge in the path.But without knowing the structure of the original graph, it's hard to say. Maybe we can assume that the original graph's maximum path is S, and each new edge has a weight X_i, so the new maximum path could be S plus the sum of some X_i's that improve the path.But again, it's not straightforward because the new edges might not necessarily be on the optimal path.Alternatively, perhaps we can model the change as the expected maximum of the original maximum and the maximum over all possible paths that include the new edges.But this is getting too vague. Maybe the expected change is approximately kμ, but adjusted by the structure of the graph and the variance.Wait, but if the new edges are added between arbitrary pairs, the expected number of new edges that are part of the maximum path is small, maybe proportional to something like (n-1)/m, but I'm not sure.Alternatively, perhaps the expected maximum increases by roughly kμ divided by the number of possible edges, but this is speculative.I think the key points are:1. Adding edges with higher μ increases the expected maximum score because the new edges are more likely to have higher weights, potentially improving the path.2. Higher σ² introduces more variability, meaning there's a higher chance that some new edges have much higher weights, which could significantly increase the maximum score. However, it also means some edges could be much lower, but since we're taking the maximum, only the high ones matter.So, overall, the expected maximum score increases with both μ and σ², but the exact relationship is complex and depends on the graph structure and the number of added edges k.I might need to look into some literature or think about probabilistic methods for maximum paths, but I don't have a precise formula in mind.So, to sum up part 2: The expected maximum total influence score increases with the addition of k edges. The increase depends on the mean μ and variance σ² of the distribution. Higher μ directly increases the expected weights of the new edges, potentially improving the maximum path. Higher σ² increases the likelihood of having edges with significantly higher weights, which can lead to larger improvements in the maximum path. Therefore, both μ and σ² positively impact the empathy spread, with μ contributing a linear effect and σ² contributing a more probabilistic, potentially larger effect.But I'm not entirely confident about the exact mathematical relationship here. It might require more detailed probabilistic analysis or simulations.Final Answer1. Finding the Hamiltonian path with maximum total influence is NP-hard, and the best-known approach is an exponential-time algorithm. 2. The expected maximum influence increases with higher μ and σ², enhancing empathy spread.The final answers are:1. boxed{text{NP-hard}} 2. The expected maximum influence score increases with both (mu) and (sigma^2).</think>

🧠:<think>Okay, so I have this problem about a sequence of numbers from a comic book series. The sequence given is 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and so on. Hmm, let me think about what this sequence could be.First, looking at the numbers: 2, 3, 5, 7, 11, 13... These are all prime numbers, right? Yeah, primes are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, this sequence is just the list of prime numbers in order.Now, the first part of the problem asks me to identify the sequence and prove that the n-th term can be expressed as a function of n. Then, I need to discuss the limitations of this expression in terms of its applicability to all natural numbers.Alright, so if it's the sequence of prime numbers, then the n-th term is the n-th prime number. But is there a formula that can express the n-th prime number as a function of n? I remember that primes are kind of irregular and don't follow a simple arithmetic or geometric progression. So, is there a known formula for the n-th prime?I think there are some formulas, but they are quite complicated. For example, I recall something called the Wilson's theorem, which says that a number p is prime if and only if (p-1)! ≡ -1 mod p. But that's more of a test for primality rather than a formula to generate primes.There's also the prime-counting function, denoted as π(n), which gives the number of primes less than or equal to n. But that's not directly giving the n-th prime; it's the inverse function. So, if π(n) is the number of primes up to n, then the n-th prime would be the value of k such that π(k) = n. But that's more of a definition rather than a formula.I also remember that there are some explicit formulas for primes, like the one using the floor function and some complex expressions, but they are not very practical. For example, I think there's a formula involving the sum of cosines or something like that, but it's not straightforward.Wait, maybe the