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