How find this $\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$

How find this
$\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$

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:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=547000&p=3185087#p3185087
and I very interesting This problem,But I can't solve it.Thank you
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*}
and $$