Find the value $$\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}.$$
This problem is from this and I am interested in this problem, but I can't solve it.
Here is my idea:
$$(-1)^{k-1}\dfrac{1}{n+k}=(-1)^{-n}\int_{0}^{-1}x^{n+k-1}dx$$ so \begin{align*}\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}&=\sum_{n=0}^{\infty}(-1)^n\left(\sum_{k=1}^{\infty}(-1)^n\int_{0}^{-1}x^{n+k-1}dx\right)^2\\ &=\sum_{n=0}^{\infty}(-1)^n\left(\int_{0}^{-1}\sum_{k=1}^{\infty}x^{n+k-1}dx\right)^2\\ &=\sum_{n=0}^{\infty}(-1)^n\left(\int_{0}^{-1}\dfrac{x^n}{1-x} dx \right)^2. \end{align*}