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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 $\frac{1}{n^2}$.

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    $\begingroup$ Split it into an alternating and an absolutely convergent part. $\endgroup$ – Daniel Fischer Jan 21 '14 at 19:48
  • $\begingroup$ Sum of two convergent sequence (each convergent by a different criterion). $\endgroup$ – mjqxxxx Jan 21 '14 at 19:51
  • $\begingroup$ I put your hints into the initial post, thanks for the help. However, could you tell me where my first try failed trying to show that $|a_n|≤\frac{1}{n}$? $\endgroup$ – Nhat Jan 21 '14 at 19:57
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    $\begingroup$ Your final form isn't quite right; the latter term (which should be added, though I assume that's just a typo) should be $\sum_{n=1}^\infty \frac{n}{n^3+1}$. The second half still converges by the limit comparison test, of course, though you need $1/n^2$ for that. $\endgroup$ – Steven Stadnicki Jan 21 '14 at 20:05
  • $\begingroup$ Corrected it. I also forgot a $+$, thanks. $\endgroup$ – Nhat Jan 21 '14 at 20:15
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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

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  • $\begingroup$ The hints Daniel and mjq gave me in the comment section gave me the same result, which I also put in my initial post. However, could you tell me why my first try went wrong? (Limit comparison with $\frac{1}{n}$) $\endgroup$ – Nhat Jan 21 '14 at 19:56
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    $\begingroup$ @KitKat The problem is that since the series is alternating, you can't use the limit comparison; you need that $a_n\gt b_n$, not just $|a_n|\gt b_n$, for that test. $\endgroup$ – Steven Stadnicki Jan 21 '14 at 19:57
  • $\begingroup$ I just looked it up in my lecture notes, you're right: "Let $\sum_{k=1}^{\infty}b_k$ be a divergent series. (...) If $|a_k|≥b_k≥0$ then $\sum_{k=1}^{\infty}|a_k|$ is divergent." Sneaky. $\endgroup$ – Nhat Jan 21 '14 at 20:00
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    $\begingroup$ Yep, and actually the series is not absolute convergent. $\endgroup$ – user127.0.0.1 Jan 21 '14 at 20:05
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Let denote $$\frac{(-1)^nn^2+n}{n^3+1}=\underbrace{\frac{(-1)^nn^2}{n^3+1}}_{=u_n}+\underbrace{\frac{n}{n^3+1}}_{=v_n}$$ The series $\displaystyle \sum_n v_n$ is convergent since $v_n\sim_\infty \frac 1 {n^2}$.

We have $$u_n=\frac{(-1)^nn^2}{n^3+1}=\frac{(-1)^n}{n}\frac{1}{1+\frac{1}{n^3}}=\frac{(-1)^n}{n}\left(1-\frac{1}{n^3}+o\left(\frac{1}{n^3}\right)\right)$$ so we can see that $\displaystyle\sum_n u_n$ is a sum of convergent series hence it's convergent. Conclude.

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    $\begingroup$ You know your way Sami.:+) $\endgroup$ – mrs Jan 22 '14 at 14:30
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Here is a different approach.

The sum of each pair of terms is $$ \begin{align} a_{2n-1}+a_{2n} &=\frac{-(2n-1)^2+(2n-1)}{(2n-1)^3+1}+\frac{(2n)^2+2n}{(2n)^3+1}\\ &=\frac{4n^3-6n^2+5n-1}{n(4n^2-2n+1)(4n^2-6n+3)}\\ &=\frac1{4n^2}\frac{1-\frac3{2n}+\frac5{4n^2}-\frac1{4n^3}}{\left(1-\frac1{2n}+\frac1{4n^2}\right)\left(1-\frac3{2n}+\frac3{4n^2}\right)}\\ &=\frac1{4n^2}+O\left(\frac1{n^3}\right) \end{align} $$ Thus, the sum of $a_{2n-1}+a_{2n}$ converges by comparison to $\frac1{n^2}$.

Since the terms go to $0$, convergence of the partial sums is the same as convergence of every other partial sum.

Let $s_n=\sum\limits_{k=1}^n a_k$ and $p_n=\sum\limits_{k=1}^n(a_{2k-1}+a_{2k})$. Then $s_{2n}=p_n$ and $s_{2n+1}=p_n+a_{2n+1}$

Since $\lim\limits_{n\to\infty}a_n=0$, if $p_n$ converges, then $s_n$ converges.

Since $p_n$ is a subsequence of $s_n$, if $s_n$ converges, then $p_n$ converges.

That is, if $\lim\limits_{n\to\infty}a_n=0$, then $$ \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty(a_{2n-1}+a_{2n}) $$

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    $\begingroup$ ♦ So instead of looking at $\sum a_n$ we're considering $\sum a_{2n-1} + a_{2n}$. This series converges by a limit comparison test with $\frac{1}{n^2}$. Can you elaborate on your last sentence, it makes sense in my head but I'm kind of dumbfounded right now, I guess. $\endgroup$ – Nhat Jan 21 '14 at 23:18
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    $\begingroup$ @KitKat: I have done so. $\endgroup$ – robjohn Jan 22 '14 at 0:15
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While the split-terms approach is easier, it's also worth noting that a little bit of massaging will let you apply the Leibniz criterion directly. Consider $a_n=\dfrac{n^2+(-1)^nn}{n^3+1}$; then your series is of the form $\sum_{n=1}^\infty (-1)^na_n$. Now, you should be able to prove that $a_n\gt 0$ for all $n$, and that $a_{n+1}\lt a_n$ for all sufficiently large $n$: for the latter, (fairly) obviously $a_{2k+1}\lt a_{2k}$, so you just have to show that $a_{2k}\lt a_{2k-1}$ for sufficiently large $k$. Write out the terms explicitly, cross-multiply (since the denominators are manifestly positive) and cancel; you should find criteria on $k$ that force the inequality.

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