Alternating sum of reciprocals Find the infinite sum: $\frac{1}{(1)(2)}+\frac{1}{(3)(4)}+\frac{1}{(5)(6)}+\cdots$. 
I thought of expanding it as $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}- \cdots$, but it would yield nothing useful as this runs into the non-convergence issue.

We have,
$$\sum_{n \geq 1, \text{odd}} \frac{1}{n(n+1)}$$
$$=\sum_{n \geq 1, \text{odd}} (\frac{1}{n}-\frac{1}{n+1} )$$
$$=\sum_{n \geq 1} \frac{1-(-1)^n}{2} (\frac{1}{n}-\frac{1}{n+1})$$
$$=\frac{1}{2}\sum_{n \geq 1} (\frac{1}{n}-\frac{1}{n+1})-\frac{1}{2} \sum_{n \geq 1} \frac{(-1)^n}{n}+\frac{1}{2} \sum_{n \geq 1} \frac{(-1)^n}{n+1}$$

Now note,
$$\int_{0}^{1} x^{n-1} dx=\frac{1}{n}$$
So that,
$$\sum_{n \geq 1} \frac{(-1)^n}{n}$$
$$=\sum_{n \geq 1} (-1)^n \int_{0}^{1} x^{n-1} dx$$
$$=\int_{0}^{1} \sum_{n \geq 1} (-1)^n x^{n-1} dx$$
$$=-\int_{0}^{1} \sum_{n \geq 1} (-x)^{n-1} dx$$
$$=-\int_{0}^{1} \frac{1}{1+x} dx$$
$$=-\ln 2$$
And the sum clearly converges by the alternating series test as $\frac{1}{n} \to 0$ and $\frac{1}{n}$ is decreasing for $n>0$.

Utilizing this result we may continue as,
$$=\frac{1}{2}\left(1+\ln 2+(\ln 2-1) \right)$$
$$=\ln 2$$
