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I am currently doing a proof which requires me, given some constant $c$ and some constant $k$, to evaluate the following: $$\sum_{n=1}^k\frac{1}{n^3-c}$$ Is there any known way to do such? If so, can it be generalized to: $$\sum_{n=1}^k\frac{1}{n^m-c}$$

Wolfram Alpha gives me a close form expression (for the former equation), but doesn't explain itself. It would be nice of one could give me an explanation of their reasoning.

  • Given that the special case of Apery's constant has no known closed form, finding a closed form for this seems non-trivial to say the least. – John Barber Sep 08 '17 at 03:57
  • True, but you are forgetting that the length of the sequence is $k$, which is finite. The length of the sequence that sums to Apery's Constant is infinite. –  Sep 08 '17 at 03:59
  • Yes, but if one were to find a closed-form expression for general $k$ and $c$, surely one could set $c = 0$ and take the limit as $k\rightarrow\infty$ and immediately recover an expression for Apery's constant. – John Barber Sep 08 '17 at 04:01
  • Well, not necessarily, because that closed form could not have a closed form limit. For example, the euler-mascheroni constant $\gamma$. Closed form function $\Gamma(k)-\frac{1}{k}$ as $k\rightarrow0$ is $\gamma$, but there is no closed form for $\gamma$. –  Sep 08 '17 at 04:05
  • Well, if we split up the summand using partial fractions, it can be evaluated in terms of polygamma functions and harmonic numbers with complex arguments. But this isn't much of a solution, since these functions are simply defined to be equal to the sums involved. – John Barber Sep 08 '17 at 04:16
  • Actually, if you were to tell me how you would do such, that would be great. Polygamma functions aren't nontrivial in this situation. –  Sep 08 '17 at 04:18

1 Answers1

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$$S(c,m,k)=\sum_{n=1}^k\frac{1}{n^m-c}\qquad (m\text{ integer})$$ FIRST : Solve the equation $\quad\nu^m=c\quad$ for $\nu$.

They are $m$ roots (real real and complex), namely $\quad\nu_j\quad$ with $\quad 1\leq j \leq m$.

SECOND : Convert $\quad\frac{1}{n^m-c}\quad$ to partial fractions : $$\frac{1}{n^m-c}=\sum_{j=1}^m \frac{a_j}{n-\nu_j}$$ Compute explicitly the $m$ coefficients $a_j$.

THIRD : Express $\quad\sum_{n=1}^k\frac{1}{n-\nu_j}\quad$ on closed form :

In the general case (arbitrary $\nu_j$) it is not possible with a finite number of elementary function. This requires a special function, namely the digamma function $\psi(x)$ : http://mathworld.wolfram.com/DigammaFunction.html

A property of this function is : $\quad\sum_{n=1}^k\frac{1}{n-C}=\psi(m+1-C)-\psi(1-C)\quad$.

Thus, for each $\nu_j$ : $$\sum_{n=1}^k\frac{1}{n-\nu_j}=\psi(m+1-\nu_j)-\psi(1-\nu_j)$$

RESULT : $$\sum_{n=1}^k\frac{1}{n^m-c} = \sum_{n=1}^k a_j\left(\psi(m+1-\nu_j)-\psi(1-\nu_j) \right)$$

You can do it in the case $\sum_{n=1}^k\frac{1}{n^3-c}$ : Compute $\nu_1\:,\:\nu_2\:,\:\nu_3$ , then $a_1\:,\:a_2\:,\:a_3$.

NOTE : For some particular values of $m$ the function digamma can be reduced to elementary functions (for example $m=4$). As a matter of fact, in those particular cases, the conversion to partial fractions leads then directly to the solution with no need for the function digamma.

JJacquelin
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