**The problem:**

Define $\displaystyle a_n=\frac{n+(-1)^n}{n(n+1)}$.

Determine whether $\displaystyle \sum a_n$ converges or diverges.

**My solution:**

Define

$\displaystyle b_n=\frac{1}{2n}$.

We will use the comparison test to show that since $\displaystyle \sum b_n$ diverges, so too does $\displaystyle \sum a_n$ diverge.

Observe that if $\displaystyle 3<n$, we may divide both sides by $\displaystyle 2$ to yield

$\displaystyle \frac{3}{2}<\frac{n}{2}$.

Adding $\displaystyle \frac{n}{2}-1$ to both sides gives us

$\displaystyle \frac{n+1}{2}=\frac{3}{2}+\frac{n}{2}-1<\frac{n}{2}+\frac{n}{2}-1=n-1$.

Dividing by $\displaystyle n(n+1)$, we have

$\displaystyle \frac{1}{2n}<\frac{n-1}{n(n+1)}$.

Since $\displaystyle b_n=\frac{1}{2n}$ and $\displaystyle a_n=\frac{n-1}{n(n+1)}$ when $\displaystyle n$ is odd, then we have

$\displaystyle b_n<a_n$

when $\displaystyle n$ is odd.

Now observe that since $\displaystyle \frac{1}{2}<1$, then dividing by $\displaystyle n$ gives us

$\displaystyle \frac{1}{2n}<\frac{1}{n}=\frac{n+1}{n(n+1)}$.

Since $\displaystyle b_n=\frac{1}{2n}$ and $\displaystyle a_n=\frac{n+1}{n(n+1)}$ when $\displaystyle n$ is even, then we have

$\displaystyle b_n<a_n$

when $\displaystyle n$ is even.

Thus, whether $\displaystyle n$ is odd or even, we have $\displaystyle 0<b_n<a_n$, with $\displaystyle \sum b_n$ divergent. By the comparison test, therefore $\displaystyle \sum a_n$ is likewise divergent.

**The textbook tells me I am wrong, and that this series is actually convergent.**

Any ideas what went wrong, here?

Thanks!