Sum of integrals probably converges to $\ln(2)$ (Seemous 2018-2019)

Question

I was trying to solve the fourth problem of SEEMOUS 2019. It goes as follows:

(a) Let $n\geq1$ be an integer. Calculate $$ \int_0^1 x^{n-1}\ln x\,dx $$

I have proved that $$ \int_0^1 x^{n-1}\ln x \,dx= -\displaystyle \frac{1}{n^2} .$$ I need help (or some guidance) on

(b) Evaluate $$ \sum_{n=0}^{\infty}(-1)^n\left( \frac{1}{(n+1)^2}- \frac{1}{(n+2)^2}+ \frac{1}{(n+3)^2}-\ldots\right). $$

I tested the sum and for $n=10000$ and it was really close to $\ln2$.

To calculate $\int_0^1x^{n-1}\ln x \ dx$, we write $x=e^{-t}$ for $t\in[0,\infty)$. So $$I=\int_0^1x^{n-1}\ln x\ dx =-\int_0^\infty e^{-(n-1)t}\ t\ e^{-t}\ dt\,.$$ That is, $$I=-\int_0^\infty t e^{-nt}\ dt = -\frac{1}{n^2}\int_0^\infty s^{2-1} e^{-s}\ ds\,,$$ where $s=nt$. Thus, $$I=-\frac1{n^2}\Gamma(2)=-\frac1{n^2}(2-1)!=-\frac1{n^2}\,.$$

Answer 1

\begin{eqnarray*} S&=& \sum_{n=0}^{\infty}(-1)^n(\displaystyle \frac{1}{(n+1)^2}-\displaystyle \frac{1}{(n+2)^2}+\displaystyle \frac{1}{(n+3)^2}-...) \\ &=& \sum_{n=0}^{\infty} (-1)^n \sum_{m=0}^{\infty} (-1)^{m} \frac{1}{(n+m+1)^2} \end{eqnarray*} Let $i=n+m+1$ we get $i$ ... $i$ times ... so \begin{eqnarray*} S=\sum_{i=1}^{\infty} (-1)^{i+1} \frac{i}{i^2} = \cdots. \end{eqnarray*}

Answer 2

Using the identity $\int_0^1 x^{n+k-1}\,dx=\frac{1}{n+k}$ we have

$$\begin{align} \sum_{n=0}^\infty (-1)^n\sum_{k=1}^\infty (-1)^{k-1}\frac{1}{(n+k)^2}&=\sum_{n=0}^\infty (-1)^n\sum_{k=1}^\infty (-1)^{k-1}\int_0^1 x^{n+k-1}\,dx\,\int_0^1 y^{n+k-1}\,dy\\\\ &=\int_0^1 \int_0^1 \sum_{n=0}^\infty (-xy)^n\,\sum_{k=0}^\infty (-xy)^k\,dx\\\\ &=\int_0^1\int_0^1 \frac{1}{(1+xy)^2}\,dx\,dy\\\\\ &=\int_0^1 \frac{1}{1+y}\,dy\\\\ &=\log(2) \end{align}$$

as was to be shown!