# Checking whether $\sum_{k=1}^{+\infty}\frac{1}{(-1)^kk+2}$ is convergent

Here's a sequence the convergence of which I'm trying to examine:

$$S_n=\sum_{k=1}^{n}\frac{1}{(-1)^kk+2}$$

The problem asks as whether this sequence is bounded and if its a Cauchy sequence. It looks similair to $\sum_{k=1}^{+\infty}(-1)^k\frac{1}{k}$ which we know is convergent so the good way of tackling this problem seems showing that $\sum_{k=1}^{+\infty}\frac{1}{(-1)^kk+2}$ converges which would mean that answer to both question is yes. Unfortunately all convergence tests I know fail on this example - clearly the series is not absolutely convergent so only Dirichlet test (more general version of alternating sequence test) seem like a good idea. Unfortunately I can't see how to use it in that series - it's clearly not monotonic... So what to do with that sequence?

• \begin{align*}& & \sum_{k=1}^\infty \frac{1}{(-1)^kk+2} \\ & = &\frac{5}{4}+\sum_{k=3}^\infty \frac{(-1)^kk-2}{k^2-2} \\ &= & \frac{5}{4}+\sum_{k=3}^\infty \frac{(-1)^kk}{k^2-2}-\sum_{k=3}^\infty \frac{2}{k^2-2} \\ &= & \frac{5}{4}+\frac{7}{24}-\ln(2)-\frac{25}{24} \\ & = & \frac{1}{2}-\ln(2)\end{align*} – Mason Jan 30 at 5:47

$$\frac{1}{(-1)^k k + 2} = \frac{(-1)^k k - 2}{k^2-4}$$
if $k \neq 2$ (or $-2$, but we don't care about $-2$). Now you can split into two sums. The first one is alternating and monotone for large enough $k$. The second is absolutely convergent.
$$\frac1{(-1)^kk+2}=\frac{(-1)^k}k\left(1+\frac{2}{(-1)^kk}\right)^{-1}=\frac{(-1)^k}k+\mathcal O\left(\frac{1}{k^2}\right)$$ so we conclude the convergence of the given series since it's sum of the series $\sum \frac{(-1)^k}k$ and $\sum \frac{1}{k^2}$ which are convergent: the first by the Leibniz criteria and the second by comparison with a Riemann series.
Let $f(k)=\frac1{(-1)^kk+2}$, and note that $f(1)=1, f(3)=-1, f(2k)=\frac{1}{2k},f(2k+3)=-\frac{1}{2k+1}$ for $k\geq 1$. Therefore the sum is $$s=-\sum_{k=2}^{\infty}\frac{(-1)^{k}}{k}.$$