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I need help with finding sum of this: $$ \sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)} $$ First, I tried to telescope it in some way, but it seems to be dead end. The only other idea I have is that this might have something to do with logarithm, but really I don't know how to proceed. Any hint would be greatly appreciated.

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The easiest thing to do is to further decompose the decomposition; i.e., $$\frac{1}{n(4n^2 - 1)} = \frac{1}{2n-1} - \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n},$$ and look at the alternating harmonic series $$\log (x+1) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k}.$$

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  • $\begingroup$ Fast and clever (+1) $\endgroup$ – tired Jan 5 '16 at 16:59
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    $\begingroup$ Partial fractions, what you'd first reach for if it was $\int$ instead of $\sum$... $\endgroup$ – vonbrand Jan 5 '16 at 17:33
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    $\begingroup$ @vonbrand Actually, if I were to integrate $$\int \frac{dx}{x(4x^2-1)},$$ I would first select the substitution $u = 4x^2 - 1$, $du = 8x \, dx$, and write $$\int \frac{dx}{x(4x^2-1)} = \int \frac{8x}{2(4x^2)(4x^2-1)} = \int \frac{du}{2u(u+1)},$$ and then use partial fractions on the simpler rational polynomial instead. $\endgroup$ – heropup Jan 5 '16 at 22:58
  • $\begingroup$ @heropup, spoilsport. $\endgroup$ – vonbrand Jan 5 '16 at 23:06
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1.) Use partial fractions $a_n=\frac{1}{n(4n^2-1)}=\frac{1}{2 n-1}+\frac{1}{2 n+1}-\frac{1}{n}$

2.) Rewrite the sum as $S =\sum_{n=1}^{\infty}a_n\equiv \lim_{N\rightarrow \infty} \sum_{n=1}^N a_n$

3.) Use the series representation of the digamma function and their relation to the harmonic numbers to obtain

$$ S=\lim_{N\rightarrow \infty}\left(-H_N+H_{N-\frac{1}{2}}+\frac{1}{2 N+1}-1+2\log (2) \right) $$

4.) It is clear that for $N\rightarrow \infty$ , $H_N\rightarrow H_{N-1/2}$ and $\frac{1}{2N+1}\rightarrow 0$ .

The result therefore is

$$ S=2 \log(2)-1\approx 0.386294 $$

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    $\begingroup$ The digamma function and the harmonic numbers with half-integer indexes can (should?) be avoided. $\endgroup$ – Did Jan 5 '16 at 16:54
  • $\begingroup$ @Did i suppose (and think i know how to do) so, but for me this was the fastest route so i took it :) $\endgroup$ – tired Jan 5 '16 at 16:57
  • $\begingroup$ +1 ... Using the digamma function, I also obtained $$\begin{align}S&=-\frac12(\psi(-1/2)+\psi(+1/2))-\gamma\\\\&=(\log(2)+\gamma/2)+( \log(2)+\gamma/2-1)-\gamma\\\\&=2\log(2)-1\end{align}$$ $\endgroup$ – Mark Viola Jan 5 '16 at 19:08
  • $\begingroup$ @Did Happy New Year. Good to see you back. Just curious, but why should one avoid the digamma function evaluated at $\pm 1/2$? The Gauss digamma theorem expresses the digamma function for rational arguments. $\endgroup$ – Mark Viola Jan 5 '16 at 19:10
  • $\begingroup$ @Dr.MV Because simpler arguments exist. To describe one which is not on the page yet, note that, neglecting a term $1/(2N+1)$ which goes to zero, $1$ plus the partial sum up to $N$ of the series to be evaluated equals $$2\sum_{n=N+1}^{2N}\frac1n=\frac2N\sum_{n=1}^N\frac1{1+\frac{n}N}\to2\int_0^1{}{}\frac{dx}{1+x}=2\log2.$$ Tools used: Riemann sums and logarithm function. $\endgroup$ – Did Jan 5 '16 at 19:23
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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)} = \sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)} = \sum_{n=1}^{\infty}\int_0^1 \int_0^1 \frac{x^{2n}y^{2n-2}}{n}\,dy \,dx $

