Mathematics

Mathematics

Problem

For which set of real values of pp does the series ∑n=1∞(−1)nnpen\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\frac{n^{p}}{e^{n}} converge? a) Only p<0p<0 b) Only p>0p>0 c) All real pp d) Only −1<p<1-1<p<1
Intuitive explanation

This is an alternating series, so you need to check two things: whether the terms decrease to zero, and whether absolute convergence holds. The key insight is that the exponential ene^n in the denominator dominates any polynomial npn^p as n→∞n \to \infty, which handles the limit condition easily. The real question is whether ∑npen\sum \frac{n^p}{e^n} converges absolutely—and this boils down to comparing growth rates: does the exponential decay overpower the polynomial growth for your value of pp?

Intuitive explanation

This is an alternating series, so you need to check two things: whether the terms decrease to zero, and whether absolute convergence holds. The key insight is that the exponential ene^n in the denominator dominates any polynomial npn^p as n→∞n \to \infty, which handles the limit condition easily. The real question is whether ∑npen\sum \frac{n^p}{e^n} converges absolutely—and this boils down to comparing growth rates: does the exponential decay overpower the polynomial growth for your value of pp?

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