Evaluation of sum of digamma function.

Question

while solving a summation problem I got stuck at this:

$$\sum _{ n={ 2 }^{ k-1 }+1 }^{ { 2 }^{ k } }{ \left( \psi \left( 2n \right) \right) } $$

The limits of this summation are weird and hence I'm not able to apply the properties of summation of digamma function. Please help.

Answer 1

This is not an answer since based on numerical simulation.

Unable to explicit $$S_k=\sum _{ n={ 2 }^{ k-1 }+1 }^{ { 2 }^{ k } }{ \left( \psi \left( 2n \right) \right) }$$ I compute the numerical values for $0\leq k\leq 20$ and observed that $\log(S_k)$ seems to follow a power law ($a k^b+c$).

A quick and dirty nonlinear regression leads to $$\log(S_k)=1.27064\, x^{0.861389}-0.991638$$ ($R^2=0.999994$) with very significant parameters from a statistical point of view (as shown below). $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 1.27064 & 0.0174181 & \{1.23371,1.30756\} \\ b & 0.861389 & 0.0040875 & \{0.852723,0.870054\} \\ c & -0.991638 & 0.0356172 & \{-1.06714,-0.916133\} \\ \end{array}$$

May be, this would give you some ideas.

Edit

Afterwards, inspired by lordoftheshadows's answer and using a CAS, what was found is $$T_k=\sum _{ n=1 }^{ k}{ \left( \psi \left( 2n \right) \right) }=\frac{1}{4} \left(-2 H_k-H_{k+\frac{1}{2}}-4 k+4 (k+1) \psi ^{(0)}(2 k+2)+4 \gamma -2-\log (4)\right)$$ which would then give (hoping no mistake) $$4S_k=2 H_{2^{k-1}}-2 \left(2^k+3\right) H_{2^k}-H_{\frac{1}{2}+2^k}+H_{\frac{1}{2} \left(1+2^k\right)}+4 H_{1+2^{k+1}}-\frac{2}{2^k+1}+2^{k+1} \left(2 \psi ^{(0)}\left(2+2^{k+1}\right)+\gamma -1\right)-2$$

Answer 2

This is not going to be an explicit answer because there is some annoying algebra that I can't do at 10:30 pm but this should help a lot.

We know that $\psi(n) = H_{n-1} - \gamma$ where $\gamma$ is the Euler-Mascheroni constant. From here we can say some things about the sum:

1. We can take out the $\gamma$s because it only appears once in each term of the series. Therefor we will have a $*2^{k-1} - 1)*\gamma$ plus some function of the harmonic numbers.

2. We now need to find some way to simplify the remain bits we have left. The remaining stuff can be expressed as $\sum _{ n={ 2 }^{ k-1 }+1 }^{ { 2 }^{ k } }{ \left (H_{n-1}\right)}$. Where $H_n$ is the nth harmonic number. I believe there is a way to simplify this summation here to a much more manageable sum.

If you are only looking to approximate it then $H_n$ is approximated quite well by $ln(x) + \gamma$. If you use this approximation then you get to cancel some $\gamma$s and you get a pretty nice approximate of the sum that you want.