Does $\sum_{n=2}^\infty \frac{(-1)^n}{n+(-1)^n}$ converge?

I try to study the convergence or divergence of the series $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n}{n+(-1)^n}$. We can see that this series diverges absolutely, because $\displaystyle\sum_{n=2}^\infty\frac{1}{|n+(-1)^n|}$ is a rearrangement of $\displaystyle\sum_{n=2}\frac{1}{n}$ which equals to $+\infty$. Consequently, we can't say for sure that the rearrangement $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n}{n+(-1)^n}$ of $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n}{n}$ converges. That's the part where I am stuck. How exactly can I study the convergence (or not) of the first series?

Should I apply Cauchy's criterion and prove that $s_n=\displaystyle\sum_{k=2}^n \frac{(-1)^k}{k+(-1)^k},n\in\mathbb{N} $ is a basic sequence?

Thanks.

Solution 1:

Without giving a formal proof of it (or even a formal statement of the theorem), the following is true:

You can do any bounded amount of rearranging of an infinite series without affecting its (conditional) convergence/divergence.

For the series at hand, note that

$$\begin{align} \sum_{n=2}^\infty{(-1)^n\over n+(-1)^n} &={-1\over3}+{1\over2}+{-1\over5}+{1\over4}+{-1\over7}+{1\over6}+\cdots\\ &={1\over2}-{1\over3}+{1\over4}-{1\over5}+{1\over6}-{1\over7}+\cdots \end{align}$$

where the "bounded" rearranging is to swap pairs of adjacent terms. The rearranged sequence now meets all the criteria of Leibniz's test, so the series is conditionally convergent. (The rearrangement makes it easy to see that the sum is $1-\ln2$.)

A somewhat more formal definition of "bounded" rearrangement is that no term is the rearranged series is ever more than $B$ positions from where it appears in the original series, for some fixed positive number $B$. (In this example, $B=1$.) Somewhat more formally, a rearrangement $\sum a_{\pi(n)}$ of $\sum a_n$, where $\pi:\mathbb{N}\to\mathbb{N}$ is a bijection (aka permutation), is "bounded" (by $B$) if $|\pi(n)-n|\le B$ for all $n$.

Solution 2:

Note that

$$\sum_{n=2}^m \frac{(-1)^n}{n+(-1)^n}= \sum_{n=2}^m \frac{(-1)^n}{n+(-1)^n}\frac{n-(-1)^n}{n-(-1)^n}= \sum_{n=2}^m \frac{(-1)^nn}{n^2-1}-\sum_{n=2}^m \frac{1}{n^2-1},$$

and $\frac{n}{n^2-1} \searrow 0$ monotonically. Try to supply the details in showing that both series on the RHS are convergent.

Solution 3:

Here is a slightly different approach: For large $n$, we kind of expect $\frac{(-1)^n}{n+(-1)^n}$ to behave similarly as $\frac{(-1)^n}{n}$, whose sum we know how to control. In light of this, we study the difference:

$$ \sum_{n=2}^{N} \left( \frac{(-1)^n}{n+(-1)^n} - \frac{(-1)^n}{n} \right) = - \sum_{n=2}^{N} \frac{1}{(n + (-1)^n)n}.$$

Now it is not hard to verify that the partial sum on the right-hand side converges absolutely as $N\to\infty$. Either the direct comparison or limit comparison will work.