# Closed form for $\sum_{n=1}^\infty\frac{\psi(n+\frac{5}{4})}{(1+2n)(1+4n)^2}$

This question came up in the process of finding solution to another problem. Eventually, the problem was solved avoiding calculation of this sum, but it looks quite interesting on its own. Is there a closed form for $$\sum_{n=1}^\infty\frac{\psi(n+\frac{5}{4})}{(1+2n)(1+4n)^2},$$ where $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$ is the digamma function?

Yes, there is!

The derivation, which involves a lot of computer-assisted manipulations, is a little bit long, so I won't put it here just now, but the answer is $$-2 C-C \gamma-\frac{1}{2}C \pi+\frac{1}{4}\gamma \pi-\frac{\pi ^2}{6}-\frac{\gamma \pi ^2}{8}+\frac{11}{64} \pi ^3-6 \Im\text{Li}_3\left(\frac{1+i}{2}\right)-3 C \log 2+\frac{1}{2} \gamma \log 2+\pi\log 2+\frac{3}{2} (\log 2)^2+\frac{3}{16} \pi (\log 2)^2+\frac{7}{4} \zeta(3) - \psi\left(\frac54\right),$$ where $C$ is the Catalan constant, $\gamma$ is Euler's gamma, and $\text{Li}_3$ is a polylogarithm, and (edited) the last term $\psi(5/4)=4-\gamma-\frac\pi2-\log 8$ is there because everything below computes the sum over $n\geq0$, so I need to subtract $n=0$ from the answer below to get the sum over $n\geq1$.

Edit to explain the calculation. The way to derive this is to write the sum as $$\frac18 S\left(\frac54\middle| \frac14,\frac14\right) - \frac18 S\left(\frac54\middle|\frac14,\frac12\right),$$ where I've introduced the notation $$S(\beta|\alpha_1,\alpha_2,\ldots) = \sum_{n\geq0} \frac{\psi(n+\beta)}{(n+\alpha_1)(n+\alpha_2)\cdots}.$$

The intermediate sums can be found in closed form mainly using the integral representation of the digamma function, in the form $$\psi(z_1)-\psi(z_2) = \int_0^1 \frac{dt}{1-t}(t^{z_2-1}-t^{z_1-1}),$$ a sum representation in the form $$\psi(z+1)+\gamma = \sum_{k\geq1} \frac{1}{k} - \frac{1}{k+z},$$ and using also the definition of the Lerch transcendent and its relation to the hypergeometric function and the incomplete beta function: $$\Phi(z,s,a) = \sum_{n\geq0}\frac{z^n}{(n+a)^s},$$ $$\Phi(z,1,a) = z^{-a}B(z,a,0) = \frac{1}{a}F(1,a;1+a;z).$$

The intermediate sum $S(\frac54;\frac14,\frac14)$ can be found using $$\sum_{n\geq0}\frac{\psi(n+\frac54)+\gamma}{(n+\frac14)^2} = \sum_{n\geq0}\sum_{k\geq1} \frac{1}{(n+\frac14)^2}\left(\frac1k - \frac1{k+n+\frac14}\right) = \sum_{k\geq1} \frac{\psi(k+\frac14)-\psi(\frac14)}{k^2} = 4F\left(\begin{array}{c}1,1,1,1\\2,2,\frac54\end{array}\right) + \int_0^1 \frac{\log t\log(1-t)}{t(1-t)^{3/4}}\,dt,$$ where $$\text{Li}_2(1) - \text{Li}_2(t) = \text{Li}_2(1-t) + \log t\log(1-t),$$ so $$S\left(\frac54\middle| \frac14,\frac14\right) = -\gamma\psi_1(1/4) + \sum_{n\geq0}\frac{\psi(n+\frac54)+\gamma}{(n+\frac14)^2}.$$

Using the above relationship between incomplete beta function, the fact that this hypergeometric is a repeated integral of the incomplete beta function (which you know because of the ones on top and matching twos on the bottom) and the fact that there is a nice closed form for the incomplete beta function with a rational first parameter and zero second parameter, like so: $$B(z,p/q,0) = -\sum_{0\leq l < q} e^{-2\pi i l p/q}\log(1-z^{1/q}e^{2\pi i l/q}), \qquad p,q\in\mathbb{Z},$$ it can be simplified to $$F\left(\begin{array}{c}1,1,1,1\\2,2,\frac54\end{array}\right) = \frac{7}{32}\pi^3 - 12\Im\text{Li}_3\left(\frac{1+i}{2}\right) + \frac32\pi(\log 2)^2 + \pi^2\log 8 - 14\zeta(3).$$

