\(\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\)
\(\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\)
Evaluate \(\displaystyle\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\).
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Re: \(\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\)
We begin by observing that \[ \frac{1}{(2n-1)(3n-1)(4n-1)} = \frac{2}{2n-1} - \frac{9}{3n-1} + \frac{8}{4n-1}.\] We now define \[ f(x) = \sum_{n=1}^{\infty} \left( \frac{2x^{12n-6}}{2n-1} - \frac{9x^{12n-4}}{3n-1} + \frac{8x^{12n-3}}{4n-1} \right).\] This is a power series which converges absolutely for \(|x| < 1\). So we can differentiate term by term to get \begin{align*}
f'(x) &= \sum_{n=1}^{\infty} (12x^{12n-7} - 36x^{12n-5} + 24x^{12n-4}) = \frac{12x^5(1-3x^2+2x^3) }{1-x^{12}} \\
& = \frac{12x^5(1+x-2x^2)}{(1+x)(1-x+x^2)(1+x+x^2)(1+x^2)(1-x^2+x^4)} \,.
\end{align*} What remains is now straightforward but tedious. (I didn't do it.): Find the partial fraction decomposition of \(f'\) (one must factor \(1-x^2+x^4\) into a product of two quadratics) then take its antiderivative to determine \(f\) (all integrals are standard giving logarithms and inverse tangents) and finally compute \(f(1)\) and invoke Abel's theorem to justify that this is equal to the required sum.
f'(x) &= \sum_{n=1}^{\infty} (12x^{12n-7} - 36x^{12n-5} + 24x^{12n-4}) = \frac{12x^5(1-3x^2+2x^3) }{1-x^{12}} \\
& = \frac{12x^5(1+x-2x^2)}{(1+x)(1-x+x^2)(1+x+x^2)(1+x^2)(1-x^2+x^4)} \,.
\end{align*} What remains is now straightforward but tedious. (I didn't do it.): Find the partial fraction decomposition of \(f'\) (one must factor \(1-x^2+x^4\) into a product of two quadratics) then take its antiderivative to determine \(f\) (all integrals are standard giving logarithms and inverse tangents) and finally compute \(f(1)\) and invoke Abel's theorem to justify that this is equal to the required sum.
Re: \(\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\)
Thank you Demetres. There is a somewhat general method dealing with sums of fractions with a product as denominator whose number of factors depends on a parameter \(k\).
Using this method, one can show for example, as a generalization of this problem, that
\[\displaystyle\sum_{n\geq1}\frac{1}{(2n+1)(3n+2)\cdots(kn+k-1)}=\] \[\displaystyle\frac{1}{k!}\sum_{m=2}^{k}(-1)^{m-1}\binom{k}{m}(m-1)m^{k-2}\left(\frac{\pi}{2}\cot\frac{\pi}{m}-\ln m+\sum_{\ell=1}^{m-1}\cos\frac{2\ell\pi}{m}\cdot\ln\left(\sin\frac{\ell\pi}{m}\right)\right),\qquad k\geq3\].
Using this method, one can show for example, as a generalization of this problem, that
\[\displaystyle\sum_{n\geq1}\frac{1}{(2n+1)(3n+2)\cdots(kn+k-1)}=\] \[\displaystyle\frac{1}{k!}\sum_{m=2}^{k}(-1)^{m-1}\binom{k}{m}(m-1)m^{k-2}\left(\frac{\pi}{2}\cot\frac{\pi}{m}-\ln m+\sum_{\ell=1}^{m-1}\cos\frac{2\ell\pi}{m}\cdot\ln\left(\sin\frac{\ell\pi}{m}\right)\right),\qquad k\geq3\].
Re: \(\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\)
A somewhat different approach would be using the digamma function. As Demetres shows we have that:akotronis wrote:Evaluate \(\displaystyle\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\).
$$\frac{1}{(2n-1)(3n-1)(4n-1)} = \frac{2}{2n-1} - \frac{9}{3n-1} + \frac{8}{4n-1}$$
Thus:
\begin{align*}
\sum_{n \geq 1} \frac{1}{(2n-1) (3n-1) \left ( 4n-1 \right )} &=\sum_{n \geq 1} \left [ \frac{2}{2n-1} -\frac{9}{3n-1} +\frac{8}{4n-1} \right ] \\
&= \sum_{n \geq 1} \left [ \frac{2}{2\left ( n- \frac{1}{2} \right )} - \frac{9}{3\left ( n - \frac{1}{3} \right )} + \frac{8}{4\left ( n-\frac{1}{4} \right )} \right ]\\
&= \sum_{n \geq 1} \left [ \frac{1}{n-\frac{1}{2} } - \frac{3}{n- \frac{1}{3}} + \frac{2}{n -\frac{1}{4}}\right ]\\
&=-\psi\left(-\frac{1}{2} \right) + 3\psi \left ( -\frac{1}{3} \right ) -2\psi \left ( -\frac{1}{4} \right ) +1 \\
&= -\pi + \frac{\sqrt{3} \pi}{2}+ 8 \log 2 - \frac{9 \log 3 }{2}
\end{align*}
since in general it holds that $\displaystyle \psi(1+x)=\frac{1}{x} + \psi(x)$. So evaluating those negative values one just plugs them in the functional equation above and then uses the special values presented here.
I suppose that the same technic also works for akotronis other proposal. I have not taken a look though.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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