1+2015 = 2016

Calculus Level 5

∑ n = 1 ∞ 1 + 201 5 n 201 6 n n 2 = π A B + ln ⁡ ( C ) ln ⁡ ( D ) − ( ln ⁡ ( C ) ) 2 \large{\displaystyle \sum^{\infty}_{n=1} \frac{1+2015^{n}}{2016^{n}n^{2}}=\frac{\pi^{A}}{B}+\ln (C) \ln(D)-(\ln(C))^2}

If the equation is true for positive integers A , B , C A,B,C and D D , find A + B + C + D A+B+C+D .


The answer is 4039.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

∑ n = 1 ∞ 1 + 201 5 n 201 6 n n 2 = ∑ n = 1 ∞ 1 201 6 n n 2 + ∑ n = 1 ∞ 201 5 n 201 6 n n 2 = Li 2 ( 1 2016 ) + Li 2 ( 2015 2016 ) Li 2 ( z ) is dilogarithm function = Li 2 ( 1 2016 ) + Li 2 ( 1 − 1 2016 ) By Abel identity with x = y = π 2 6 − ln ⁡ ( 1 2016 ) ln ⁡ ( 1 − 1 2016 ) = π 2 6 − ln ⁡ ( 1 2016 ) ln ⁡ ( 2015 2016 ) = π 2 6 + ln ⁡ 2016 ln ⁡ 2015 − ln ⁡ 2 2016 \begin{aligned} \sum_{n=1}^\infty \frac{1+2015^n}{2016^n n^2} & = \sum_{n=1}^\infty \frac{1}{2016^n n^2} + \sum_{n=1}^\infty \frac{2015^n}{2016^n n^2} \\ & = \text{Li}_2 \left(\frac{1}{2016} \right) + \text{Li}_2 \left(\frac{2015}{2016} \right) \quad \quad \quad \quad \small \color{#3D99F6}{\text{Li}_2(z) \text{ is dilogarithm function}} \\ & = \text{Li}_2 \left(\frac{1}{2016} \right) + \text{Li}_2 \left(1 - \frac{1}{2016} \right) \quad \quad \small \color{#3D99F6}{\text{By Abel identity with } x =y} \\ & = \frac{\pi^2}{6} - \ln\left(\frac{1}{2016} \right) \ln\left(1-\frac{1}{2016} \right) \\ & = \frac{\pi^2}{6} - \ln\left(\frac{1}{2016} \right) \ln\left(\frac{2015}{2016} \right) \\ & = \frac{\pi^2}{6} + \ln 2016 \ln 2015 - \ln^2 2016 \end{aligned}

⇒ A + B + C + D = 2 + 6 + 2016 + 2015 = 4039 \Rightarrow A+B+C+D = 2 + 6 + 2016 + 2015 = \boxed{4039}

More about polylogarithm function and Abel identy .

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