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I'd like to write the following sum in a closed form:

$$\sum_{i=1}^{N} \frac{i^n}{k + i^2}$$

where $k$ is a positive real. Is this possible?

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  • $\begingroup$ What is $n$? It appears neither as an index, nor as an index bound. $\endgroup$ – avs Aug 18 '16 at 17:01
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Let me use $j$ for the summation index, because I'll want $i$ for $\sqrt{-1}$. I'll assume $m$ is a nonnegative integer.

$$ \dfrac{1}{k+j^2} = \dfrac{i}{2 \sqrt{k}(j + i \sqrt{k})} - \dfrac{i}{2\sqrt{k} (j - i \sqrt{k})} $$

Now if $S(m)$ is your sum, we have

$$\eqalign{S(m) &= \sum_{j=1}^{N} \text{Re} \left(\dfrac{ij^m}{\sqrt{k}(j + i \sqrt{k})} \right)\cr &= \sum_{j=1}^N \text{Re} \left( \dfrac{i(-i\sqrt{k})^m}{\sqrt{k}(j+i\sqrt{k}) } + \sum_{n=0}^{m-1} (-1)^{m-n+1} i^{m-n}k^{(m-n-2)/2}j^n \right) }$$ Note that $$ \sum_{j=1}^N (j+i\sqrt{k})^{-1} = \Psi(N+1 +i \sqrt{k}) - \Psi(1+i\sqrt{k})$$ while for $n \ge 0$, $\sum_{j=1}^N j^n$ is a polynomial in $N$ given by Faulhaber's formula. So for each $m$ we will get a "closed-form" formula involving $\Psi$ and a polynomial. For example,

$$ S(3) = \frac{N^2 + N -k\Psi \left( N+1-i\sqrt {k} \right) -k\Psi \left( N+1+i\sqrt { k} \right) +k\Psi \left( 1-i\sqrt {k} \right) +k\Psi \left( 1+i\sqrt { k} \right) }{2}$$

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  • $\begingroup$ Thanks so much! Could you explain how you got the expression in the second equation for S(m), where you introduce the second summation? $\endgroup$ – user2452976 Aug 18 '16 at 20:14
  • $\begingroup$ $$ \dfrac{x^m}{x-y} = \dfrac{y^m}{x-y} + \dfrac{x^m-y^m}{x-y} = \dfrac{y^m}{x-y} + y^{m-1} + y^{m-2} x + \ldots + x^{m-1}$$ Here $x = j$ and $y = -i\sqrt{k}$ (and we have an extra factor of $i/\sqrt{k}$). $\endgroup$ – Robert Israel Aug 18 '16 at 21:24
  • $\begingroup$ @user2452976 Geometric series. $\endgroup$ – Simply Beautiful Art Aug 18 '16 at 21:39
  • $\begingroup$ Hmm. I seem to have left out some factors of $-1$. Corrected. $\endgroup$ – Robert Israel Aug 18 '16 at 22:06

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