Polylog

Calculus Level 5

$\large \sum _{k=1}^{\infty }\left[\text{Li}_{2k}\left(\frac 14 \right) - \frac 14\right]=\frac{A}{B}-\frac{C}{D}\ln 2+\frac{C}{E}\ln 3$

The equation above holds true for positive integers $A,B,C,D$ and $E$ , where $\gcd(A,B)=\gcd(C,D)=\gcd(C,E)=1$ .

Find $A+B+C+D+E$ .

Notation : ${ \text{Li} }_{ n }(a)$ denotes the polylogarithm function, ${ \text{Li} }_{ n }(a)=\displaystyle\sum _{ k=1 }^{ \infty }{ \dfrac { { a }^{ k } }{ { k }^{ n } } }$ .

The answer is 52.

1 solution

Mark Hennings
Jun 11, 2016

$\begin{array}{rcl} \displaystyle \sum_{k=1}^\infty \left(\mathrm{Li}_{2k}\big(\tfrac14\big) - \tfrac14\right) & = & \displaystyle \sum_{k=1}^\infty \left(\sum_{n=1}^\infty \frac{1}{n^{2k}4^n} - \tfrac14\right) \\ & = & \displaystyle \sum_{k=1}^\infty \sum_{n=2}^\infty \frac{1}{n^{2k}4^n} \; =\; \sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{n^{2k}4^n} \\ & = & \displaystyle \sum_{n=2}^\infty \frac{1}{(n^2-1)4^n} \; = \; \sum_{n=2}^\infty \frac{1}{2^{2n+1}}\left(\frac{1}{n-1} - \frac{1}{n+1}\right) \\ & = &\displaystyle \sum_{n=1}^\infty \frac{1}{n2^{2n+3}} - \sum_{n=3}^\infty \frac{1}{n2^{2n-1}} \; = \; -\tfrac18\ln\tfrac34 + \tfrac12 + \tfrac{1}{16} + 2\ln\tfrac34 \\ & = & \tfrac{9}{16} - \tfrac{15}{4}\ln2 + \tfrac{15}{8}\ln3 \end{array}$ making the answer $9+16+15+4+8 = \boxed{52}$ .

Polylog

The answer is 52.

1 solution

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