Sum of "alternating-like" series I tried to prove that the following series converges.
$$1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4}+ \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} +\frac{1}{9} - ...$$
I found a expression for this sum in sigma form by summing pair of numbers with the same sign:
$$1 + \sum_{n=1}^{\infty} \frac{8n +1}{16n^{2}+4n} - \frac{8n -3}{16n^{2} - 12n +2} = 1 + \sum_{n=1}^{\infty} \frac{(8n +1)(16n^{2}+4n) - (8n -3)(16n^{2}+4n)}{(16n^{2}+4n)  (16n^{2} - 12n +2)}$$
But I'm getting a very long expression that can't find a way to prove a convergence of it , I feel that there is more elegant way to prove convergences of this series.
I will thank for your help.

$$S_p=1 + \sum_{n=1}^{p} \frac{8n +1}{16n^{2}+4n} - \sum_{n=1}^{p}\frac{8n -3}{16n^{2} - 12n +2} $$
Notice that
$$ \frac{8n +1}{16n^{2}+4n}=\frac{1}{4 n+1}+\frac{1}{4 n}$$
$$\frac{8n -3}{16n^{2} - 12n +2}=\frac{1}{4 n-1}+\frac{1}{4n-2}$$ Computing the four summations
$$\sum_{n=1}^{p}\frac{1}{4 n+1}=\frac{1}{4} \left(\psi \left(p+\frac{5}{4}\right)-\psi \left(\frac{5}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{4 n}=\frac{1}{4} H_p$$
$$\sum_{n=1}^{p}\frac{1}{4 n-1}=\frac{1}{4} \left(\psi \left(p+\frac{3}{4}\right)-\psi \left(\frac{3}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{4n-2}=\frac{1}{4} \left(\psi\left(p+\frac{1}{2}\right)-\psi\left(\frac{1}{2}\right)\right)$$ Summing and using harmonic numbers in place of digamma's
$$S_p=\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{1}{4} \left(-H_{p-\frac{1}{2}}-H_{p-\frac{1}{4}}+H_p+H_{p+\frac{1}{4}}\right)$$ Using the asymptotics
$$S_p=\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{1}{4 p}-\frac{3}{32 p^2}+O\left(\frac{1}{p^3}\right)$$gives the limit, shows how the limit is approached and provides a shortcut method for the evaluation of the partial sums.
Using $p=10$
$$S_{10}=\frac{101405260644075743}{219060189739591200}=0.462910\cdots$$ while the truncated expansion gives
$$S_{10}\sim\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{77}{3200}=0.462887\cdots$$ to which corresponds a relative error of $0.005$%.
