# [Calculation 20] Simple Series Calculation for Σ1/(n²ᵏ+c)

Edited by Leun Kim

(revision of 2012)  In this post, we will evaluate the closed form of the series $\sum_n (n^{2k}+x^k)^{-1}$ for $k=2,4,6,\cdots$ and $x>0$. We are going to show the following formula.

 Calculation 20. For any $k \in 2 \mathbb N$, and $x >0$, $$\sum_{n=-\infty}^{\infty} \frac{1}{n^{2k} +x^k} = \frac{2 \pi }{k} x^{\frac{1-2k}{2}} \sum_{r=1}^{k/2} \frac{ \sinh \left( 2 \pi b_{k,r} \sqrt{x} \right) \sin \frac{2r-1}{2k} \pi – \sin \left( 2 \pi a_{k,r} \sqrt{x} \right) \cos \frac{2r-1}{2k} \pi }{\cos \left ( 2 \pi a_{k,r} \sqrt{x} \right) – \cosh \left( 2 \pi b_{k,r} \sqrt{x} \right)}$$ holds true where $$a_{k,r} := \sin \left( \frac{k-1+2r}{2k} \pi \right),\qquad b_{k,r} := \cos \left( \frac{k-1+2r}{2k} \pi \right).$$

Before the proof, we introduce several simple examples of Calculation 20 in the case of $k=2,4$:

 Examples. For any $x>0$, we have \begin{align*} &\text{(i) } \sum_{n=1}^\infty \frac{1}{n^4 + x^2} = -\frac{1}{2x^2} + \frac{\sqrt{2} \pi}{4x\sqrt{x}}\cdot \frac{\sinh \left( \pi \sqrt{2x} \right) + \sin \left( \pi \sqrt{2x} \right)}{\cosh \left( \pi \sqrt{2x} \right) – \cos \left( \pi \sqrt{2x} \right)}\\ &\text{(ii) } \sum_{n=1}^{\infty}\frac{1}{n^8 + x^4} \\ &\quad\;\; = -\frac{1}{2x^4} + \frac{\pi}{8x^{\frac{7}{2}}} \left[\frac{\sqrt{2-\sqrt{2}} \sinh l + \sqrt{2+\sqrt{2}} \sin t}{\cosh l – \cos t} + \frac{ \sqrt{2 + \sqrt{2}} \sinh t + \sqrt{2 – \sqrt{2}} \sin l}{\cosh t – \cos l} \right] \end{align*} where $l=l(x) = \pi \sqrt{(2 – \sqrt{2})x}$ and $t=t(x) = \pi \sqrt{(2 + \sqrt{2}) x}$.

Proof. The proof follows just a usual application of the residue theorem. Let $k\in 2\N$ be given and assume $x>0$. First, we define $f(z) := (z^{2k} +x^k)^{-1} \pi \cot \pi z$ as usual and $C_N$ as below:

Then we have simple poles at the points

$$z_p = \sqrt{x} e^{\frac{p}{2k} \pi i }, \quad z_{q+1} = \sqrt{x} e^{- \frac{q}{2k} \pi i },\quad 1, \cdots, N$$

in $C_N$ for sufficiently large $N$, where $p,q = 1,3,5,\cdots, 2k-1$.

#### 1. Residues.

First, we can easily check that $$\text{Res}(f,n) = \lim_{z \to n} (z-n) \frac{ \pi \cot \pi z }{z^{2k} + x^k} = \lim_{z \to n} \frac{\pi}{z^{2k} + x^k} \lim_{z \to n} \frac{(z-n) \cos \pi z }{\sin \pi z} = \frac{1}{n^{2k} + x^k}$$

