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.
 A: 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.
A: Define $b_k := a_{2k+1}$. Then
$$b_k \leq a_{2k} + (-1)^{2k}\frac{1}{2k} \leq b_{k-1} + (\frac{1}{2k} - \frac{1}{2k-1}) \leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k \to b$.
Suppose $a_n \nrightarrow b$. Then there exists an $\varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > \varepsilon$.
Assume that $|a_{2m+1}-a_m| > \frac{\varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m \leq \frac{1}{2m}$ we have that 
\begin{align}
a_{2m+1} - a_m < - \frac{\varepsilon}{2}
\end{align}
for infinitely many $m$. Let $M := \{m \geq 1 : a_{2m+1} - a_m < - \frac{\varepsilon}{2} \text{ is fulfilled for } m \}$
\begin{align*}
d_m := 1_M (m)
\end{align*}
This implies
\begin{align*}
0 \leq a_{2m+1} = a_1 + \sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + \sum_{k=1}^m (a_{2k+1} - a_{2k}) + \sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \\
\leq a_1 + \sum_{k=1}^m (-1)^{2k} \frac{1}{2k}- \frac{\varepsilon}{2} d_k + \sum_{k=1}^m (-1)^{2k-1}\frac{1}{2k-1} \to a_1 - \sum_{k=1}^\infty \frac{\varepsilon}{2} d_k + \sum_{i=1}^\infty (-1)^i \frac{1}{i} = - \infty
\end{align*}
since $|M| = \infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $K\geq 1$ s.t. for all $k\geq K$ it holds: $|a_{2k+1} - a_k| \leq \frac{\varepsilon}{2}$. We can conclude that
\begin{align*}
|a{2n+1} - b| \geq |a_{2n} - b| - |a_{2n+1} - a_n| \geq \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}
\end{align*}
for infinitely $n \geq K$. Contradiction. Thus $a_n \to b$.
