Conversion of $\int_0^1\frac{\log( 1+x)}{x}dx$ to summation We know,

$$\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^nf\left(\frac rn\right)=\int_0^1f(x)dx$$

Now,we have $\int_0^1\frac{\log( 1+x)}{x}dx$
So,$f(x)=\frac{\log (1+x)}{x}$.So we find $f(r/n)$ by replacing $x$ with $r/n$. Then putting in the summation form on LHS.
Converting to summation,
$$\lim_{n\to\infty}\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}$$
Now this should evaluate to $1-\frac{1}{2^2}+\frac1{3^3}-\cdots$.How do I carry on the simplification?
Thanks for any help!
 A: HINT:
Write \begin{align}\frac1r\log\left(1+\frac rn\right)&=\frac1r\left[\frac rn-\frac12\left(\frac rn\right)^2+\frac13\left(\frac rn\right)^3-\cdots\right]\\&=\frac1n-\frac r{2n^2}+\frac{r^2}{3n^3}-\cdots\end{align} so \begin{align}\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}&=1-\frac{\sum r}{2n^2}+\frac{\sum r^2}{3n^3}-\cdots\\&=1-\frac12\int_0^1r\,dr+\frac13\int_0^1r^2\,dr-\cdots\end{align} as each term is essentially a Riemann sum.
A: HINT
Following the previous suggestion by TheSimpliFire
$$\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}=1-\frac{\sum r}{2n^2}+\frac{\sum r^2}{3n^3}-\cdots\to1-\frac1{2\cdot 2}+\frac1{3\cdot 3}\ldots=$$$$=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^2}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12}$$
indeed from the well known results


*

*Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)

*Sum of Reciprocals of Squares of Odd Integers
we have
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}=\sum_{k=1}^\infty \frac{1}{(2k-1)^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}=\sum_{k=1}^\infty \frac{1}{(2k-1)^2}-\frac14\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}8-\frac{\pi^2}{24}=\frac{\pi^2}{12}$$
