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I was playing around on desmos and made the function $a(k) = \frac {k^k} {e^k k!}$. I was then curious if $\lim_{n \to \infty} \left[ \sum_{k=1}^n \left[ \frac {k^k} {e^k k!} \right] \right]$ converged or diverged.

I first used the divergence test, but got stuck after $\ln \left( \lim_{n \to \infty} \left[ \frac {n^n} {e^n n!} \right] \right) = x\left(\ln\left(x\right)-1\right)-\ln(x!)$

I then tried the ratio test, but got inconclusive results as I found $\lim_{n \to \infty} \left[ \frac {a_{n+1}} {a_n} \right] = 1$

I also considered Root Test, Integral Test, & Abel's Test but got stuck. How should I find the convergence/divergence of the series?

amWhy
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MathEnthusiastThatsBadAtMath
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1 Answers1

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The $ k $-th term of the series is $$ a_k = \frac{k^k}{e^k k!} $$

Stirling's approximation states that for large $ k $ $$ k! \sim \sqrt{2 \pi k} \left( \frac{k}{e} \right)^k $$ Substituting this into $ a_k $, we get $$ a_k = \frac{k^k}{e^k k!} \sim \frac{k^k}{e^k \cdot \sqrt{2 \pi k} \left( \frac{k}{e} \right)^k} $$ Simplify $$ a_k \sim \frac{k^k}{e^k \cdot \sqrt{2 \pi k} \cdot \frac{k^k}{e^k}} = \frac{1}{\sqrt{2 \pi k}} $$

Thus, for large $ k $, the general term behaves as $$ a_k \sim \frac{1}{\sqrt{2 \pi k}} $$

To determine whether the series converges or diverges, we analyze the asymptotic behavior of $ a_k $

From the above, $ a_k \sim \frac{1}{\sqrt{2 \pi k}} $. The series $$ \sum_{k=1}^\infty \frac{1}{\sqrt{k}} $$ is a divergent $ p $-series (since $ p = \frac{1}{2} \leq 1 $).

Since $ a_k \sim \frac{1}{\sqrt{2 \pi k}} $ and the comparison series $ \sum_{k=1}^\infty \frac{1}{\sqrt{k}} $ diverges, the original series also diverges by the limit comparison test.

The partial sums $ S_n = \sum_{k=1}^n \frac{k^k}{e^k k!} $ grow without bound as $ n \to \infty $. Therefore, the limit $$ \lim_{n \to \infty} \left[ \sum_{k=1}^n \frac{k^k}{e^k k!} \right] $$ diverges.

Blue Cat Blues
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