Another series I found I'm struggling with.
Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$
Ratio test and n-th root test are both inconclusive, Leibniz – criterion cannot be applied since the sequence given is not in the form of $(-1)^na_n$. I am sure the problem can be solved with the limit comparison test, though, the $n^{(…)}$ look pretty inviting after all. Let $a_n:=\frac{(-1)^nn^2+n}{n^3+1}$ then $|a_n|= \frac{n^2+n}{n^3+1}$. A try showing that the series diverges using the divergence of $\sum\frac{1}{n}$. For $n≥1$ it is clear that
$$ \frac{n^3+n^2}{n^3+1} ≥ 1 $$ dividing by $n$ yields:
$$ \frac{n^2+n}{n^3+1} = |a_n| ≥ \frac{1}{n}$$
Thus the series diverges. (?)
EDIT: Hints you gave me yield:
$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1} = \sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} +
\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ With the first series converging by Leibniz-theorem and the second by limit comparison test with $\frac{1}{n^2}$.
Best Answer
The series $$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1}$$ converges by Leibniz criterion, in addition the series $$\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ converges by the comparison test. Hence $$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} + \sum_{n=1}^{\infty}\frac{n}{n^3+1} = \sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$ also converges