$\{a_n\}$ is a sequence of nonnegative real numbers satisfying $a_{n+1} \leq a_n + \frac{(-1)^n}{n}$. Prove convergence. 
Let $\{a_n\}$ be a sequence of non-negative real numbers satisfying $a_{n+1} \leq a_n + \frac{(-1)^n}{n}$. Prove $\{a_n\}$ converges.

It is easy to show $a_n$ is bounded by $a_1$ plus the alternating harmonic series.  By similar means, it is easy to provide a small upperbound on the difference $a_m - a_n$ for $m > n$.  However, I'm having difficulty giving a lower bound for this difference to give us an absolute bound.  I'm now searching for alternative approaches.
 A: Define $s_n = \sum_{k=1}^n \frac{(-1)^{n}}{n}$, and then define $b_n = a_{n+1}-s_n$. Note that $s_n$ is bounded (because it converges), and $a_n$ is also bounded (since $0 \leq a_n \leq a_1+s_n$). Therefore $b_n$ is bounded.
Note that $b_n = a_{n+1}-s_n \leq a_n +\frac{(-1)^n}{n}-s_n = a_n-s_{n-1}=b_{n-1}$. Thus $b_n$ is decreasing.
Since $b_n$ is bounded and decreasing, it converges. Since $b_n$ and $s_n$ are convergent, their sum $a_{n+1}=b_n+s_n$ also converges. Thus $a_n$ converges.
A: Probably it is easier to show $\{a_n\}$ converges by showing that $\limsup_{n\to\infty}a_n = \liminf_{n\to\infty}a_n$.  I'll give a sketch below: I think you'll be able to fill in the details.
Since $a_n$ consists of nonnegative real terms, the $\liminf$ is finite, so for every $\epsilon > 0$ and for every $N$, you can find an $n>N$ such that $a_n - \liminf_{n\to\infty} a_n < \frac{\epsilon}{2}$.  Also, by the convergence theory for alternating series, you can choose an $N$ such that $\left|\sum_{k=n}^m \frac{(-1)^k}{k}\right|  < \frac{\epsilon}{2}$ for all $n > N$.  
Now, using your reasoning, $a_m \le a_n + \left|\sum_{k=n}^m \frac{(-1)^k}{k}\right| < \liminf_{n\to\infty} a_n + \epsilon$.  Taking the $\limsup$ now and since $\epsilon > 0$ was arbitrary, you can find that $\limsup_{n\to\infty}a_n \le \liminf_{n\to\infty} a_n$.  This implies that the $\limsup$ and $\liminf$ are equal, so $a_n$ converges.
