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.

Your last equation has a couple of bugs in it: you should write $\limsup$ instead of $\lim$ because you haven't yet shown that the limit exists, and the index of summation on the righthand side should match the index in the other RHS term.
Once you fix it up, though, it'll give you what you want. Since $\sum \frac{1}{k^2}$ converges, you can make the quantity
$(\limsup_n x_n) - x_k$
as small as you like by choosing $k$ large enough.