The second intermediate sum can be simplified using partial fractions on $\frac1{(n+\frac14)(n+\frac12)}$, and the fact that $$\sum_{n\geq0}\frac{1}{(n+\alpha_1)(n+\alpha_2)} = \frac{\psi(\alpha_1)-\psi(\alpha_2)}{\alpha_1-\alpha_2}.$$ Then $$S\left(\frac54\middle|\frac14,\frac12\right) + \gamma\frac{\psi(\frac12)-\psi(\frac14)}{\frac12-\frac14} = \sum_{n\geq 0}\frac{\psi(n+\frac54)+\gamma}{(n+\frac14)(n+\frac12)} = \sum_{k\geq1}\frac{\psi(k+\frac14)-\psi(\frac12)}{k(k-\frac14)},$$ where I've expanded difference of digamma functions as an infinite sum over $k$ and performed the sum over $n$. This expression can be handled using the integral representation for $\psi$ mentioned above, giving $$\int_0^1\frac{dt}{1-t}\left( (12\log2-2\pi)t^{-\frac12} - t^{-\frac34}\left( 4\log(1-t)+\frac{16}{3}tF\left(1,\frac34;\frac74;t\right)\right)\right) \\= 16C+\frac43\pi^2-8\pi\log2-12(\log2)^3.$$

Putting everything together, and using Mathematica to evaluate the easier sums, gives the expression I gave above.

Basically, the most important steps in the derivation are the integral representation for $\psi(x)-\psi(y)$ and the explicit formula for $B(z,\beta,0)$ when $\beta$ is a rational number. By luck all the integrals then simplify to something Mathematica can do in closed form.

I don't know any standard reference for this; the whole question is basically just computing $$\frac{d}{d\epsilon}\big|_{\epsilon=0}F\left(\begin{array}{c}1,\frac12,\frac14,\frac14,\frac54+\epsilon\\ \frac32,\frac54,\frac54,\frac54 \end{array}\right),$$ and in general derivatives of hypergeometric functions are given by Kampe de Feriet functions, which don't have a closed form in terms of hypergeometric functions. So in general there should be no closed form, but in this case the parameters are just right that all integrals can be done.

When $Q(n)$ is some rational function having a partial fraction expansion $\sum_k q_k (n+\alpha_k)^{-s}$, the sum $$\sum_{n\geq0}Q(n)(\psi(n+\beta)-\psi(\beta))$$ can be expressed in terms of sums $$\sum_{n\geq0} \frac{\psi(n+\beta)-\psi(\beta)}{(n+\alpha)^s} = \int_0^1 B(v,\beta,0)\Phi'(v,s,\alpha)\,dv,$$ where $\Phi'$ is the Lerch transcendent's derivative. If the sum on the left diverges, it is necessary to consider the asymptotic expansion $$\sum_{n\geq0} \frac{\psi(n+\beta)-\psi(\beta)}{n+\alpha}z^n \\= \Theta(\log(1-z))^2 + \Theta(\log(1-z)) + S(\beta,\alpha) + o(1), \qquad z\to1-$$ and only take the non-divergent term $S(\beta,\alpha)$ as the value of the sum at $z\to1$ (the divergent terms should be independent of $\alpha$ and cancel out when the partial fractions of $Q(n)$ are added together).

So the problem of evaluating the sum $$\sum_{n\geq0}Q(n)(\psi(n+\beta)-\psi(\beta))$$ reduces to calculating asymptotic expansions of integrals of the form $$z \int_0^1 B(u,\beta,0)\Phi'(z u,s,\alpha)\,du, \qquad z\to1-.$$

• Can you please refer to some place which elaborates a bit more on this? Jun 14, 2013 at 5:31
• +1 : For the powerful work ! A numerical check (to more than $200$ digits) returned all your terms at the beginning right with just some first order terms missing : $$-4+\gamma+\frac{\pi}2+3\log(2)$$ (this is merely $-\psi(5/4)$ so that there is probably a problem with an additional $n=0$ term...) Jun 20, 2013 at 20:28
• @Kirill The answer is absolutely amazing! Thank you very much, Kirill! Jul 11, 2013 at 20:56
• @LailaPodlesny No problem, it was an interesting question. Jul 13, 2013 at 2:22
• @user153012 Bad transcription: 1/4 ((7 \[Pi]^3)/8 - 48 Im[PolyLog[3, 1/2 + I/2]] + 3/2 \[Pi] Log[2]^2 + \[Pi]^2 Log[8] - 14 Zeta[3]) Oct 27, 2014 at 22:19