Show that the following sequence converges. Please Critique my proof. The problem is as follows:

Let $\{a_n\}$ be a sequence of nonnegative numbers such that 
  $$
a_{n+1}\leq a_n+\frac{(-1)^n}{n}.
$$
  Show that $a_n$ converges.

My (wrong) proof:
Notice that 
$$
|a_{n+1}-a_n|\leq \left|\frac{(-1)^n}{n}\right|\leq\frac{1}{n}
$$
and since it is known that $\frac{1}{n}\rightarrow 0$ as $n\rightarrow \infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.
My question:
This is a question from a comprehensive exam I found and am using to review.
Should I argue that we should select $N$ so that $n>N$ implies $\left|\frac{1}{n}\right|<\epsilon$ as well?
Notes: Currently working on the proof.

Consider $b_n = a_n + \sum_{k=1}^{n-1} \frac{(-1)^{k-1}}{k}$. Then
$$ b_{n+1}
= a_{n+1} + \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k}
\leq a_n + \frac{(-1)^n}{n} + \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k}
= b_n, $$
which shows that $(b_n)$ is non-increasing. Moreover, since $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.
