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There is post here asking for $$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$ And the answer is $18-24\ln 2$.

It is easy to elvaluate that $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1+2+\cdots+n}=2$ and it is not difficult to justify the convergence of $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$ for all $k>1$.

Here comes my question: How to evaluate $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^3+2^3+\cdots+n^3}$ or in general, what is $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$ if $k$ is a positive integer.

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  • $\begingroup$ could probably follow similar logic for $k=3,4$, doubt you can find for $k=5$. $\endgroup$ – gt6989b Feb 28 '13 at 23:57
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    $\begingroup$ @gt6989b in fact $k=3$ and $k=5$ is possible, $k=4$ is not a nice number $\endgroup$ – Dominic Michaelis Mar 1 '13 at 0:04
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I guess the way is pretty much the same, as there are always formulas to evaluate $$\sum_{i=1}^n i^k, $$ For $k=3$ the result of the sum above is $$\frac{4}{3} (-9+\pi^2)$$ For higher terms you will probably not always get a nice result, but for $k=5$ you have $$ 4\left(15-\pi^2 + 2\sqrt{3} \pi \tan\left(\frac{\sqrt{3}\pi}{2}\right)\right)$$

I will calculate the $k=3$ case.
As $$\sum_{i=1}^n i^3 = \frac{1}{4} n^2 (1+n)^2$$ we have \begin{align*} \sum_{n=1}^\infty \frac{1}{\sum_{i=1}^n n^3 } &= 4 \sum_{n=1}^\infty \frac{1}{m^2(1+m)^2}\\ &=\sum_{n=1}^\infty \frac{1}{n^2} - \frac{2}{n} + \frac{2}{n+1} + \frac{1}{(n+1)^2}\\ &=4\sum_{n=1}^\infty \frac{1}{n^2} +4\sum_{n=1}^\infty \frac{1}{(1+n)^2} - 8 \cdot \sum_{n=1}^\infty \frac{1}{n} -\frac{1}{n+1}\\ &=4\left(\frac{\pi^2}{6} +\frac{\pi^2}{6}-1-2\right)\\ &=\frac{4}{3}(\pi^2-9)\end{align*}

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  • $\begingroup$ Mh where ? the zeta function is $$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s} $$ $\endgroup$ – Dominic Michaelis Feb 28 '13 at 23:52
  • $\begingroup$ OOps. I meant the Bernoulli function! I suppose $k=3$ refers to OP's question. $\endgroup$ – Maesumi Feb 28 '13 at 23:58
  • $\begingroup$ @DominicMichaelis thanks for your answer. For $k=3$, I was thinking of the same approach as well. I believe that you are using the similar method for $k=5$. Just curious that why $k=4$ was missing? $\endgroup$ – pipi Mar 1 '13 at 0:53
  • $\begingroup$ Mathematica result for $k=4$ is $$\frac{30}{7} \left(16 \log (2)-3 \left(H_{\frac{1}{6} \left(3-\sqrt{21}\right)}+H_{\frac{1}{6} \left(3+\sqrt{21}\right)}+3\right)\right)$$ after a Fullsimplify, for $k=4$ I have a Denominator which i can't simplify as much as i liek $\endgroup$ – Dominic Michaelis Mar 1 '13 at 0:59
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Though it may be tricky to find a general formula, we get with Faulhaber's formula that

$$\sum_{n=1}^\infty \frac{1}{\sum_{m=1}^n m^k}= \sum_{n=1}^\infty \frac{1}{\sum_{j=0}^k\binom{k}{j}\frac{B_{k-j}}{j+1}n^{j+1}}=$$

where $B_k$ are Bernoulli numbers.

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    $\begingroup$ Is it just me or does it look more complicated that way? Could you explain what that may help ? $\endgroup$ – Dominic Michaelis Mar 1 '13 at 0:06
  • $\begingroup$ @DominicMichaelis The number of the terms in the sum is only $k$. This is the formula used to obtain the sum of the squares from the post being referenced by the OP. $\endgroup$ – Argon Mar 1 '13 at 0:08
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Once the sum over the $k$-th powers is written as a polynomial in $n$, the generic partial fraction decomposition leads to some manageable result, see http://psychedelic-geometry.blogspot.de/2009/04/square-pyramidal-numbers-reciprocals.htm . The differences appear mainly due to the problem of finding the roots of $(k+1)$st order polynomials.

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