problem is expecting me to recognize that while primes can be defined, there isn't a simple closed-form expression for the n-th prime. So, perhaps the function is defined as p(n) = the n-th prime number, but without a closed-form formula. So, in terms of limitations, it's that we can't express p(n) in a simple algebraic expression; it's more of a piecewise or recursively defined function.Alternatively, maybe it's expecting me to use the prime number theorem, which approximates the distribution of primes. The prime number theorem says that the number of primes less than a number x is approximately x / ln x. So, the n-th prime is roughly around n ln n for large n. But that's just an approximation, not an exact formula.So, putting this together, the sequence is the prime numbers, and while we can define the n-th term as p(n), there's no known simple closed-form formula for p(n). Therefore, the limitations are that we can't express p(n) with a straightforward function; it's more of a definition based on primality rather than an explicit formula.Moving on to the second part. The high schooler finds that the comic book sequence is related to a complex mathematical object. They define an infinite series based on the reciprocal of the sequence's terms. So, the series would be 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + ... and so on.The question is to prove whether this series converges or diverges and explain the implications in the context of reading all issues of the comic book series.Alright, so the series is the sum of reciprocals of primes. I remember that the harmonic series, which is the sum of reciprocals of natural numbers, diverges. But what about the sum of reciprocals of primes?I think this is a known result. I recall that the sum of reciprocals of primes diverges. Let me try to remember the proof or the reasoning behind it.One approach is to use the fact that the primes are infinite, and their reciprocals, although decreasing, do not decrease fast enough for the series to converge.Alternatively, I think there's a proof using the Euler product formula for the Riemann zeta function. The zeta function is ζ(s) = sum_{n=1}^∞ 1/n^s, which converges for Re(s) > 1. Euler showed that this can be expressed as a product over primes: ζ(s) = product_{p prime} 1/(1 - 1/p^s).If we take the logarithm of both sides, we get ln ζ(s) = sum_{p prime} ln(1/(1 - 1/p^s)) = sum_{p prime} sum_{k=1}^∞ 1/(k p^{ks}).For s = 1, the left-hand side becomes ln ζ(1), but ζ(1) is the harmonic series, which diverges to infinity. So, ln ζ(1) is also infinity.On the right-hand side, for s = 1, we have sum_{p prime} sum_{k=1}^∞ 1/(k p^{k}).But wait, for each prime p, sum_{k=1}^∞ 1/(k p^{k}) is a convergent series because it's similar to the expansion of ln(1/(1 - 1/p)). Specifically, ln(1/(1 - x)) = sum_{k=1}^∞ x^k / k for |x| < 1.So, for each prime p, sum_{k=1}^∞ 1/(k p^{k}) converges to ln(1/(1 - 1/p)).Therefore, the right-hand side becomes sum_{p prime} ln(1/(1 - 1/p)).But the left-hand side is infinity, so the sum of ln(1/(1 - 1/p)) over primes must also diverge. Since ln(1/(1 - 1/p)) is approximately 1/p for small 1/p (using the approximation ln(1/(1 - x)) ≈ x for small x), the sum over primes of 1/p must diverge.Therefore, the series sum_{p prime} 1/p diverges.So, coming back to the problem, the series is the sum of reciprocals of primes, which diverges. The implications are that if the high schooler wants to "read all issues" of the comic book series, which would correspond to summing all terms of the series, they would have to read an infinite number of issues because the series doesn't converge to a finite sum. In other words, the total sum would grow without bound as they read more issues, meaning they can't finish reading all issues in finite time or with a finite effort.Alternatively, in the context of the comic book, if each issue corresponds to a term in the series, and the hidden code is related to this series, then the fact that the series diverges might imply that the codes become increasingly complex or that there's no finite point where the codes stop being meaningful or challenging.So, summarizing, the sequence is the prime numbers, and while we can define the n-th term, there's no simple closed-form expression. The series of reciprocals of primes diverges, meaning the sum grows without bound, which implies that the comic book series has an infinite number of issues with increasingly challenging codes.Final Answer1. The sequence is the list of prime numbers. The (n)-th term, while defined, lacks a simple closed-form expression. Thus, it cannot be easily expressed for all natural numbers. (boxed{p(n)}) represents the (n)-th prime number.2. The series of reciprocals of primes diverges. This implies that reading all issues would require an infinite effort. (boxed{text{The series diverges}}).</think>