Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – user 1618033 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.^%28x%2B1%29. Maybe from here is easier. I think of using the squeeze theorem. – user 1618033 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. – user 1618033 Sep 15 '12 at 11:00
up vote 5 down vote accepted

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 $\lim\limits_{n\to\infty}S_n=\frac12$.

share|cite|improve this answer
thanks for your nice proof! – user 1618033 Sep 15 '12 at 12:36



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

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


share|cite|improve this answer
@y zhao: thank you for your solution (+1) – user 1618033 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 $$\lim_{n\to\infty}\sum_{k=1}^n \left(\frac k{n^2}\right)^{\frac k{n^2}+1}=\frac12.$$

share|cite|improve this answer
I don't think the sum is $0$. You may check this with wolfram alpha. – user 1618033 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) – user 1618033 Sep 15 '12 at 12:31

Let $x: = \frac {k} {n^2}$. Then, we have $$\lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\left(\frac{k}{n^2} + 1\right)} = \lim_{n\to\infty} n^2 \int_{1/n^2}^{1/n} x^{x + 1}\, dx.$$ Trapezoidal rule (in quadrature) gives $$\int_{1/n^2}^{1/n} x^{x + 1}\, dx = \frac {1} {2} \left(\frac {1} {n} - \frac {1} {n^2} \right) \left(\left(\frac {1} {n}\right)^{1 + \frac {1} {n}} + \left(\frac {1} {n^2}\right)^{1 + \frac {1} {n^2}}\right) + R_n,$$ where $R_n \to 0$.

This is $$= \frac {1} {2 n^2} (n - 1) \left (\left(\frac {1} {n}\right)^{1 + \frac {1} {n}} + \left(\frac {1} {n}\right)^{2 + \frac {2} {n^2}}\right) + R_n = \frac {n - 1} {2 n^3} \left (\left(\frac {1} {n}\right)^{\frac {1} {n}} + \left(\frac {1} {n}\right)^{1 + \frac {2} {n^2}}\right) + R_n.$$ Since $$\left (\left(\frac {1} {n}\right)^{\frac {1} {n}} + \left(\frac {1} {n}\right)^{1 + \frac {2} {n^2}}\right) \to 1 \qquad \text{and} \qquad R_n \to 0,$$ we have $$\int_{1/n^2}^{1/n} x^{x + 1}\, dx \to \frac {n - 1} {2 n^3}.$$ Hence, $$\sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\left(\frac{k}{n^2} + 1\right)} \to \frac {n - 1} {2 n} \to \frac {1} {2}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.