How to Evaluate $ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} $ How can I evaluate
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} \approx - 0.198909 $$
The Sum can be given also as
$$ \frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)\sqrt[4]{(-x)^{3}}}\,\left(\,\tan^{-1}\left(\sqrt[4]{-x}\right)-\tanh^{-1}\left(\sqrt[4]{-x}\right)\,\right) $$
Unfortunately i have not been able to evaluate either the Sum or the Integral using methods I know. Mathematica gives really weird results for the integral.
Is there a closed form for this Sum/Integral?
Thank you kindly for your help and time.
EDIT
For those of you that still care about the question i was able to find the following closed form. I will let the above $ sum = S $
and as such
$$ S =  C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} $$
Where $C$ denotes Catalan's constant.
Thank you very much once again to those who provided answers!
EDIT #2 (Proof as Requested )
I will not show this one (too much typing) but ,
$$S= \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} = 4 \sum_{n=1}^{\infty} (-1)^n \sum_{k=0}^{\infty} \frac{1}{(4k+3)} \frac{1}{(4k+(4n+3))} $$
next expand the terms on the RHS into a Matrix as such :
$$
\begin{matrix} 
\color{red}{+(\frac13\times\frac13)} & -(\frac13\times\frac17)& +(\frac13\times\frac1{11})& -(\frac13\times\frac1{15}) \\
\color{blue}{-(\frac17\times\frac13)} & \color{red}{+(\frac17\times\frac17)} & -(\frac17\times\frac1{11}) & +(\frac17\times\frac1{15})\\
\color{blue}{+(\frac1{11}\times\frac13)} & \color{blue}{-(\frac1{11}\times\frac17)}&\color{red}{+(\frac1{11}\times\frac1{11})}&-(\frac1{11}\times\frac1{15})\\ 
\end{matrix}
$$
The black terms x 4 are our desired sum
I then added the red and blue terms to  "complete"  the
matrix
One can then see that the matrix (complete) may be given as
$$ \left(\frac13-\frac17+\frac1{11}...\right)\left(\frac13-\frac17+\frac1{11}...\right) $$
which is just
$$P= \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{4n+3}\right)^{2} = \left(\frac{\pi}{4 \sqrt{2}}+\frac{\ln(\sqrt{2}-1)}{2 \sqrt{2}}\right)^2 $$
So
$$ P = \color{red}{\sum_{n=1}^{\infty} \frac{1}{(4n-1)^2}} + \color{blue}{\text{Blue terms}} + \text{Black terms} $$
but one can see that $ \color{blue}{\text{Blue terms}} = \text{Black terms} $
Therefore :
$$ P = \frac{\pi^2}{16}-\frac{C}{2}+\frac{S}{2} $$
Solve for S to find :
$$ S =  C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} $$
where $C$ denotes Catalan's Constant.
 A: I do not believe there is a "nice" closed form to this but one way to approximate would be:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n}\sum_{k=1}^n\frac 1{4k-1}>\frac 14 \sum_{n=1}^\infty\frac{(-1)^nH_n}{n}$$
A: Not an answer ...
The sum can be rewritten as
\begin{eqnarray*}
- \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{(n+m-1)(4m-1)}.
\end{eqnarray*}
This can be expressed as the following double integral
\begin{eqnarray*}
-\int_0^1 \int_0^1 \frac{ y^2 dx dy }{(1+x)(1+xy^4)}.
\end{eqnarray*}
Partial fractions do the $x$ intergration gives
\begin{eqnarray*}
-\int_0^1  \frac{ y^2  (\ln(2) -\ln(1+y^4) )dy }{1-y^4}.
\end{eqnarray*}
Hopefully some of these forms might give some one else a better start point for this problem.
Something similar ... (where $K$ is the Catalan constant)
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{(2n-1)(n+m-1)}=K 
\end{eqnarray*}
would certainly give hope that there is a "nice" closed form for your sum.
A: The result is not so bad. If
$$S=\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1}$$ For more legibility, I shall write $S$ as
$$S=\frac {A}{96}-i\frac B 4$$ where $A$ and $B$ contain real and complex parts.
$$A=24 C-5 \pi ^2+9 \log ^2\left(3-2 \sqrt{2}\right)+(24-6 i) \pi  \log \left(3-2
   \sqrt{2}\right)$$
$$B=\text{Li}_2\left(\frac{1+i}{2+\sqrt{2}}\right)-\text{Li}_2\left(\frac{1-i}{2+\sqrt{
   2}}\right)+\text{Li}_2\left(-\frac{1+i}{-2+\sqrt{2}}\right)-\text{Li}_2\left(-\frac{1-i}{-2+\sqrt{2}}\right)+ 
i \left(\text{Li}_2\left(i \left(-1+\sqrt{2}\right)\right)+\text{Li}_2\left(-i
   \left(1+\sqrt{2}\right)\right)\right)$$
$$S=-0.19890902742911208266537143997251410413430136724348\cdots$$
It is amazing that this number is very close to
$$\frac{1}{100} \left(\psi \left(\frac{1}{15}\right)+\psi
   \left(\frac{3}{16}\right)-\psi \left(\frac{7}{10}\right)\right)$$ which is
$ -0.1989090283$
