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Computing $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1^n+2^n+\cdots \cdots +n^n}{n^n}\right)$

Attempt: $\displaystyle \lim_{n\rightarrow \infty}\bigg[\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots \cdots +\left(\frac{n-1}{n}\right)^n+\left(\frac{n}{n}\right)^n\bigg] = 1$

Is my answer right? If not, then could someone help me? Thanks.

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  • $\begingroup$ yea you right . $\endgroup$ – Saketh Malyala Apr 7 '17 at 4:18
  • $\begingroup$ No. What is $(1-\frac{1}{n})^n$ as $n$ to infinity, the last term. Do you see why your reasoning is fallacious? $\endgroup$ – Ahmed S. Attaalla Apr 7 '17 at 4:19
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    $\begingroup$ Your attempt is only doing distributive and putting "equals 1". Not sufficient effort. $\endgroup$ – Santropedro Apr 7 '17 at 4:20
  • $\begingroup$ That is not correct: wolframalpha.com/input/?i=Sum%5B(k%2Fn)%5En,%7Bk,1,n%7D%5D $\endgroup$ – parsiad Apr 7 '17 at 4:20
  • $\begingroup$ @AhmedS.Attaalla do you mean the 2nd to last term? $\endgroup$ – browngreen Apr 7 '17 at 4:25
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Using $$ \lim_{n\rightarrow \infty} \left(1-\frac{a}{n}\right)^n= e^{-a} $$ Term by term $$ \displaystyle \lim_{n\rightarrow \infty}\bigg[\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots \cdots +\left(\frac{n-1}{n}\right)^n+\left(\frac{n}{n}\right)^n\bigg] = (e^{-n+1}+e^{-n+2}+\cdots +e^{-1}+1) = \frac{1}{1-e^{-1}} = \frac{e}{e-1} \sim 1.58198 $$

Numerically

  • $n = 10$, the sum equals 1.49143
  • $n = 100$, the sum equals 1.57211.
  • $n = 200$, the sum equals 1.57702.

Edit I have to warn that the term by term step is problematic (thanks a lot to the commentators!), as it uses $$ \lim_{n\rightarrow \infty} \frac{1}{n^n} = \frac{e}{e^{n}} $$ I couldn't figure out a good way of explaining this and make the whole process more rigorous (I am from physical sciences instead of mathematics), but I decide to leave the answer here as it should still be partially useful. I will attempt to refine it in the future.

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  • $\begingroup$ HOW DID I NOT RECOGNIZE THAT LIMIT o_O $\endgroup$ – Saketh Malyala Apr 7 '17 at 4:47
  • $\begingroup$ @Taozi Term by term You really need to elaborate on that step some more. It looks like you are exchanging some limits between terms and series along the way, without giving any justification why that would happen to be valid here (which it is not necessarily valid in general). $\endgroup$ – dxiv Apr 7 '17 at 5:06
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    $\begingroup$ While it seems that your final answer is correct, there's one step in your proof that I'm not so clear about. It's true that $\lim_{n\to \infty} \left(1-\frac{a}{n}\right)^n=e^{-a}$, but does that necessarily mean that $\lim_{n\to \infty}\left(1-\frac{n-1}{n}\right)^n=e^{-n+1}$? $\endgroup$ – browngreen Apr 7 '17 at 5:15
  • $\begingroup$ @RRL I didn't downvote anything. $\endgroup$ – browngreen Apr 7 '17 at 5:22

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