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Let $$L=\lim_{n\to\infty} \frac{1^n+2^n+\cdots+n^n}{n^n}.$$ If we write: $$L=\lim_{n\to\infty}\left[\left(\frac1n\right)^n+\left(\frac2n\right)^n+\cdots+\left(\frac nn\right)^n\right].$$ Then this doesn't help due to the fact that as $n\to\infty,$ each of the individual terms $\left(\frac1n\right)^n, \left(\frac2n\right)^n,\ldots$ is of the form $0^{\infty}$ or equivalently $0×\infty$ excet for the last term $\left(\frac nn\right)^n=1^n=1.$ But later I realized that if we write: \begin{align} L&=\lim_{n\to\infty} \frac{n^n+(n-1)^n+(n-2)^n+\cdots+1^n}{n^n}\\\\ &=\lim_{n\to\infty} \left[1+\left(1-\frac1n\right)^n+\left(1-\frac2n\right)^n+\cdots+\left(\frac1n\right)^n\right]\\\\ &=1+e^{-1}+e^{-2}+\cdots,\text{ as }n\to\infty\text{ so number of terms is surely infinite}\\\\ &=\frac1{1-e^{-1}}\\\\ &\color{red}{=\frac{e}{e-1}.} \end{align} I don't know the exact answer. Tried Wolfram Alpha but it says that the input is wrong. I'm confused now. Is my approach correct? If yes, is there any other way to show the same.. Please help! Thanks in advance.

Dhrubajyoti Bhattacharjee
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  • If you just want to know if you have the correct limit, have you tried looking at the sequence of partial sums and seeing if it converges to $\frac{e}{e-1}$? Looking at the sum of the first 10,000 terms we get $1.58187710211243\dots$, where $\frac{e}{e-1} = 1.58197670686933\dots$. So if I had to guess, I would say that your limit is correct. – user413766 Jun 22 '20 at 16:46
  • Got it thank you sir. BTW how did you calculate the sum of first 10000 terms!? – Dhrubajyoti Bhattacharjee Jun 22 '20 at 17:02
  • Don’t you need to show you can exchange limit and sum – Vivaan Daga Jun 22 '20 at 17:15
  • @Vivaan Daga, Yeah! This can be shown by dominated convergence theorem. At that time, I didn't think of it. – Dhrubajyoti Bhattacharjee Jun 22 '20 at 23:04

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