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Series with roots of digamma

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Post Thu Mar 03, 2016 10:10 am

Posts: 46
Location: http://www.tolaso.com.gr/
Let $a_k$ denote the zeroes of digamma. Prove that:
$$\sum_{k=0}^{\infty} \frac{1}{a_k^2}=\gamma^2 +\frac{\pi^2}{2}$$

Post Sun Jun 05, 2016 2:30 am

Posts: 38
Sum involving roots of a function immediately suggests logarithms of a Weierstrass/Hadamard factorization. $\psi(z)$ has simple poles at nonpositive integers, same as $\Gamma(z)$, so $\psi(z)/\Gamma(z)$ is entire and has a factorization of the form
$$\frac{\psi(z)}{\Gamma(z)}=c_{1}e^{c_{2}z}\prod_{a\in A}\left(1-\frac{z}{a}\right)e^{\frac{z}{a}}$$
where A is the set of zeros of $\psi(z)$. At z=0, we have the asymptotic formulas
$$\Gamma(z)=\frac{\Gamma(z+1)}{z}=\frac{1}{z}+O(1)$$
$$\psi(z)=\psi(z+1)-\frac{1}{z}=-\frac{1}{z}-\gamma+\frac{\pi^{2}}{6}z+O(z^{2})$$
$$\psi^{(1)}(z)=\psi^{(1)}(z+1)+\frac{1}{z^{2}}=\frac{1}{z^{2}}+\frac{\pi^{2}}{6}+O(z)$$
The value of $c_{1}$ is not actually needed to evaluate the sum but it can be computed using
$$\frac{\psi(z)}{\Gamma(z)}=\frac{-\frac{1}{z}+O(1)}{\frac{1}{z}+O(1)}=-1+O(z)$$
so $c_{1}=-1$. Logarithmically differentiating the product gives
$$\frac{\psi^{(1)}(z)}{\psi(z)}-\psi(z)=c_{2}+\sum_{a\in A}\frac{1}{z-a}+\frac{1}{a}$$
Using the asymptotics again, we get
$$\frac{\psi^{(1)}(z)}{\psi(z)}-\psi(z)=2\gamma+O(z)$$
so taking $z\to 0$ we see that $c_{2}=2\gamma$. Now we can use
$$\frac{1}{z}\left(\psi(z)-\frac{\psi^{(1)}(z)}{\psi(z)}+2\gamma\right)=\sum_{a\in A}\frac{1}{a(a-z)}$$
again with the asymptotics to get
$$\begin{align*}\frac{1}{z}\left(\psi(z)-\frac{\psi^{(1)}(z)}{\psi(z)}+2\gamma\right)&=\frac{1}{z}\left(\left(-\frac{1}{z}-\gamma+\frac{\pi^{2}}{6}z+O(z^{2})\right)-\frac{\frac{1}{z^{2}}+\frac{\pi^{2}}{6}+O(z)}{-\frac{1}{z}-\gamma+\frac{\pi^{2}}{6}z+O(z^{2})}+2\gamma\right)\\
&=\gamma^{2}+\frac{\pi^{2}}{2}+O(z)\end{align*}$$
Now take $z\to 0$ to get
$$\sum_{a\in A}\frac{1}{a^{2}}=\gamma^{2}+\frac{\pi^{2}}{2}$$

EDIT:

