How to Evaluate $ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} $

This is going to be a long answer.

To start, you can use the polygamma function of order $0$ (also known as the digamma function) to shorten the expressions a little bit. This is tricky though - the series definition $$\psi^{[n]}(z)=(-1)^{n+1}n!\sum_{k=0}^\infty \frac{1}{(z+k)^{n+1}}$$ is only valid for $n>0$, whereas for $n=0$ one must use the derivative definition $$\psi^{[0]}(z)=\frac{\mathrm{d}}{\mathrm{d}z}\ln(\Gamma(z))=\frac{\Gamma'(z)}{\Gamma(z)}$$ Because for $n=0$ the series definition doesn't converge.

Despite this however, you can "abuse" the series definition for finite sums and write $$\sum_{k=1}^n \frac{1}{ak+b}=\frac{1}{a}\left(\psi^{[0]}\left(\frac{b}{a}+n+1\right)-\psi^{[0]}\left(\frac{a+b}{a}\right)\right)$$ This is difficult, but not impossible to prove. We can then write our sum as $$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1}=\sum_{n=1}^{\infty} \left[\frac{(-1)^n}{n}\frac{1}{4}\left(\psi^{[0]}\left(n+\frac{3}{4}\right)-\psi^{[0]}\left(\frac{3}{4}\right)\right)\right]$$ In order to calculate $\psi^{[0]}(3/4)$ we can use the two formulas $$\psi^{[0]}(1-z)-\psi^{[0]}(z)=\pi\cot(\pi z)$$ $$\psi^{[0]}(2z)=\frac{1}{2}\psi^{[0]}(z)+\frac{1}{2}\psi^{[0]}\left(z+\frac{1}{2}\right)+\ln 2$$ And plug in $z=1/4$ to obtain a linear system $$\begin{bmatrix} 1 & -1\\ 1/2 & 1/2 \end{bmatrix}\begin{bmatrix} \psi ^{[ 0]}( 3/4)\\ \psi ^{[ 0]}( 1/4) \end{bmatrix} =\begin{bmatrix} \pi \cot( \pi /4)\\ \psi ^{[ 0]}( 1/2) -\ln 2 \end{bmatrix}$$ If we use the well known identity $\psi^{[0]}(1/2)=-\gamma-2\ln 2$ ($\gamma$ being the Euler-Mascheroni constant) we can solve the system to obtain $$\psi^{[0]}(1/4)=-\frac{\pi}{2}-\gamma-\ln(8)$$ $$\psi^{[0]}(3/4)=\frac{\pi}{2}-\gamma-\ln(8)$$ Now, $$\mathcal{S}=\frac{1}{4}\sum_{n=1}^\infty\frac{(-1)^n\psi^{[0]}\left(n+\frac{3}{4}\right)}{n}+\frac{1}{4}\left(\gamma+\ln(8)-\frac{\pi}{2}\right)\sum_{n=1}^\infty\frac{(-1)^n}{n}$$

The latter is a well known sum and is equal to $-\ln(2)$. As for the former, perhaps you can use the asymptotic expansion of the digamma function

$$\psi^{[0]}(z)\asymp \ln(z)-\frac{1}{2z}-\sum_{n=1}^\infty\frac{B_{2n}}{2nz^{2n}}$$

$B_k$ being the $k$th Bernoulli number. The first few terms are $$\psi^{[0]}(z)\approx \ln(z)-\frac{1}{2z}-\frac{1}{12z^2}+\frac{1}{120z^4}-\frac{1}{252z^6}+\frac{1}{240z^8}+...$$

