sum of this series: $\sum_{n=1}^{\infty}(-1)^{n-1}(\frac{1}{4n-3}+\frac{1}{4n-1})$ $$\sum_{n=1}^{\infty} (-1)^{n-1}  \left(\frac{1}{4n-3}+\frac{1}{4n-1}\right)$$
What I did
$$\sum_{n=1}^{\infty} (-1)^{n-1}  \left(\frac{1}{4n-3}+\frac{1}{4n-1}\right)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{4n-3}+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{4n-1} $$
$$=\sum_{n=1}^{\infty} (-1)^{n-1}\int_0^{x}t^{4n-4}dt+ \sum_{n=1}^{\infty} (-1)^{n-1}\int_0^{x}t^{4n-2}\mathrm{d}t $$
$$=\int_0^{x}\frac{1+t^{2}}{1+t^4}dt$$
$$=\frac{1}{\sqrt{2}}\arctan{\frac{x-1/x}{\sqrt{2}}}+\frac{\pi}{2\sqrt{2}}$$
According to the Abel’s limit theorem ,the sum is $\frac{\pi}{2\sqrt{2}}$,but the answer is $-\frac{(\sqrt{2}-1) \pi}{2 (\sqrt{2}- 2)}$.
 A: A different approach
Apply the linearity (since the sums are convergent by alternating series test) $$\sum_{n\geq 1} (-1)^{n-1}\left(\frac{1}{4n-3} +\frac{1}{4n-1}\right)=\sum_{n\geq 1} \frac{(-1)^{n-1}}{4n-3}+ \sum_{n\geq 1}\frac{(-1)^{n-1}}{4n-1}=S_1+S_2$$ We part sum into even and odd parity. That is $$S_1 = \sum_{n\geq 1}\left(\frac{(-1)^{2n-1}}{4(2n)-3}+\frac{(-1)^{2n-1-1}}{4(2n-1)-3}\right)=\sum_{n\geq 1} \left(\frac{1}{8n-7}-\frac{1}{8n-3}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{8n+1}-\frac{1}{8n+5}\right)=\color{red}{\frac{1}{8}\left(\psi_0\left(\frac{5}{8}\right)-\psi_0\left(\frac{1}{8}\right)\right)}$$ Similarly, $$S_2=\sum_{n\geq 1}\left(\frac{1}{8n-5}-\frac{1}{8n-1}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{8n+3}-\frac{1}{8n+7}\right)=\frac{1}{8}\left(\psi_0\left(\frac{7}{8}\right)-\psi_0\left(\frac{3}{8}\right)\right)$$ There we used the classical result $$\sum_{n=0}^{\infty}\left(\frac{1}{an+b}-\frac{1}{an+d}\right)=\color{red}{\frac{1}{a}\left(\psi_0\left(\frac{d}{a}\right)-\psi_0\left(\frac{b}{a}\right)\right)},\; \;  a,b,d>0$$ Adding $S_1$ and $S_2$ and using the reflection formula we have $$S_1+S_2=\frac{\pi}{8}\left(\cot\frac{3\pi}{8}+\cot\frac{\pi}{8}\right)=\frac{\pi}{8}\left(\sqrt 2- 1+\sqrt 2+1\right)=\frac{\pi}{2\sqrt 2}$$
