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Use the ratio test to determine if the following series converges or diverges:

$$\begin{align*} \sum_{k=1}^{\infty}\frac{k!}{k^{k}} \end{align*}$$

Trying to solve this problem, used algebra to simplify to $\frac{k^k}{(k+1)^k},$ but from there is goes off the tracks.

Thomas Andrews
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Caleb Raes
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2 Answers2

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Use Taylor's theorem: $\log(1-x) = -x + o(x)$ as $x \to 0$.

As $k \to \infty$, $$ \log \frac{k^k}{(k+1)^k} = k \log\frac{k}{(k+1)} =k\log\left(1-\frac{1}{k+1}\right) =k\left(-\frac{1}{k+1}+o(1/k)\right) = \frac{-k}{k+1}+o(1) . $$ So $$ \log \frac{k^k}{(k+1)^k} \to -1 , \\ \frac{k^k}{(k+1)^k} \to e^{-1} . $$

GEdgar
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  • So you do not define $\mathrm{e}$ to be the limit of $\left(1+\frac{1}{k}\right)^k$. – Gary Mar 29 '23 at 01:18
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    @Gary ... Some texts may do that... I refer to Spivak, where $\log$ is defined by $$\log x = \int_1^x\frac{dt}{t}$$ and $e$ is defined by $$\log(e)=1$$ and $e^x$ is the inverse of $\log x$. – GEdgar Mar 29 '23 at 01:24
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    The solution is an overkill. Usually calculus textbooks cover the differential calculus, the Taylor expansions, of $\log (1+x)$ in particular, long after studying the limit $(1+k^{-1})^k.$ Another overkill solution can apply the Stirling formula and the ratio test. – Ryszard Szwarc Mar 29 '23 at 02:25
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Let $a_k$ equal $\frac{k!}{k^k}$.

Then we have,

$$\left| \frac{a_{k+1}}{a_k} \right| = \frac{\frac{(k+1)!}{(k+1)^{k+1}}}{\frac{k!}{k^k}} = k\cdot \frac{k^k}{(k+1)^{k+1}} = \frac{k^{k+1}}{(k+1)^{k+1}}$$

$$= \left(\frac{1}{1+\frac{1}{k}}\right)^{k+1} = \frac{1}{\left(1+\frac{1}{k}\right)^{k+1}} \to \frac{1}{e} < 1.$$

Therefore the series

$$\sum_{k\ge 1}{a_k}$$

converges.

Luca Armstrong
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