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I have the following problem: $$\sum_{k=1}^{\infty}\frac{2^{k}k!}{k^{k}}$$

I only need to find whether the series converges or diverges. My initial thinking was to use the ratio test. I hit a stump after simplifying my limit, though:


Normally I'd just derive by L'Hopital's Rule, but the derivative is pretty long and complicated. I doubt that's the correct route.

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Hint: $(k+1)^{k+1} = (k+1)(k+1)^{k}$. – Nigel Overmars Apr 13 '14 at 0:24
Hint: $\lim k\cdot \log(k/(k+1)) = \lim (1/k^2) \cdot (-1/k^2 \cdot k/(k+1))^{-1} = -1$. – Chris K Apr 13 '14 at 5:58
up vote 3 down vote accepted

$$ \frac{2k^{k}(k+1)}{(k+1)^{k+1}} = \frac {2}{ \left(1+1/k\right)^k} \to \frac 2e <1 $$

You can also find the result using the Stirling formula.

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$$\lim\frac{2^{k+1}\cdot(k+1)!\cdot k^k}{(k+1)^{k+1} \cdot 2^k\cdot k!}=\lim\frac{2\cdot (k+1)\cdot k^k}{(k+1)^{k+1}}=2 \lim \frac{k^k}{(k+1)^k}=$$

$$=2\lim \left(\frac{k}{k+1}\right)^k=2\lim \left( 1-\frac{1}{k+1}\right)^k$$

Now let $t=k+1$, so

$$2 \lim\left(1-\frac{1}{t} \right)^{t-1}= 2 \lim \left(1-\frac{1}{t} \right)^t\cdot \left(1-\frac{1}{t} \right)^{-1}=\frac{2}{e}<1$$

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As suggested by mookid, if you use Stirling approximation $$k! \simeq \sqrt{2 \pi } e^{-k} k^{k+\frac{1}{2}}$$ defining $$u_k=\frac{2^{k}k!}{k^{k}}$$ and replacing $k!$ by the above expression you have $$\frac{u_{k+1}}{u_k}=\frac{2 \sqrt{k+1}}{e \sqrt{k}}$$ the limit of which being $\frac {2}{e}$ as already given to you in other answers.

What is nice it that using the above, the summation can be approximated and $$\sum_{k=1}^{\infty}\frac{2^{k}k!}{k^{k}}\simeq \sqrt{2 \pi } \text{Li}_{-\frac{1}{2}}\left(\frac{2}{e}\right)=12.5684$$ while the exact value is equal to $12.9490$

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