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The area below $x \ln (1+x)$, $ \ x \in \left[0,1\right] $, can be written as the limit of the Riemann sum below.

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

This is trivial to convert into an integral and compute, but I was wondering whether one perhaps could spot a known series and algebraically work out the limit (i.e. not use integral calculus)?

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I somewhat doubt it; generally one goes the other way, interpreting an expression like this as an integral. –  Alex Becker Dec 9 '13 at 9:25

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It doesn't look like an easy sum to do directly, but you can do it by using the Taylor series for $\log(1+x)$, to get

$\displaystyle \sum_{k=1}^n \frac{k}{n^2} \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m} \left(\frac{k}{n}\right)^m$

and by reordering the sums (it converges absolutely if you avoid the $k=n$ term, which won't contribute alone anyway) you get

$\displaystyle \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m} \sum_{k=1}^n \frac{k^{m+1}}{n^{m+2}}.$

The sums $\sum_1^n k^p $ are known polynomials in $n$ of degree $p+1$, and in the limit as $n\to\infty$ we need only keep the leading term $(p+1)^{-1}n^{p+1}$, to get

$\displaystyle \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m(m+2)}=\frac{1}{2}\displaystyle \sum_{m=1}^\infty (-1)^{m+1}\left(\frac{1}{m}-\frac{1}{m+2}\right)$,

which is a nice telescoping sum where nearly everything cancels, leaving you with the answer $\frac{1}{4}$.

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How wonderful, so it was in fact possible. But I guess we were lucky that $x \approx 0$, I do not see how one would have been done it for, say, $x >> 0$. –  infinityreward2 Dec 9 '13 at 9:58

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