Another method of evaluating the sum is by repeatedly differentiating
$$\frac{\psi^{(1)}(z)}{\psi(z)}-\psi(z)=2\gamma+\sum_{a\in A}\frac{1}{z-a}+\frac{1}{a}$$
This doesn't require computation of $c_{2}$ but does require series expansions of higher order polygammas. Fortunately these are easy to compute, so we get
$$\begin{align*}
\sum_{a\in A}\frac{1}{a^{s}}&=\frac{1}{(s-1)!}\lim_{z\to 0}\left(\frac{d}{dz}\right)^{s}\log\frac{\psi(z)}{\Gamma(z)} \\
&=-s\operatorname{Res}\left(\frac{1}{z^{s+1}}\log\frac{\psi(z)}{\Gamma(z)},z=0\right)
\end{align*}$$
for integers s>1. With the help of mathematica, we have the following table of values:
$$\begin{array}{cc}
s & \sum_{a\in A}a^{-s} \\
2 & \gamma ^2+\frac{\pi ^2}{2} \\ \\
3 & -4 \zeta (3)-\gamma ^3-\frac{\gamma \pi ^2}{2} \\ \\
4 & 4 \gamma \zeta (3)+\gamma ^4+\frac{\pi ^4}{9}+\frac{2 \gamma ^2 \pi ^2}{3} \\ \\
5 & -5 \gamma ^2 \zeta (3)-\frac{5 \pi ^2 \zeta (3)}{6}-6 \zeta (5)-\gamma ^5-\frac{5 \gamma ^3 \pi ^2}{6}-\frac{7 \gamma \pi ^4}{36} \\ \\
6 & 3 \zeta (3)^2+6 \gamma ^3 \zeta (3)+2 \gamma \pi ^2 \zeta (3)+6 \gamma \zeta (5)+\gamma ^6+\frac{\pi ^6}{36}+\frac{19 \gamma ^2 \pi ^4}{60}+\gamma ^4 \pi ^2 \\ \\
7 & -7 \gamma \zeta (3)^2-7 \gamma ^4 \zeta (3)-\frac{49 \pi ^4 \zeta (3)}{180}-\frac{7}{2} \gamma ^2 \pi ^2 \zeta (3)-7 \gamma ^2 \zeta (5)-\frac{7 \pi ^2 \zeta (5)}{6}-8 \zeta (7)-\gamma ^7 \\
& -\frac{7 \gamma ^5 \pi ^2}{6}-\frac{7 \gamma ^3 \pi ^4}{15}-\frac{71 \gamma \pi ^6}{1080} \\ \\
8 & 12 \gamma ^2 \zeta (3)^2+\frac{4 \pi ^2 \zeta (3)^2}{3}+8 \gamma ^5 \zeta (3)+\frac{38}{45} \gamma \pi ^4 \zeta (3)+\frac{16}{3} \gamma ^3 \pi ^2 \zeta (3)+8 \zeta (5) \zeta (3)+8 \gamma ^3 \zeta (5) \\
& +\frac{8}{3} \gamma \pi ^2 \zeta (5)+8 \gamma \zeta (7)+\gamma ^8+\frac{779 \pi ^8}{113400}+\frac{8 \gamma ^2 \pi ^6}{63}+\frac{29 \gamma ^4 \pi ^4}{45}+\frac{4 \gamma ^6 \pi ^2}{3} \\ \\
9 & -3 \zeta (3)^3-18 \gamma ^3 \zeta (3)^2-\frac{9}{2} \gamma \pi ^2 \zeta (3)^2-9 \gamma ^6 \zeta (3)-\frac{71 \pi ^6 \zeta (3)}{840}-\frac{9}{5} \gamma ^2 \pi ^4 \zeta (3)-\frac{15}{2} \gamma ^4 \pi ^2 \zeta (3) -18 \gamma \zeta (5) \zeta (3) \\
& -9 \gamma ^4 \zeta (5)-\frac{7 \pi ^4 \zeta (5)}{20} -\frac{9}{2} \gamma ^2 \pi ^2 \zeta (5)-9 \gamma ^2 \zeta (7)-\frac{3 \pi ^2 \zeta (7)}{2}-10 \zeta (9)-\gamma ^9-\frac{3 \gamma ^7 \pi ^2}{2}-\frac{17 \gamma ^5 \pi ^4}{20}-\frac{271 \gamma ^3 \pi ^6}{1260}-\frac{517 \gamma \pi ^8}{25200} \\ \\
10 & 10 \gamma \zeta (3)^3+25 \gamma ^4 \zeta (3)^2+\frac{19 \pi ^4 \zeta (3)^2}{36}+10 \gamma ^2 \pi ^2 \zeta (3)^2+10 \gamma ^7 \zeta (3)+\frac{20}{63} \gamma \pi ^6 \zeta (3)+\frac{29}{9} \gamma ^3 \pi ^4 \zeta (3)+10 \gamma ^5 \pi ^2 \zeta (3) \\
& +30 \gamma ^2 \zeta (5) \zeta (3)+\frac{10}{3} \pi ^2 \zeta (5) \zeta (3)+10 \zeta (7) \zeta (3)+5 \zeta (5)^2+10 \gamma ^5 \zeta (5)+\frac{19}{18} \gamma \pi ^4 \zeta (5)+\frac{20}{3} \gamma ^3 \pi ^2 \zeta (5) \\
& +10 \gamma ^3 \zeta (7)+\frac{10}{3} \gamma \pi ^2 \zeta (7)+10 \gamma \zeta (9)+\gamma ^{10}+\frac{229 \pi ^{10}}{136080}+\frac{2087 \gamma ^2 \pi ^8}{45360}+\frac{253 \gamma ^4 \pi ^6}{756}+\frac{13 \gamma ^6 \pi ^4}{12}+\frac{5 \gamma ^8 \pi ^2}{3} \\ \\
\vdots & \vdots
\end{array}$$