$\displaystyle =\int_0^1 \int_0^1\sum_{n=1}^{\infty} \frac{x^{2n}y^{2n-2}}{n} \,dy \,dx = \int_0^1 \int_0^1 \frac{\log(1-x^2y^2)}{y^2}\,dy \,dx \\ \displaystyle = \int_0^1 2 x \tanh^{-1}(x)+\log(1-x^2)\,{dx} = \log(4)-1. $

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    $\begingroup$ (+1) it is nice to see so many clever solutions :) $\endgroup$ – tired Jan 5 '16 at 17:03
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Let $$ f(x)=\sum_{n=1}^\infty\frac{1}{n(4n^2-1)}x^{2n+1}. $$ Clearly $\sum_{n=1}^\infty\frac{1}{n(4n^2-1)}=f(1)$. Note $$ f'(x)=\sum_{n=1}^\infty\frac{1}{n(2n-1)}x^{2n}, f''(x)=2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}, f'''(x)=2\sum_{n=1}^\infty x^{2n-2}=\frac{2}{1-x^2}$$ So \begin{eqnarray} f(1)&=&\int_0^1\int_0^x\int_0^y\frac{2}{1-z^2}dzdydx\\ &=&\int_0^1\int_z^1\int_z^1\frac{2}{1-z^2}dydxdz\\ &=&\int_0^1\frac{1-z}{1+z}dz\\ &=&2\log2-1. \end{eqnarray}

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One way forward is to note that

$$\begin{align} \sum_{n=1}^{N}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)&=\sum_{n=1}^{N}\left(\frac{1}{2n-1}+\frac{1}{2n}\right)-\sum_{n=1}^{N}\frac1n \tag 1\\\\ &=\sum_{n=1}^{2N}\frac1n -\sum_{n=1}^{N}\frac1n \tag 2\\\\ &=\sum_{n=N+1}^{2N}\frac1n \\\\ &=\sum_{n=1}^{N}\frac{1}{n+N} \\\\ &=\frac1N \sum_{n=1}^{N}\frac{1}{1+n/N} \tag 3 \end{align}$$

In going from $(1)$ to $(2)$ we simply noted that the sum, $\sum\limits_{n=1}^{2N}\frac1n$, can be written in terms of sums of even and odd indexed terms.

Now, we observe that limit of $(3)$ is the Riemann sum for the integral $$\int_0^1 \frac{1}{1+x}\,dx=\log(2).$$ Similarly, we see that

$$\begin{align} \sum_{n=1}^{N}\left(\frac{1}{2n+1}-\frac{1}{2n}\right)&=-1+\frac1N \sum_{n=1}^{N}\frac{1}{1+n/N} \end{align}$$

is the Riemann sum for $$-1+\int_0^1\frac{1}{1+x}\,dx=-1+\log(2).$$ Putting all of this together, we recover the expected result

$$\sum_{n=1}^\infty \frac{1}{n(2n-1)(2n+1)}=\sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\sum_{n=1}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n}\right)=2\log(2)-1.$$

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Hint:

$$\frac{1}{n(4n^2-1)}=-\frac{1}{n}+\frac{1}{2n+1}+\frac{1}{2n-1}$$

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    $\begingroup$ If you had followed your own hint, you would have seen it is definitely not the end of the story to solve the question... (Add here some castigating words about posts titled hints and parading as answers.) $\endgroup$ – Did Jan 5 '16 at 16:31
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    $\begingroup$ @Did: IMHO, answers with only hint in it gives the OP space to think and continue working on the problem. I would prefer this way to giving a complete solution to the problem. By the way, the answer can be obtained using the hint. $\endgroup$ – KittyL Jan 5 '16 at 16:35
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    $\begingroup$ @user276387: The harmonic series does not converge. But this becomes alternating harmonic series. $\endgroup$ – KittyL Jan 5 '16 at 16:37
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    $\begingroup$ Not when, like here, there are still some significant difficulties involved after one applies the hint. There is a substantial literature about this aspect of "Hints" answers (pseudo-hints, really) on meta, and the serious dysfunctionality they amount to. $\endgroup$ – Did Jan 5 '16 at 16:38
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    $\begingroup$ @Did: I agree. I should probably have pointed out some potential difficulties. However, this is still a valid and crucial hint, instead of a pseudo-hint. (I know it falls on the borderline though. Thanks for the reminder.) The OP can still ask for more help if needed, and either I or other people would answer. $\endgroup$ – KittyL Jan 5 '16 at 16:43

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