for all $n \in \Z$. Also we calculate
\begin{eqnarray*}
\text{Res}(f, z_1)
&=&
\lim_{z \to z_1} \frac{ \pi \cot \pi z }{(z-z_2) \cdots (z-z_{2k})} \\ &=&
\frac{\pi \cot \left ( \pi \sqrt{x} e^{\frac{\pi i }{2k}} \right)}{(2i \sqrt{x})^{2k-1} (-1)^{k-1} e^{\frac{k-1}{2k} \pi i } \prod_{m=1}^{k-1} \sin^2 \frac{m \pi }{2k}} \\
&=&
\frac{\pi e^{- \frac{2k-1}{2k} \pi i} \cot \left( \pi \sqrt{x} e^{\frac{\pi i }{2k}} \right)}{2k x^{\frac{2k-1}{2}}},
\end{eqnarray*}

where we used the trigonometric identity
\begin{align}
2^{2k-1} \prod_{m=1}^{k-1} \sin^2 \frac{m \pi }{2k} = 2k, \qquad k=2,3,4,\cdots
\end{align}
for the last equality (For the proof of (1), see Appendix below). Similarly, we see that
$$\text{Res}(f, z_p) = \frac{\pi e^{- \frac{2k – p}{2k} \pi i } \cot \left( \pi \sqrt{x} e^{\frac{p}{2k} \pi i } \right)}{2 k x^{\frac{2k-1}{2}}},$$ $$\text{Res}(f, z_{q+1}) = \frac{\pi e^{\frac{2k – q}{2k} \pi i } \cot \left( \pi \sqrt{x} e^{- \frac{q}{2k} \pi i } \right)}{2 k x^{\frac{2k-1}{2}}}$$
for $p,q = 1,3,5,\cdots , 2k-1$. Since $\int_{C_N} f(z)dz$ vanishes as $N\to\infty$, by the Residue Theorem, we have $$0 = 2 \pi i \left( \sum_{n= – \infty}^{\infty} \frac{1}{n^{2k} + x^k} + \sum_{r=1}^{2k} \text{Res} (f, z_r) \right).$$

#### 2. Calculations.

Let $y = \pi \sqrt{x}$, and define
$$A:= e^{y \exp \left( \frac{k-1+2r}{2k} \pi i \right) },\;B:= e^{-y \exp \left( \frac{k-1+2r}{2k} \pi i \right) }, \; C:= e^{y \exp \left( \frac{k+1-2r}{2k} \pi i \right) }, \; D:= e^{-y \exp \left( \frac{k+1-2r}{2k} \pi i \right) }$$

so that
$$AC = \exp \left ( 2y i \sin \frac{k-1+2r}{2k} \pi \right),\quad AD = \exp \left ( 2y \cos \frac{k-1+2r}{2k} \pi \right),$$
$$BC = \exp \left (- 2y \cos \frac{k-1+2r}{2k} \pi \right), \quad BD = \exp \left ( -2y i \sin \frac{k-1+2r}{2k} \pi \right).$$

