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Convergent, absolutely convergent or divergent: $$\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}$$

I have used ratio test because we got a fraction here and I think I did alright till the end:

$$\lim_{k\rightarrow \infty}\frac{\frac{2^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}}{\frac{2^{k}\cdot k!}{k^{k}}} = \lim_{k\rightarrow \infty}\frac{2^{k+1}\cdot (k+1)! \cdot k^{k}}{(k+1)^{k+1}\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2^{k}\cdot 2^1\cdot k! \cdot (k+1)\cdot k^{k}}{(k+1)^{k}\cdot (k+1)\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2k^{k}}{(k+1)^{k}}$$ $$=2\lim_{k\rightarrow \infty} \left( \frac{k}{k+1}\right)^{k}$$

Now I don't know (without a calculator) to what this would converge / diverge to... In the exam we are not allowed to use a calculator... So what to do?

The denominator will be greater than the enumerator by 1, so dividing each other we got something $<1$. We take exponent $k$ which is $\geq 1$, so we will end up with $<1$ again. Multiply this with 2 we get something $< 1$ but $> 0$ and thus the series is convergent...?

I hope I have described it well? Would you give me full points to this task? :D

Edit: I haven't described it well in the end. See the accepted answer and its comments! Thanks a lot to everyone - from every question I ask here, I always learn new things :-)

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    $\begingroup$ Use Stirling's formula. $\endgroup$ Sep 18, 2016 at 9:25
  • $\begingroup$ @Jean-ClaudeArbaut Isn't that a bit overkill? $\endgroup$
    – Arthur
    Sep 18, 2016 at 9:37
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    $\begingroup$ @Arthur When I see $k!/k^k$, it's a reflex... And given that with it the conclusion follows immediately, why not? But yes, the usual ratio test works too. $\endgroup$ Sep 18, 2016 at 9:42

2 Answers 2

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Hint:

Use that $$\lim_{k\to\infty}\left(1+\frac{1}{k}\right)^k=e$$

So, $$2\lim_{k\to\infty}\left(\frac{k}{k+1}\right)^k=\frac{2}{e}\lt1$$

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  • $\begingroup$ Oh to be honest I could have never seen that but I know that the thing above you wrote equals $e$. The way I solved is still correct or you think they want see it using your way? Because your way seems more precise than mine. $\endgroup$
    – cnmesr
    Sep 18, 2016 at 9:24
  • $\begingroup$ @cnmesr Apart from everything else, how do you say that two times something less than one is still less than one? $\endgroup$
    – GoodDeeds
    Sep 18, 2016 at 9:27
  • $\begingroup$ I got no proof but I've always thought: Something divided by something greater (by 1) will ALWAYS be lower than 1. And then, take the result and make it $^{k}$ where you know that $k \geq 1$. Here we also know, if the base is lower than 1, the complete result will also be, no matter how great $k$, at least greater or equal 1. Or that's not always the case? $\endgroup$
    – cnmesr
    Sep 18, 2016 at 9:30
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    $\begingroup$ @cnmesr Your argument says "Multiply this with 2 we get something <1 but >0" $\endgroup$
    – GoodDeeds
    Sep 18, 2016 at 9:34
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    $\begingroup$ Oh damn I forgot about the 2, sorry you are absolutely right! If there was just not this 2.. : / Alright then my end was wrong.. $\endgroup$
    – cnmesr
    Sep 18, 2016 at 9:35
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Observe \begin{align} \lim_{k\rightarrow \infty}\left(\frac{k}{k+1} \right)^k = \lim_{k \rightarrow \infty}\left(1-\frac{1}{k+1} \right)^{k+1}\left( 1-\frac{1}{k+1}\right)^{-1} = e^{-1}. \end{align}

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    $\begingroup$ I think it's easier to see $\left(\frac k{k+1}\right)^k = \frac{1}{\left(1 + \frac{1}{k}\right)^k}$, but it's the same thing in the end. $\endgroup$
    – Arthur
    Sep 18, 2016 at 9:26

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