# Prove limn ↦ ∞4nψ(2)(2n)−ψ(2)(1)=2ζ(3)−1\lim_{n\ \mapsto\ \infty}4^n \psi^{(2)}(2^n)-\psi^{(2)}(1)=2\zeta(3)-1

Prove

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

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

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

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

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

and by setting $$k=2k=2$$ and $$x=2^nx=2^n$$ then letting $$n\mapsto \inftyn\mapsto \infty$$, 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 series page 90.

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

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

twice, we get

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

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

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

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

Solution 2. Alternatively, starting from the identity

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

we get

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

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

$$z^2 \psi^{(2)}(z) \stackrel{(t=zs)}= – \int_{0}^{\infty} \frac{t/z}{1 – e^{-t/z}} \, t e^{-t} \, \mathrm{d}t \xrightarrow[z\to\infty]{} – \int_{0}^{\infty} t e^{-t} \, \mathrm{d}t = -1 z^2 \psi^{(2)}(z) \stackrel{(t=zs)}= - \int_{0}^{\infty} \frac{t/z}{1 - e^{-t/z}} \, t e^{-t} \, \mathrm{d}t \xrightarrow[z\to\infty]{} - \int_{0}^{\infty} t e^{-t} \, \mathrm{d}t = -1$$

by the dominated convergence theorem. This completes the proof.

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

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

Then the repeated application of this identity yields

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

Differentiating both sides twice followed by plugging $$z = 2^nz = 2^n$$ and multiplying $$4^n4^n$$ to both sides, we get

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

As $$n\to\inftyn\to\infty$$, this converges to

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