Convergence of a sequence of non-negative real numbers $x_n$ given that $x_{n+1} \leq x_n + 1/n^2$. Let $x_n$ be a sequence of the type described above. It is not monotonic in general, so boundedness won't help. So, it seems as if I should show it's Cauchy. A wrong way to do this would be as follows (I'm on a mobile device, so I can't type absolute values. Bear with me.) $$\left|x_{n+1} - x_n\right| \leq \frac{1}{n^2}.$$ So, we have $$
\left|x_m -x_n\right| \leq \sum_{k=n}^{m} \frac{1}{k^2}$$ which is itself Cauchy, etc., etc. But, of course, I can't just use absolute values like that. One thing I have shown is that $x_n$ is bounded. Inductively, one may show $$ \limsup_{n \to \infty} x_n \leq x_k + \sum_{k=n}^{\infty} \frac{1}{k^2},$$ although I'm not sure this helps or matters at all. Thanks in advance.
Disclaimer I've noticed that asking a large number of questions in quick succession on this site is often frowned upon, especially when little or no effort has been given by the asker. However, I am preparing for a large test in a few days and will be sifting through dozens of problems. Therefore, I may post a couple a day. I will only do so when I have made some initial, meaningful progress. Thanks.

I think the key is to use the liminf and limsup. Call them $a$ and $b$ respectively. Suppose they are not equal. Then their difference is some $3 \varepsilon > 0$ (You'll see why I multiplied by 3 in a moment). Now I need to use the fact that the series $\frac{1}{n^2}$ converges. Choose a natural number $i$ such that $\Sigma_{n=1}^{i} \frac{1}{n^2}$ is within $\varepsilon$ of it's limit ( I.E. $< \frac{6}{\pi ^2}$ for those who care). 
From here we can say the following. Some subsequence of $x_n$ converges to $a$, so we can choose some natural $j$ such that $x_j$ is within $\varepsilon$ of the liminf. Now, the sequence of $x_n$ starting at $i+j$ is bounded above by $a+2\varepsilon$. In other words, always outside of $b$ by at least $\varepsilon$. This contradicts the fact that $b$ is the limsup. So $a=b$ thus the sequence converges.
Trying to do cauchy might be hard because it's difficult to exclude a large drop in the value of any particular $x_n$. Using the fact that the series converges is crucial. I'm pretty sure you can come up with a counter example using $\frac{1}{n}$ instead.
