Have you heard of reciprocal power sum?

Algebra Level 5

$\frac 1 1 + \frac 1 {1^3 + 2^3}+ \frac 1 {1^3 + 2^3 + 3^3 }+ \frac 1 {1^3 + 2^3 + 3^3 +4^3 } + \ldots$

If the series above can be expressed as $\frac {a \pi^2}{b} - c$ for positive integers $a,b,c$ with $a,b$ coprime. Find $a+b+c$ .

The answer is 19.

1 solution

Brian Charlesworth
Mar 31, 2015

Since $\sum_{k=1}^{n} k^{3} = (\frac{n(n + 1)}{2})^{2}$ this series can be written as

$\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{2}{n(n + 1)}\right)^{2} = 4*\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)^{2} =$

$4*\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{2}} + 4*\sum_{n=1}^{\infty} \dfrac{1}{(n + 1)^{2}} - 8*\sum_{n=1}^{\infty} \dfrac{1}{n(n + 1)} =$

$4*\dfrac{\pi^{2}}{6} + 4*(\dfrac{\pi^{2}}{6} - 1) - 8*\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right) = \dfrac{4\pi^{2}}{3} - 4 - 8 = \dfrac{4\pi^{2}}{3} - 12.$

(Note that the sum in this last line is a telescoping sum, where all terms cancel pairwise except the very first term, namely $1$ . Note also that $\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ is the well-known Basel sum.)

Thus $a + b + c = 4 + 3 + 12 = \boxed{19}.$

Moderator note:

Fantastic! Is it possible to state $\displaystyle \sum_{k=1}^\infty \frac {1}{1^4 + 2^4 + \ldots + k^4 }$ in terms of elementary terms?

Small erratum: I think you mean $\displaystyle \sum_{k=1}^{n} k^{3} = (\frac{n(n+1)}{2})^{2}$ in the first line.

Jake Lai - 6 years, 2 months ago

Edit made; thanks for catching that. :)

Brian Charlesworth - 6 years, 2 months ago

Have you heard of reciprocal power sum?

The answer is 19.

1 solution

Moderator note:

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