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Let $f(x) = 2 \psi^{(1)}(x+1) + x \psi^{(2)}(x+1) $

for $ x > 0 $, where $\psi^{(i)}(x)$ is the $i^{th}$ derivative of the digamma function $\psi(x)$.

The goal is to prove that $ f(x) < \frac{1}{x} $ for $x > 0$.

As I understand:

$\psi^{(1)}(x+1) = \sum_{k=1}^{\infty} \frac{1}{(x+k)^2} $ and

$\psi^{(2)}(x+1) = \sum_{k=1}^{\infty} \frac{-2}{(x+k)^3} $

and hence $f(x) = \sum_{k=1}^{\infty} \frac{2k}{(x+k)^3}$

All numerical evidence suggests that the conjecture is true. I have attempted (in vain) several approaches to prove this, including operations on a partial fraction decomposition of an upper bound to the series:

$ f(x) < g(x) = \sum_{k=1}^{\infty} \frac{2k}{(x+k)(x+k-1)(x+k-2)}$

which gave me a seemingly nice (and correct) bound, but unfortunately it is not tight enough to validate the hypothesis as it turns out that $g(x)$ is not less than $\frac{1}{x}$ for all $x$. It appears that $\frac{1}{x}$ converges very quickly from above to $f(x)$ and so I am guessing that some new insight that I am missing is required.

Any ideas? The problem arose with me in the context of statistical estimation.

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Hint. You may compare your series with the related integral $$ \int_1^{+\infty}\frac{2t}{(x+t)^3}dt=\frac{1}{2 (1+x)^2}+\frac{1}{2 (1+x)}<\frac1x,\quad x>0. $$

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  • $\begingroup$ Thanks, this is a natural approach, but I compute your integral as double what you list. Another thing is $\frac{2t}{(x+t)^3}$ is not monotone (it decreases only for $t>x/2$). But for sake of argument, suppose $x$ is an even integer. Then using the integral, we can bound the tail of the series from $x/2 + 1$ onwards as $< \frac{8}{9x} $. Unfortunately, I found that even if we use the exact values of the series from 1 to $x/2$ this bound can be too loose against $1/x$ (numerically) so there was no point in trying to bound the initial part of the series with an integral. $\endgroup$ – H.F. May 26 '15 at 17:16

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