Polylogarithm is connected to the infinite geometric progression sum
Li0(x)=n=1∑∞xn=1−xx.
We can divide by x and differentiate with respect to x to get
Li−1(x)=n=1∑∞nxn=(1−x)2x.
We can keep doing this to get closed form of Li−s(x), but if we were given something like 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:
In general, we can express Li−n(z) as Li−n(z)=(z∂z∂)n1−zz=k=0∑nk!S(n+1,k+1)(1−zz)k+1, where n is a non-negative integer, and S(n,k) denote the Stirling's number of the second kind. □
Try to solve the following problem using polylogarithms!
n=1∑∞2nn5
Let the value of the above summation be A. What is the value of A+7?