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I found this problem in my deceased grandpa's note today when I was visiting my grandma's home. \begin{equation} \lim_{j\to0}\lim_{k\to\infty}\frac{k^j}{j!\,e^k} \end{equation} I asked my brother and he said the answer is $\cfrac{1}{2}$, but as usual, he didn't give me any explanation why the answer is $\cfrac{1}{2}$. I only get the indeterminate form $\infty^0$ in the numerator part if I substitute $j=0$ and $k=\infty$. I have no idea how to answer this problem. Could anyone here help me to answer it? I really appreciate for your help. Muchas gracias!

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  • $\begingroup$ Who did give my question a bounty? $\endgroup$ – Anastasiya-Romanova 秀 Apr 19 '14 at 14:25
  • $\begingroup$ I did .$\phantom$ $\endgroup$ – user135041 Apr 28 '14 at 23:08
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The sum was probably transcribed wrong somewhere along the way,because it doesn't make much sense as written (the limit as $j \rightarrow 0$ of something involving $j!$ ?). But since the answer is supposed to be $1/2$, I'm guessing that the intended formula was something like $$ \lim_{k \rightarrow \infty} \sum_{j=0}^k \frac{k^j} {j! e^k}, $$ which I'm sure I've seen before (possibly on this forum or on mathoverflow) but can't easily locate a reference. If we had replaced the $\sum_{j=0}^k$ by $\sum_{j=0}^\infty$ then the sum would equal $1$ for all $k$, because on writing $$ \sum_{j=0}^\infty \frac{k^j} {j! e^k} = \frac1{e^k} \sum_{j=0}^\infty \frac{k^j}{j!} $$ we recognize the sum as the power series for $e^k$. So the problem is asking in effect to prove that as $k\rightarrow\infty$ the $\sum_{j=0}^k$ part of the sum is asymptotically half of the entire $\sum_{j=0}^\infty$ sum. To see that this is at least plausible, observe that the $j=k$ term is the largest, but still accounts for only $O(1/\sqrt k)$ of the total, and the $j=k+1, \, k+2, \, k+3, \, \ldots$ terms are approximately equal to the $j=k-1, \, k-2, \, k-3, \, \ldots$ terms respectively. There are various ways to finish the proof, and your brother can probably point you towards one of them :-)

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  • $\begingroup$ Hmm, this is supposed to be the complete problem. It seems make sense. I've learnt Taylor series by myself, the answer should be 1. OK, I'll accept your answer if I can prove the intended formula that you proposed equals half. $\endgroup$ – Anastasiya-Romanova 秀 Apr 20 '14 at 8:50
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    $\begingroup$ I forgot to say thanks. Thank you Sir for helping me out and responding my message. (^-^) $\endgroup$ – Anastasiya-Romanova 秀 Apr 20 '14 at 8:52
  • $\begingroup$ I understand now Prof! Although I had lots of trouble to prove that it equals half. Thank you so much for your help. ≧◠◡◠≦✌ $\endgroup$ – Anastasiya-Romanova 秀 Apr 21 '14 at 15:32
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Unfortunately, your brother is incorrect. Note that $$\lim_{k\to \infty} \dfrac{k^j}{j! e^k} = 0$$ Hence, $$\lim_{j \to 0}\lim_{k\to \infty} \dfrac{k^j}{j! e^k} = \lim_{j\to0}0 = 0$$

Even by mistake, if he had interchanged the order of the limiting process, we still get the limit as $0$, since $$\lim_{j\to 0} \dfrac{k^j}{j! e^k} = \dfrac1{e^k}$$ and therefore $$\lim_{k\to \infty} \lim_{j \to 0}\dfrac{k^j}{j! e^k} = \lim_{k\to \infty} \dfrac1{e^k} = 0$$

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  • $\begingroup$ Luckily, you've edited your answer. I almost lost my temper and voted down your answer when you said "my deceased grandpa is incorrect". Just kidding, I won't do such a bad thing. Definitely, I appreciate your answer and also every answer. Hmm, your answer seem correct that was also my first thought, but why did my brother say the answer is $\cfrac{1}{2}$? Did he trick me again?? $\endgroup$ – Anastasiya-Romanova 秀 Apr 19 '14 at 14:22

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