EDIT 2:

This method also works for higher order polygammas. $\psi^{(n)}(z)$ has poles of order n+1 at nonpositive integers, so $\psi^{(n)}(z)/\Gamma(z)^{n+1}$ is entire. Exactly the same method works, and you get
$$\sum_{a\in A_{n}}\frac{1}{a^{s}}=\frac{1}{(s-1)!}\lim_{z\to 0}\left(\frac{d}{dz}\right)^{s}\log\frac{\psi^{(n)}(z)}{\Gamma(z)^{n+1}}$$
where $A_{n}$ is the set of zeros of $\psi^{(n)}(z)$ for integer n >= 0, and s is an integer >= 2. Interestingly, the case with n=2k-1, s=2k, k>0, always seems to equal zero.
$$\sum_{a\in A_{2k-1}}\frac{1}{a^{2k}}=0$$

EDIT 3:

We can rewrite the sum using the previously mentioned residue formula:
$$\sum_{a\in A_{2k-1}}\frac{1}{a^{2k}}=-2k\operatorname{Res}\left(\frac{1}{z^{2k+1}}\log\frac{\psi^{(2k-1)}(z)}{\Gamma(z)^{2k}},z=0\right)$$
So we need to show that the coefficient of $z^{2k}$ in the series expansion of $\log\frac{\psi^{(2k-1)}(z)}{\Gamma(z)^{2k}}$ is zero. Around z=0, we have the series expansion
$$\log\Gamma(z)=-\log(z)-\gamma z+\sum_{n=2}^{\infty}\frac{(-1)^{n}\zeta(n)}{n}z^{n}$$
So the coefficient of $z^{2k}$ in $\log\Gamma(z)$ is $\frac{\zeta(2k)}{2k}$. Now use
$$\begin{align*}
\psi^{(2k-1)}(z)&=\psi^{(2k-1)}(z+1)+\frac{(2k-1)!}{z^{2k}} \\
&=\frac{(2k-1)!}{z^{2k}}+\psi^{(2k-1)}(1)+O(z) \\
&=\frac{(2k-1)!}{z^{2k}}+(2k-1)!\zeta(2k)+O(z)
\end{align*}$$
to get
$$\begin{align*}
\log(\psi^{(2k-1)}(z))&=\log\left(\frac{(2k-1)!}{z^{2k}}+(2k-1)!\zeta(2k)+O(z)\right) \\
&=\log(2k-1)!-2k\log{z}+\log\left(1+z^{2k}\zeta(2k)+O(z^{2k+1})\right) \\
&=\log(2k-1)!-2k\log{z}+z^{2k}\zeta(2k)+O(z^{2k+1}) \\
\end{align*}$$
So the coefficient of $z^{2k}$ in $\log\psi^{(2k-1)}(z)$ is $\zeta(2k)$, and so the coefficient of $z^{2k}$ in $\log\frac{\psi^{(2k-1)}(z)}{\Gamma(z)^{2k}}$ is $\zeta(2k)-2k\left(\frac{\zeta(2k)}{2k}\right)=0$, so the series $\sum_{a\in A_{2k-1}}a^{-2k}$ is equal to zero.

Post Tue Jun 07, 2016 12:35 am

Posts: 12
Heres an outline of how I'd do it
Image


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