I need the power series (starting with index $k=1$) of a difference between two digamma functions. $$\psi(t+2)-\psi\left(\frac{t+3}{2}\right)=\sum_{k=1}^\infty a_kt^k$$ In otherwords I want $a_k$ in the equation above.
I know how to find the $t$-expansion of a related function, $$\psi\left(\frac{t+2}{2}\right)-\psi\left(t+1\right)=\sum_{n=1}^\infty\left(\frac{1}{n+t}-\frac{1}{n+t/2}\right)$$ where for $|t|<1$, $$\frac{1}{n+t}=\sum_{k=0}^\infty\frac{(-t)^k}{n^{k+1}},\quad\frac{1}{n+t/2}=\sum_{k=0}^\infty\frac{(-t)^k}{n^{k+1}2^k}$$ substituting and writing out the first term gives, $$\sum_{n=1}^\infty\left(\frac{1}{n+t}-\frac{1}{n+t/2}\right)=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^k}{n^{k+1}}\left(1-\frac{1}{2^k}\right)t^k$$ interchanging the double summation and recognizing the Riemann zeta function gives me the expansion I am after, $$\psi\left(\frac{t+2}{2}\right)-\psi\left(t+1\right)=\sum_{k=1}^\infty\underbrace{(-1)^k\left(1-\frac{1}{2^k}\right)\zeta(k+1)}_{a_k}t^k.$$ However (with a slight abuse of notation) simply taking $t=t+1$ gives me an expansion around $t=-1$ and not $t=0$.
Any ideas? Thanks for the help.
The identity $\psi(1+z)=\psi(z)+1/z$ gives $$ f(t):=\psi(t+2)-\psi\left(\frac{t+3}2\right)=\psi(1+t)-\psi\left(\frac{1+t}2\right)-\frac1{1+t}. $$ Now the duplication formula for $\Gamma(2z)$ at $z=(1+t)/2$ reads $$ \Gamma(1+t)=2^t\pi^{-1/2}\,\Gamma\left(\frac{1+t}2\right)\Gamma\left(1+\frac t2\right) $$ which, after taking the logarithmic derivative w.r.t. $t$, yields $$ \psi(1+t)=\log 2+\frac12\psi\left(\frac{1+t}2\right)+\frac12\psi\left(1+\frac t2\right), $$ hence $$ f(t)=2\log 2-\frac1{1+t}+\psi\left(1+\frac t2\right)-\psi(1+t). $$ It remains to use the known series $\psi(1+z)=-\gamma-\sum_{n=1}^\infty\zeta(n+1)(-z)^n$ to get $$ f(t)=2\log 2-1+\sum_{n=1}^\infty\big[(1-2^{-n})\zeta(n+1)-1\big](-t)^n. $$