Alternating sum of reciprocals

Question

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.

Answer 1

You can see proof here. $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=\ln 2$$

Answer 2

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

$$ \sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k} = \sum_{k=1}^{\infty} \left( \frac{1}{2k-1} - \frac{1}{2k} \right) $$

Answer 3

We have,

$$\sum_{n \geq 1, \text{odd}} \frac{1}{n(n+1)}$$

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

$$=\sum_{n \geq 1} \frac{1-(-1)^n}{2} (\frac{1}{n}-\frac{1}{n+1})$$

$$=\frac{1}{2}\sum_{n \geq 1} (\frac{1}{n}-\frac{1}{n+1})-\frac{1}{2} \sum_{n \geq 1} \frac{(-1)^n}{n}+\frac{1}{2} \sum_{n \geq 1} \frac{(-1)^n}{n+1}$$

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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 $n>0$.

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Utilizing this result we may continue as,

$$=\frac{1}{2}\left(1+\ln 2+(\ln 2-1) \right)$$

$$=\ln 2$$