Then by the direct calculations, we get
\begin{align*}
&\sum_{n= -\infty}^{\infty} \frac{1}{n^{2k} +x^k}
=
– \sum_{r=1}^{2k} \text{Res} (f, z_r) \\
&=
– \left[ \sum_{p=1,3,\cdots, 2k-1} \text{Res} (f, z_p) + \sum_{q=1,3, \cdots, 2k-1} \text{Res}(f, z_{q+1}) \right] \\
&=
– \sum_{r=1}^{k/2} \left[ \text{Res} \left( f,z_r \right) + \text{Res} \left( f, z_{k-r+1}\right) \right] \\
&=
– \frac{ \pi}{2k x^{\frac{2k-1}{2}}} \sum_{r=1}^k  \left( e^{- \frac{2k-(2r-1)}{2k} \pi i } – e^{\frac{2r-1}{2k} \pi i} \right) \cot \left (y e^{\frac{2r-1}{2k} \pi i } \right)\\
&=
\frac{\pi}{k x^{\frac{2k-1}{2}}} \sum_{r=1}^k e^{\frac{2r-1}{2k} \pi i} \cot \left( y e^{\frac{2r-1}{2k} \pi i } \right) \\
&=
\frac{\pi i}{k x^{\frac{2k-1}{2}}} \sum_{r=1}^k e^{\frac{2r-1}{2k} \pi i} \frac{e^{y \exp ( \frac{2r+k-1}{2k} \pi i )} + e^{-y \exp ( \frac{2r+k-1}{2k} \pi i )} }{e^{y \exp ( \frac{2r+k-1}{2k} \pi i )} – e^{-y \exp ( \frac{2r+k-1}{2k} \pi i )}} \\
&=
\frac{ \pi i }{k x^{\frac{2k-1}{2}}} \sum_{r=1}^{k/2} \left( e^{\frac{2r-1}{2k} \pi i} \cdot \frac{ A+B}{A-B} + e^{ – \frac{2r-1}{2k} \pi i} \cdot \frac{C+D}{C-D} \right) \\
&=
\frac{\pi i}{k x^{\frac{2k-1}{2}}} \sum_{r=1}^{k/2} \frac{ e^{\frac{2r-1}{2k} \pi i} (A+B)(C-D) + e^{- \frac{2r-1}{2k} \pi i } (C+D)(A-B)}{(A-B)(C-D)} \\
&=
\frac{\pi}{kx^{\frac{2k-1}{2}}} \sum_{r=1}^{k/2} \frac{ (AC-BD)i \left( e^{\frac{2r-1}{2k} \pi i} + e^{- \frac{2r-1}{2k} \pi i} \right) + (AD – BC)i \left( e^{ – \frac{2r-1}{2k} \pi i } –  e^{\frac{2r-1}{2k} \pi i} \right)}{(AC + BD) – (BC + AD)} \\
&=
\frac{2 \pi }{k} x^{\frac{1-2k}{2}} \sum_{r=1}^{k/2} \frac{ \sinh \left( 2 \pi b_{k,r} \sqrt{x} \right) \sin \frac{2r-1}{2k} \pi – \sin \left( 2 \pi a_{k,r} \sqrt{x} \right) \cos \frac{2r-1}{2k} \pi }{\cos \left ( 2 \pi a_{k,r} \sqrt{x} \right) – \cosh \left( 2 \pi b_{k,r} \sqrt{x} \right)},\end{align*}
which proves our claim.$\square$

#### 3. Appendix : Proof of (1)

Here we are going to show that
$$2^{2k-1} \prod_{m=1}^{k-1} \sin^2 \frac{m \pi}{2k} = 2k$$
for $k=2,3,4,\cdots$. First we note that

\begin{eqnarray*} \prod_{m=1}^{k-1} \sin \frac{m \pi}{2k}
&=&
\prod_{m=1}^{k-1} \frac{e^{\frac{m \pi}{2k} i} – e^{- \frac{m \pi}{2k} i}}{2i} \\
&=&
(2i)^{1-k} e^{-\sum_{m=1}^{k-1} \frac{m \pi}{2k} i} \prod_{m=1}^{k-1} \left( e^{ \frac{m \pi i}{k}} -1 \right) \\
&=&
(-2i)^{1-k} e^{- \frac{\pi i (k-1)}{4}} \prod_{m=1}^{k-1} \left ( 1 – e^{ \frac{m \pi i }{k}} \right) \\
&=&
(-2i)^{1-k} e^{- \frac{\pi i (k-1)}{4}} \sqrt{k}e^{- \frac{\pi i (k-1)}{4}} \\
&=&
\sqrt{k} 2^{1-k}.
\end{eqnarray*}

Thus we get
$$2^{2k-1} \prod_{m=1}^{k-1} \sin^2 \frac{m \pi}{2k} = 2^{2k-1} \left( \prod_{m=1}^{k-1} \sin \frac{m \pi}{2k} \right)^2 = 2k$$
for all $k=2,3,4,\cdots$.$\square$

#### Leun Kim

Ph.D Candidate at The University of Tokyo
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.