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.
Best Answer
Note that $$ \lim_{n\to\infty}\sum_{k=n}^\infty\frac1{k^2}=0 $$ and for $m\gt n$, $$ a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ First, take the $\limsup\limits_{m\to\infty}$: $$ \limsup_{m\to\infty}a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ which must be non-negative. Then take the $\liminf\limits_{n\to\infty}$: $$ \limsup_{m\to\infty}a_m\le\liminf_{n\to\infty}a_n $$ Thus, the limit exists.