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.
$\endgroup$ 13 Answers
$\begingroup$I'm guessing your question is about whether rearranging the series in this fashion can change its sum.
The simple answer as to why not is that the sequence of partial sums
$$ \lim_{n \to \infty} \sum_{k=1}^{n} (-1)^{k+1} \frac{1}{k} $$
is a convergent limit, which means every subsequence converges to the same limit:
$$ \sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k} = \lim_{n \to \infty \\ n \mathrm{\ even}} \sum_{k=1}^{n} (-1)^{k+1} \frac{1}{k} $$
and so you can group pairs of terms to get
$$ \sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k} = \sum_{k=1}^{\infty} \left( \frac{1}{2k-1} - \frac{1}{2k} \right) $$
$\endgroup$ $\begingroup$You can see proof here. $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=\ln 2$$
$\endgroup$ $\begingroup$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,
$$=\frac{1}{2}\left(1+\ln 2+(\ln 2-1) \right)$$
$$=\ln 2$$
$\endgroup$
