# Compute $\lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\left(\frac{k}{n^2} + 1\right)}$

Compute the limit

$$\lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\left(\frac{k}{n^2} + 1\right)}$$

At a first look, I only thought of Riemann sums, but I don't see how I may apply it. What else could I do? I need some hints, suggestions. Thanks!

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Where is the problem from? –  Davide Giraudo Sep 15 '12 at 10:37
@Davide Giraudo: I have it from one of my colleagues that asked me help him with this. Honestly, nothing seems to work here. –  Chris's sis Sep 15 '12 at 10:41
It would be great here a feedback from did. By the way, if we use Taylor expansion of $x^{x+1}$, and take into account the first 2 terms then we ge that the limit is 1/2. http://www.wolframalpha.com/input/?i=series+x^%28x%2B1%29. Maybe from here is easier. I think of using the squeeze theorem. –  Chris's sis Sep 15 '12 at 10:53
The above sounds like my general idea, which was to show that each summand is equal to $k/n^2$ plus some error. I suppose it remains to bound the sum of the errors. –  Christopher A. Wong Sep 15 '12 at 10:59
@Christopher A. Wong: I only want to make rigurous that point such that I may have a valid proof. I'm thinking of it right. –  Chris's sis Sep 15 '12 at 11:00

## 3 Answers

1. Upper bound For every $1\leqslant k\leqslant n$, $$\left(\frac{k}{n^2}\right)^{1+\frac{k}{n^2}}\leqslant\frac{k}{n^2}.$$ Summing up these and using the fact that the sum of the $n$ first positive integers is $\frac12n(n+1)$, one sees that the $n$th sum $S_n$ is such that $$S_n\leqslant\sum_{k=1}^n\frac{k}{n^2}=\frac{n+1}{2n}.$$ 2. Lower bound For every $1\leqslant k\leqslant n$, $$\left(\frac{k}{n^2}\right)^{1+\frac{k}{n^2}}\geqslant\left(\frac{k}{n^2}\right)^{1+\frac1{n}}.$$ The usual comparison of a sum with an integral, plus the fact that the function $u:x\mapsto x^{1+1/n}$ is increasing on $(0,1)$, yield $$S_n\geqslant n^{-1-1/n}\sum_{k=1}^nu(k/n)\geqslant n^{-1/n}\int_0^1u(x)\,\mathrm dx=\frac{n^{1-1/n}}{2n+1}.$$ 3. Coda The upper and lower bounds of $S_n$, which are valid for every $n\geqslant1$, both converge to $\frac12$ when $n\to\infty$ hence the gendarmes theorem ensures that $\lim\limits_{n\to\infty}S_n=\frac12$.

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thanks for your nice proof! –  Chris's sis Sep 15 '12 at 12:36

Let

\begin{align}x=\frac{k}{n^2}=O(\frac{1}{n})\end{align}

and when $n$ is sufficiently large,

\begin{align} (\frac{k}{n^2})^{\frac{k}{n^2}}=e^{x \log{x}}=1+O(x\log x) \end{align}

where the big O constant is absolute.

\begin{align} O(\sum_{k=1}^{n}\frac{k}{n^2}\frac{k}{n^2}\log{\frac{k}{n^2}})=O(\sum_{k=1}^{n}\frac{k}{n^2}\frac{k}{n^2}\log{k})+O(\sum_{k=1}^{n}\frac{k}{n^2}\frac{k}{n^2}\log{n^2})=o(1) \end{align}

Hence the principle part of the sum is

\begin{align} \sum_{k=1}^{n}\frac{k}{n^2}=\frac{1}{2}+o(1) \end{align}

Q.E.D.

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@y zhao: thank you for your solution (+1) –  Chris's sis Sep 15 '12 at 12:31

Yes, Riemann looks like a good idea. Using $n$ equidistant points with distances $\frac1{n^2}$ in the interval $\left[0,\frac1n\right]$, we expect $$\tag{1}\int_0^{\frac1n} x^{x+1} dx\approx \frac1{n^2}\sum_{k=1}^n \left(\frac k{n^2}\right)^{\frac k{n^2}+1}.$$ More specifically, the derivative of the integrand $f(x):=x^{x+1}=\exp((x+1)\ln x)$ is $f'(x)=(\ln x + 1+\frac1x)x^{x+1}$. Note that $\ln x + 1 + \frac1x=-y+1+e^y$ with $y=-\ln x$ and $y\to+\infty$ as $x\to 0^+$. For sufficiently small $x$ the exponential in $y$ will dominate the polynomial in $y$, i.e. $f'$ will be positive and hence $f$ strictly increasing. Therefore, the $\approx$ in $(1)$ can be gotten rid of for sufficiently big $n$ as follows: $$\tag{2}\int_0^{\frac1n} x^{x+1} dx\le \frac1{n^2}\sum_{k=1}^n \left(\frac k{n^2}\right)^{\frac k{n^2}+1}\le\int_{\frac1{n^2}}^{\frac1n+\frac1{n^2}} x^{x+1} dx.$$ We may additionally assume that $\frac1n+\frac1{n^2}\le 1$ and hence that the integrand is $\le x^1$. This makes the right hand side integral of $(2)$ $$\le \int_{\frac1{n^2}}^{\frac1n+\frac1{n^2}} x dx=\frac12\left(\left(\frac1n+\frac1{n^2}\right)^2-\left(\frac1{n^2}\right)^2\right)=\frac{n+2}{2n^3}$$ Thus for almost all $n$ $$\tag{3}\sum_{k=1}^n \left(\frac k{n^2}\right)^{\frac k{n^2}+1}\le\frac{n+2}{2n}=\frac12+\frac1n.$$

For $0\le x\le \frac1n<1$ we have $x^{x+1}\ge x^{1+\frac1n}$, therefore the left hand side of $(2)$ is $$\ge \int_0^{\frac1n}x^{1+\frac1n}dx=\frac1{2+\frac1n}\cdot\left(\frac1n\right)^{2+\frac1n}$$ and hence for almost all $n$ $$\tag{4}\sum_{k=1}^n \left(\frac k{n^2}\right)^{\frac k{n^2}+1}\ge \frac1{2+\frac1n}\sqrt[n]n.$$ Since $\sqrt[n]n\to 1$, the bounds in $(3)$ and $(4)$ both converge to $\frac12$, hence finally $$\lim_{n\to\infty}\sum_{k=1}^n \left(\frac k{n^2}\right)^{\frac k{n^2}+1}=\frac12.$$

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I don't think the sum is $0$. You may check this with wolfram alpha. –  Chris's sis Sep 15 '12 at 11:44
Hm, did I lose a factor of $n$ somewhere? –  Hagen von Eitzen Sep 15 '12 at 12:07
@Chris's sister: Yes, apparently I dropped a factor of $\frac1n$ right in the beginning. Thanks for the hint. –  Hagen von Eitzen Sep 15 '12 at 12:19
your approach of things is somehow special. Thank you! (+1) –  Chris's sis Sep 15 '12 at 12:31