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Find the sum $$\sum_{n=1}^{\infty} \dfrac{4n}{n^4+2n^2+9}.$$

By calculator, we can predict that its sum is equal to $\dfrac{5}{6}$ so I think we should use inequalities to prove it. And I found that

$\dfrac{5}{6(n^4+n^2)} < \dfrac{4n}{n^4+2n^2+9}< \dfrac{5}{6(n^2+n)}$ for all $n\ge n_0$, $n_0$ is large enough.

And $\sum_{n=1}^{\infty} \dfrac{5}{6(n^4+n^2)}= \sum_{n=1}^{\infty}\dfrac{5}{6(n^2+n)}=\dfrac{5}{6}$.

But it is not enough to confirm that the given series converges to $\dfrac{5}{6}$. Can someone help me, please? Thanks in advanced.

Najib Idrissi
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liverpool29
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    How can $\sum_{n=1}^{\infty} \dfrac{5}{6(n^4+n^2)}= \sum_{n=1}^{\infty}\dfrac{5}{6(n^2+n)}$ be correct??? The former is strictly smaller than the latter. – guestDiego Oct 11 '16 at 14:48
  • These two series both converges to $\dfrac{5}{6}$ thus I wrote so. Isn't it right? Thank you for caring about my problem. – liverpool29 Oct 11 '16 at 14:51
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    See also : http://math.stackexchange.com/questions/449510/how-to-find-the-sum-of-the-sequence-frac111214-frac212224 and http://math.stackexchange.com/questions/304851/evaluate-sum-limits-k-1-infty-frack2-1k4k21 – lab bhattacharjee Oct 11 '16 at 14:56
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    @liverpool29: It is not true that $\sum _{n \ge 1} \frac 5 {6(n^4 + n^2)} = \frac 5 6$. – Alex M. Oct 11 '16 at 14:58
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    @liverpool29, the second series indeed converges to $5/6$. But the first series converges to something less than $5/6$, because, as guestDiego pointed out, each term in the first series (beyond $n=1$) is less than the corresponding term in the second series. – Barry Cipra Oct 11 '16 at 14:58
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    It is known that $\sum_{n=1}^{\infty}\dfrac{1}{(n^2+n)}= \sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=1$. However $\sum_{n=1}^{\infty}\dfrac{1}{(n^4+n^2)}= \sum_{n=1}^{\infty}\left(\dfrac{1}{n^2}-\dfrac{1}{n^2+1}\right)<1$ – guestDiego Oct 11 '16 at 14:58
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    Thank you all again. I understand now. Thanks so much. – liverpool29 Oct 11 '16 at 15:04

2 Answers2

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HINT:

$$(n^2)^2+3^2+2n^2=(n^2+3)^2-(2n)^2=(n^2+2n+3)(n^2-2n+3)$$

$$(n^2+2n+3)-(n^2-2n+3)=?$$

Now if $f(m)=m^2-2m+3,$

$f(m+2)=(m+2)^2-2(m+2)+3=m^2+2m+3$

imranfat
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lab bhattacharjee
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}{4n \over n^{4} + 2n^{2} + 9}}$.

\begin{align} &\sum_{n = 1}^{\infty}{4n \over n^{4} + 2n^{2} + 9} = \sum_{n = 0}^{\infty}{4n \over \pars{n^{2} - 2n + 3}\pars{n^{2} + 2n + 3}} = \sum_{n = 0}^{\infty}\pars{{1 \over n^{2} - 2n + 3} - {1 \over n^{2} + 2n + 3}} \\[5mm] = &\ \lim_{N \to \infty}\pars{\sum_{n = 0}^{N}{1 \over \pars{n - 1}^{2} + 2} - \sum_{n = 0}^{N}{1 \over \pars{n + 1}^{2} + 2}} = \lim_{N \to \infty}\pars{\sum_{n = -1}^{N - 1}{1 \over n^{2} + 2} - \sum_{n = 1}^{N + 1}{1 \over n^{2} + 2}} \\[5mm] = &\ \lim_{N \to \infty}\pars{{1 \over 3} + {1 \over 2} + \sum_{n = 1}^{N - 1}{1 \over n^{2} + 2} - \sum_{n = 1}^{N - 1}{1 \over n^{2} + 2} - {1 \over N^{2} + 2} - {1 \over \pars{N + 1}^{2} + 2}} \\[5mm] = &\ \lim_{N \to \infty}\pars{{5 \over 6} - {1 \over N^{2} + 2} - {1 \over \pars{N + 1}^{2} + 2}} = \bbx{\ds{5 \over 6}} \end{align}

Felix Marin
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