Reciprocal sums

Calculus Level 5

If 1 1 + 1 1 + 2 + 1 1 + 2 + 3 + ⋯ = 2 \dfrac{1}{1}+ \dfrac{1}{1+2}+\dfrac{1}{1+2+3} +\cdots = 2 , what is

1 1 2 + 1 1 2 + 2 2 + 1 1 2 + 2 2 + 3 2 + ⋯ = ? \large \frac{1}{1^2} +\frac{1}{1^2 + 2^2} +\frac{1}{1^2+2^2+3^2} +\cdots =\,?

Input your answer to 3 decimal places.


The answer is 1.364.

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Mark Hennings
Aug 15, 2017

We have ∑ n = 1 ∞ 6 n ( n + 1 ) ( 2 n + 1 ) = ∑ n = 1 ∞ ( 6 n + 6 n + 1 − 24 2 n + 1 ) = ∑ n = 1 ∞ ( 6 n + 1 − 6 n + 24 2 n − 24 2 n + 1 ) = ∑ n = 1 ∞ ( 6 n + 1 − 6 n ) + 24 ∑ n = 2 ∞ ( − 1 ) n n = − 6 + 24 ( 1 − ln ⁡ 2 ) = 18 − 24 ln ⁡ 2 = 1.36447 \begin{aligned} \sum_{n=1}^\infty \frac{6}{n(n+1)(2n+1)} & = \; \sum_{n=1}^\infty \left(\frac{6}{n} + \frac{6}{n+1} - \frac{24}{2n+1}\right) \; = \; \sum_{n=1}^\infty \left(\frac{6}{n+1} - \frac{6}{n} + \frac{24}{2n} - \frac{24}{2n+1}\right) \\ & = \; \sum_{n=1}^\infty \left(\frac{6}{n+1} - \frac{6}{n}\right) + 24\sum_{n=2}^\infty \frac{(-1)^n}{n} \; = \; -6 + 24(1 - \ln2) \; = \; 18 - 24\ln2 \; = \; \boxed{1.36447} \end{aligned}

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