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Please help me to find the sum of $\sum\limits_{n=1}^\infty \left[ \frac{\left(\frac{3 - \sqrt 5}{2}\right)^n}{n^3}\right]$

Is there any special technique to solve this one ?

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closed as off-topic by Travis, Claude Leibovici, 6005, Mark Fantini, RE60K Jan 14 '15 at 11:03

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  • $\begingroup$ I think it's a duplicate :3 unless $[.]$ means a box function in this context! $\endgroup$ – r9m Jan 14 '15 at 10:18
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I don't know anything about the polylogarithm function myself, so if I were forced to solve the problem I would use an elementary technique. Here, through a series of differentiations and multiplication by $x$, I can deduce an integral equation for a function with the given series expansion:

Assume $f(x) = \sum_{n=1}^{\infty} {x^n \over n^3}$. Then we want $f\left({3-\sqrt{5} \over 2} \right)$. This series is dominated by the geometric series so it must converge absolutely at the given point (which has modulus less than 1). Now, term-by-term we have

$$ x{d\over dx} \left[x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^3} \right] \right] \right] = x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^2} \right] \right] = ... = x^n, $$ so $$ \sum_{n=1}^{\infty} x{d\over dx} \left[x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^3} \right] \right] \right] = \sum_{n=1}^{\infty} x^n = {x\over 1-x}.$$ Moving the sum through the derivatives, we find $x(x(xf'(x))')' = {x\over 1-x}$. Solving this differential equation gives an expression for $f$ which can be used to evaluate the sum.

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  • $\begingroup$ Thank you. The idea was innovative as well as beautiful indeed. $\endgroup$ – Anjan3 Jan 14 '15 at 6:14
  • $\begingroup$ ... but you won't get an explicit answer except in terms of a polylogarithm or related special function. BTW, you're missing a factor of $x$: the right side of your differential equation should be $x/(1-x)$ $\endgroup$ – Robert Israel Jan 14 '15 at 6:18
  • $\begingroup$ No escape from using polylogarithm as I said. $\endgroup$ – Mhenni Benghorbal Jan 14 '15 at 6:25
  • $\begingroup$ @Robert Thank you, fixed. You are right, this problem is drawn directly from the definition of a polylogarithm. I'll look at it again. $\endgroup$ – Titus Jan 14 '15 at 6:43
  • $\begingroup$ As Robert has posted, the closest thing to a closed form expression involves $\zeta(3)$ and the expression for $f$ above leaves one with an integral equation (describing $Li_3$). So there doesn't appear to be a good closed-form description of your series without using polylogarithms or zeta functions (which are just placeholders for certain series). That being said, if one were to study the properties of such sums this seems like the most obvious inroad. $\endgroup$ – Titus Jan 14 '15 at 7:07
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Practically speaking, considering the terms in $$S_n=\sum_{i=1}^n \frac{a^n}{n^3}$$ with $a=\frac{1}{2} \left(3-\sqrt{5}\right)$, you can notice that $a$ is rather small ($\approx 0.382$) which means that the numerator of the fraction $a^n$ will decrease quite fast while the denominator $n^3$ will increase quite fast; this makes that each term will be significantly smaller than the previous.

Let us compute the partial sums and get for $6$ significant figures $$S_1=0.381966$$ $$S_2=0.400203$$ $$S_3=0.402267$$ $$S_4=0.402600$$ $$S_5=0.402665$$ $$S_6=0.402679$$ $$S_7=0.402683$$ $$S_8=0.402684$$ Now, the question is : how many terms $k$ have to be added in order to reach an accuracy such that $$\frac{a^k}{k^3} \leq \epsilon$$

The answer is given by the solution of $a^k =k^3 \epsilon$ which can be expressed in terms of Lambert function (another special function) $$k=-\frac{3 }{\log (a)}W\left(\frac{1}{3 \sqrt[3]{-\frac{\epsilon }{\log ^3(a)}}}\right)$$ which can seem very complex. However, very good approxations exist for Lambert function such as $$W(x)\approx L_1-L_2+\frac{L_2}{L_1}$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.

Applied to the case $a=\frac{1}{2} \left(3-\sqrt{5}\right)$ and $\epsilon=10^{-6}$, this would give for the value of the argument of Lambert function $x=32.0808$ from which $W(x)=2.58319$ and then $k=8.05213$ which is what we saw ealier.

If you want to change the tolerance to $\epsilon=10^{-p}$, quick and dirty fit would show that the number of terms to be added is approximately given by $$k=0.0148695 p^2+1.67776 p-2.80166$$

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hint: Check polylogarithm function. See related problems I.

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  • $\begingroup$ Sorry sir, I really don't know that stuff. Is it possible to solve this problem through some basic technique ? $\endgroup$ – Anjan3 Jan 14 '15 at 5:51
  • $\begingroup$ when do we come across polylogarithm? btw sorry if my answer hurt(ed) you $\endgroup$ – RE60K Jan 14 '15 at 5:55
  • $\begingroup$ @AnjanDebnath: it is a special function. You should learn it $\endgroup$ – Mhenni Benghorbal Jan 14 '15 at 5:59
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According to Maple, $$ \text{polylog}\left(3, \dfrac{3-\sqrt{5}}{2}\right) = \dfrac{4}{5} \zeta(3) + \dfrac{\pi^2}{15} \ln \left(\dfrac{3-\sqrt{5}}{2}\right) - \dfrac{1}{12} \ln\left(\dfrac{3-\sqrt{5}}{2}\right)^3 $$

I don't know where it gets this rather remarkable identity.

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    $\begingroup$ for the polylog$(3,\frac{3 -\sqrt{5}}{2})$ you may check here, I think it's a dupe unless $[.]$ means a box function in this context. $\endgroup$ – r9m Jan 14 '15 at 10:21
  • $\begingroup$ seethis $\endgroup$ – RE60K Jan 14 '15 at 11:03

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