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


\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


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 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)

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.

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

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