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?

  • $\begingroup$ What is $n$? It appears neither as an index, nor as an index bound. $\endgroup$ – avs Aug 18 '16 at 17:01

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,

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.