Hopefully $\psi^{[0]}(z)\approx \ln(z)-\frac{1}{2z}-\frac{1}{12z^2}$ will already give us a reasonable approximation. So now, $$\mathcal{S}\approx \frac{1}{4}\underbrace{\sum_{n=1}^{\infty}\frac{(-1)^n\ln\left(n+\frac{3}{4}\right)}{n}}_{S_1}-\frac{1}{8}\underbrace{\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+3/4)}}_{S_2}-\frac{1}{48}\underbrace{\sum_{n=1}^\infty\frac{(-1)^n}{n(n+3/4)^2}}_{S_3}-\frac{\ln(2)}{4}\left(\gamma+\ln(8)-\frac{\pi}{2}\right)$$ With some work, the second sum can be decomposed into partial fractions to obtain $$S_2=\sum_{n=1}^\infty\frac{(-1)^n}{n(n+3/4)}=4\left(\frac{4}{9}-\frac{\pi}{3\sqrt{2}}-\frac{\ln(2)}{3}+\frac{\sqrt{2}}{3}\ln(1+\sqrt{2})\right)$$ Or perhaps one could use the properties of the Lerch transcendent: $$\sum_{n=1}^\infty\frac{(-1)^n}{n^2+an}=\frac{\Phi(-1,1,a+1)-\ln(2)}{a}$$ To compute the Lerch transcendent one can use an integral identity $$\Phi(z,s,\alpha)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-ze^{-x}}\mathrm{d}x ~~~|~~~\operatorname{Re}(s),\operatorname{Re}(\alpha)>0~;~z\in\mathbb{C}~\backslash~ [1,\infty)$$ I'm not the greatest at integration, but Mathematica produces $$\Phi\left(-1,1,\frac{7}{4}\right)=\int_0^\infty\frac{e^{-7x/4}}{1+e^{-x}}$$ $$=\frac{4}{3}-2(-1)^{1/4}\arctan((-1)^{1/4})+2(-1)^{1/4}\operatorname{arctanh}((-1)^{1/4})$$ Which, by using the complex definitions of arctan, arctanh and ln, one can arrive at the form we got before.

The first sum, $S_1$, is the most bothersome. Not only does it not have any reasonable closed form representations, but it converges quite slowly. So, I'm going to use an Euler Transform to accelerate the convergence of the series. For an alternating series we can use the transform $$\sum_{n=0}^\infty (-1)^n a_n=\sum_{n=0}^\infty (-1)^n\frac{\Delta^n a_0}{2^{n+1}}$$ Using the forward difference operator: $$\Delta^n a_0=\sum_{k=0}^n(-1)^k~{}_{n}\mathrm{C}_k ~a_{n-k}$$ First, an index shift: $$S_1=\sum_{n=1}^\infty\frac{(-1)^n\ln\left(n+\frac{3}{4}\right)}{n}=-\sum_{n=0}^\infty\frac{(-1)^n\ln\left(n+\frac{7}{4}\right)}{n+1}$$ Let $a_n=\ln(n+7/4)/(n+1).$ Euler's transform tells us that $$S_1=-\lim_{N\to\infty}\sum_{n=0}^N\left[\frac{(-1)^n}{2^{n+1}}\left(\sum_{k=0}^n(-1)^k~{}_n\mathrm{C}_k\frac{\ln\left(n-k+\frac{7}{4}\right)}{n-k+1}\right)\right]$$ Which converges to 5 decimal precision with only $N=11$. With $N=35$ it converges to 12 decimal precision. See my implementation on Desmos. So approximately speaking $$S_1\approx −0.288525102601$$ Now for the third sum. According to Mathematica, it does actually have a "closed form", but it's pretty horrific. I can't be bothered to typeset it all, so I'll just post a screenshot.

It makes use of the Hurwitz zeta function. Anyway, the numerical value is $$S_3\approx -0.276850451954$$ So, finally, $$\mathcal{S}\approx \frac{−0.288525102601}{4}+\frac{0.276850451954}{48}-\frac{1}{2}\left(\frac{4}{9}-\frac{\pi}{3\sqrt{2}}-\frac{\ln(2)}{3}+\frac{\sqrt{2}\ln(1+\sqrt{2})}{3}\right)-\frac{\ln(2)}{4}\left(\gamma+\ln(8)-\frac{\pi}{2}\right)\approx -0.198728103723$$ You might ask yourself: why did we do all this work? The answer: speed of convergence. If we look at the partial sums of the original: $$\mathcal{S}_N=\sum_{n=1}^N\left[\frac{(-1)^n}{n}\sum_{k=1}^n\frac{1}{4k-1}\right]$$ It converges very poorly. See my implementation on Desmos. Even at $N=40$, with respect to the approximate sum that I found, the partial sums of the above jump around with a relative error of $\mathbf{15\%}$ (!) So yes, our work wasn't all pointless :)

I do not believe there is a "nice" closed form to this but one way to approximate would be: $$\sum_{n=1}^\infty\frac{(-1)^n}{n}\sum_{k=1}^n\frac 1{4k-1}>\frac 14 \sum_{n=1}^\infty\frac{(-1)^nH_n}{n}$$

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.