Prove

\lim_{n\ \mapsto\ \infty}4^n \psi^{(2)}(2^n)-\psi^{(2)}(1)=2\zeta(3)-1\tag{1}

where \psi^{(m)}(x) is the polygamma function and \zeta is the Riemann zeta function.

This problem was proposed by a friend and solved using the series expansion of digamma function:

\psi(x)=\ln x+O\left(\frac1x\right) and by differentiating both sides k times we get

\psi^{(k)}(x)=\frac{(-1)^{k-1}(k-1)!}{x^k}+O\left(\frac1{x^{k+1}}\right)

and by setting k=2 and x=2^n then letting n\mapsto \infty, the result follows.

The question here is can we prove (1) in a different way?

UPDATE:: You can find the generalization of this problem in the book,Almost impossible integrals, sums and seriespage 90.

**Answer**

**Solution 1.** Differentiating the partial fraction decomposition of the digamma function

\psi(z) = -\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} – \frac{1}{n+z} \right)

twice, we get

\psi^{(2)}(z) = -\sum_{n=0}^{\infty} \frac{2}{(n+z)^3}.

From this, we easily find that \psi^{(2)}(1) = -2\zeta(3). Moreover, for z > 1 we have

\frac{1}{2z^2} = \int_{0}^{\infty} \frac{1}{(x+z)^3} \, \mathrm{d}x \leq \sum_{n=0}^{\infty} \frac{1}{(n+z)^3} \leq \int_{0}^{\infty} \frac{1}{(x+z-1)^3} \, \mathrm{d}x = \frac{1}{2(z-1)^2},

and so, we deduce that \lim_{z\to\infty} z^2\psi^{(2)}(z) = -1. Combining these two computations prove the desired result. This solution is essentially unwinding the derivation of the asymptotic form \psi(z) = \log z + \mathcal{O}(1/z), thus is not so different from OP’s solution.

**Solution 2.** Alternatively, starting from the identity

\psi(z) = -\gamma + \int_{0}^{\infty} \frac{1-e^{-(z-1)s}}{e^s-1} \mathrm{d}s,

we get

\psi^{(2)}(z) = – \int_{0}^{\infty} \frac{s^2 e^{-(z-1)s}}{e^s – 1} \, \mathrm{d}s,

and so, \psi^{(2)}(1) = -\Gamma(3)\zeta(3) = -2\zeta(3) and

z^2 \psi^{(2)}(z)

\stackrel{(t=zs)}= – \int_{0}^{\infty} \frac{t/z}{1 – e^{-t/z}} \, t e^{-t} \, \mathrm{d}t

\xrightarrow[z\to\infty]{} – \int_{0}^{\infty} t e^{-t} \, \mathrm{d}t

= -1

by the dominated convergence theorem. This completes the proof.

**Solution 3.** Here we assume that we already know \psi^{(2)}(1) = -2\zeta(3). By log-differentiating the Legendre duplication formula, we get

\psi(z) = \log 2 + \frac{1}{2} \left[ \psi\Big(\frac{z}{2}\Big) + \psi\Big(\frac{z+1}{2}\Big) \right].

Then the repeated application of this identity yields

\psi(z) = n \log 2 + \frac{1}{2^n} \sum_{k=0}^{2^n-1} \psi\Big(\frac{z+k}{2^n}\Big).

Differentiating both sides twice followed by plugging z = 2^n and multiplying 4^n to both sides, we get

4^n \psi^{(2)}(2^n) = \frac{1}{2^n} \sum_{k=0}^{2^n-1} \psi^{(2)}\Big(1+\frac{k}{2^n}\Big).

As n\to\infty, this converges to

\lim_{n\to\infty} 4^n \psi^{(2)}(2^n) = \int_{0}^{1} \psi^{(2)}(1+x) \mathrm{d}x = \psi^{(1)}(2) – \psi^{(1)}(1) = -1.

**Attribution***Source : Link , Question Author : Ali Shadhar , Answer Author : Sangchul Lee*