The infinite sum I want to find.
$$\frac{1}{1\cdot2} - \frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+ \cdots$$
My attempt at it
$$\begin{align} &\;\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)} - \sum_{n=1}^{\infty} \frac{1}{(2n)(2n+1)} \\[4pt] =&\;\sum_{n=1}^{\infty} \left(\frac{1}{2n-1} - \frac{1}{2n}\right) - \sum_{n=1}^{\infty} \left(\frac{1}{2n} - \frac{1}{2n+1}\right) \\[4pt] =&\;\sum_{n=1}^{\infty} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) - 2 \sum_{n=1}^{\infty} \frac{1}{2n} \\[4pt] =&\;\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots\right) + \left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right) - 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots\right) \\[4pt] =&\;1 + 2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right) - \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots\right) \\[4pt] =\;&-1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \\[4pt] =\;&\ln 2 - 1 \end{align}$$
The answer I got here is negative, while the series seems to be positive. What is the mistake here?
Yes, the sum is positive. Note that instead of $$= \sum_{n=1}^{\infty} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) - 2 \sum_{n=1}^{\infty} \frac{1}{2n}$$ which is indefinite since the series are divergent, you should have the limit as $N$ goes to $\infty$ of $$\sum_{n=1}^{N} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) - 2 \sum_{n=1}^{N} \frac{1}{2n}\\ =-1+2\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2N-1}\right) +\frac{1}{2N+1} - 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots+\frac{1}{2N}\right)$$ which is a different thing. What is its limit as $N$ goes to $\infty$?
More simply, you may consider the partial sum and then take the limit as $N$ goes to $\infty$ $$S_N=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n(n+1)}=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}+\sum_{n=1}^{N}\frac{(-1)^{n+2}}{n+1}= \sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}+\sum_{n=2}^{N+1}\frac{(-1)^{n+1}}{n}\\ =-1+2\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}+\frac{(-1)^{N+2}}{N+1}\to -1+2\ln(2)>0.$$