Polylogarithm

The polylogarithm function is an important function for integration, and finding seemingly complicated sum.

Definition

The polylogarithm function is defined as $\operatorname{Li}_s(x) = \sum_{n=1}^\infty \dfrac{x^n}{n^s}$ for all complex $s$ and $|x|\leq 1$ and can be computed for $|x|>1$ by analytical continuation. $_\square$

Connection between 2 Polylogarithms

$\operatorname{Li}_s(x)=\int_0^x \dfrac{\operatorname{Li}_{s-1}(u)}{u} du.\ _\square$

Consider $I=\int_0^x \dfrac{\operatorname{Li}_{s-1}(u)}{u} du=\int_0^x \sum_{n=1}^\infty \dfrac{u^{n-1}}{n^{s-1}} du.$ Interchanging the summation and integral sign, we end up with $I=\sum_{n=1}^\infty \int_0^x \dfrac{u^{n-1}}{n^{s-1}}du=\sum_{n=1}^\infty \dfrac{x^n}{n^s} = \operatorname{Li}_s(x).\ _\square$

Polylogarithm and Geometric Progression

Polylogarithm is connected to the infinite geometric progression sum $\operatorname{Li}_0(x)=\sum_{n=1}^\infty x^n=\dfrac{x}{1-x}.$ We can divide by $x$ and differentiate with respect to $x$ to get $\operatorname{Li}_{-1}(x)=\sum_{n=1}^\infty nx^n=\dfrac{x}{(1-x)^2}.$ We can keep doing this to get closed form of $\operatorname{Li}_{-s} (x)$ , but if we were given something like $\operatorname{Li}_{-25}(0.5)$ , it would become extremely tedious to divide by $x$ and differentiate with respect to $x$ 25 times. Similar to the binomial theorem which was a shortcut to the tedious Pascal's triangle, there is an easy way to calculate these:

$\begin{array} { | c | c | l | } \hline s & \operatorname{Li}_s(z) & \text{ Algebraic expression } \\ \hline 1 & \operatorname{Li}_1(z) & -\ln(1-z) \\ \hline 0 & \operatorname{Li}_0(z) & \dfrac z{1-z} \\ \hline -1 & \operatorname{Li}_{-1}(z) & \dfrac {z}{(1-z)^2} \\ \hline -2 & \operatorname{Li}_{-2}(z) & \dfrac {z(z+1)}{(1-z)^3} \\ \hline -3 & \operatorname{Li}_{-3}(z) & \dfrac {z(z^2+4z+1)}{(1-z)^4} \\ \hline -4 & \operatorname{Li}_{-4}(z) & \dfrac {z(1+z)(z^2+10z+1)}{(1-z)^5} \\ \hline -5 & \operatorname{Li}_{-5}(z) & \dfrac {z(z^4+26z^3+ 66z^2+26z+1)}{(1-z)^6} \\ \hline \end{array}$

In general, we can express $ \def\Li{Li\,} \Li_{-n}(z) $ as \[ \def\Li{Li\,} \Li_{-n}(z) = \left( z \; \dfrac {\partial}{ \partial z} \right)^n \dfrac z{1-z} = \sum_{k=0}^n k! \; S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k+1} \; , \] where $n$ is a non-negative integer, and $S(n,k)$ denote the Stirling's number of the second kind. $_\square$

Try to solve the following problem using polylogarithms!

$\large \displaystyle \sum_{n=1}^{\infty} \frac {n^5}{2^n}$

Let the value of the above summation be $A$ . What is the value of $\sqrt{A+7}$ ?

The correct answer is: 33

We wish to calculate $\displaystyle \sum_{n=1}^\infty \dfrac{n^5}{x^n}$ , when $x=2$ .

Let us start with the arithmetic-geometric progression, $\displaystyle \sum_{n=1}^\infty \dfrac n{x^n} = \dfrac x{(x-1)^2} = \dfrac1{x-1} + \dfrac1{(x-1)^2},$ where $|x| > 1$ .

If we differentiate the expression with respect to $x$ , we get

$\displaystyle -\sum_{n=1}^\infty \dfrac {n^2}{x^{n+1}} = -\dfrac1{(x-1)^2} - \dfrac1{2(x-1)^3} \; .$

Multiplying both sides by $-x$ gives us

$\displaystyle \sum_{n=1}^\infty \dfrac {n^2}{x^n} = \dfrac x{(x-1)^2} + \dfrac x{2(x-1)^3} = \dfrac 1{x-1} + \dfrac1{(x-1)^2} + \dfrac1{2(x-1)^2} + \dfrac1{2(x-1)^3} \; .$

Once again, let's differentiate the latest expression with respect to $x$ to obtain

$\displaystyle - \sum_{n=1}^\infty \dfrac {n^3}{x^{n+1}} = -\dfrac1{(x-1)^2} - \dfrac1{2(x-1)^3} - \dfrac1{4(x-1)^3} - \dfrac1{6(x-1)^4} \; .$

Multiplying both sides by $-x$ gives us

$\begin{aligned} && \displaystyle \sum_{n=1}^\infty \dfrac {n^3}{x^{n}} \\ &=& \dfrac x{(x-1)^2} + \dfrac x{2(x-1)^3} + \dfrac x{4(x-1)^3} + \dfrac x{6(x-1)^4} \\ &=& \dfrac1{x-1} + \dfrac1{(x-1)^2} + \dfrac1{2(x-1)^2} + \dfrac1{2(x-1)^3} + \dfrac 1{4(x-1)^2} + \dfrac1{4(x-1)^3} + \dfrac1{6(x-1)^3} + \dfrac1{6(x-1)^4} \; . \end{aligned}$

Do you see the pattern here? Every time we differentiate and multiply by $-x$ , we get closer to the desired expression.

Repeating these steps another two more times tells us that

$\displaystyle \sum_{n=1}^\infty \dfrac {n^5}{x^{n}} = \dfrac1{x-1} + \dfrac{31}{(x-1)^2} + \dfrac{180}{(x-1)^3} + \dfrac{390}{(x-1)^4} + \dfrac{360}{(x-1)^5} + \dfrac{120}{(x-1)^6} \; .$

Substituting $x=2$ gives $\displaystyle \sum_{n=1}^\infty \dfrac {n^5}{2^{n}} = 1082$ , implying $A = 1082$ and thus our answer is $\sqrt{A+7} = 33$ .

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