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Check whether $(a_n)_n$ converges if the general term is given by the formula:

$$a_n= 1-(-1)^n+\frac{1}{n^3+1}$$

I know this sequence is not monotonic because $a_1=\frac{5}{2},\ a_2=\frac{1}{9},\ a_3=\frac{57}{28},...$. But that still doesn't mean it isn't convergent. What's the next step? Should we check whether it is bounded? If it's not then that would definitely mean it isn't convergent. But if it is bounded, that still doesn't mean that it is convergent.

I think a lower bound is $0$ and the upper bound is $3$ but I'm not sure how to prove it. Do we even need a formal proof or is it enough to say $\frac{1}{n^3+1}$ is always greater than zero and $1-(-1)^n$ can be either $0$ or $2$?

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    The sequence does not converge. The term $\frac{1}{n^3+1}$ will tend to zero as $n\to\infty$. So the issue is with $1-(-1)^n$. For odd $n$, $1-(-1)^n\to2$, while for even $n$, $1-(-1)^n\to0$. – 2017-02-19

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The sum of convergent sequences is convergent. If $a_n$ was convergent then note that $b_n\colon = -1-\frac{1}{n^3+1}$ is convergent and so $a_n+b_n=-(-1)^n$ would be convergent, which it isn't.

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No, it does not converge

$a_{2n}$ converges to $0$

$a_{2n+1}$ converges to $2$

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The series $\sum_{n\ge1}(a_{n+1}-a_n)$ is divergent because the sequence $(a_{n+1}-a_n)$ doesn't have limit $0$ (in fact it doesn't have any limit). Hence the sequence $(a_n)$ is divergent.