How to prove $\frac{1}{v}-\sum_{n=0}^{\infty}\frac{1}{(v+n)^2}<0$ I want to prove $$\frac{1}{v}-\frac{\partial^2ln(\Gamma(v))}{\partial v^2}<0$$
I checked on wiki that this equals $$\frac{1}{v}-\sum_{n=0}^{\infty}\frac{1}{(v+n)^2}<0$$ I tried hard to prove it while any trials to deal with this inequality failed... I plot the function using software, it's very close to $0$ although negative when $v \rightarrow \infty$. Although on wiki if I search trigamma I can get tons of conclusion about this second derivative of $ln(\Gamma(v))$. I am wondering how to prove this. Thank you!
 A: The integral method suggested by Sangchul Lee above should work and it's a good method to apply to a wide range of similar problems.
But in this case we can also apply a smarter trick using telescoping sums.
Note that $$\frac{1}{x^2}>\frac{1}{x^2+x}=\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$$ holds for all positive $x$.
Now, let's turn to the original problem. For $ v \le 0$ this is clearly true. So, assume $v >0$. Then,
$$\sum_{n=0}^{k} \frac{1}{(v+n)^2} > \sum_{n=0}^k \frac{1}{v+n}-\frac{1}{v+n+1}=\sum_{n=0}^k \frac{1}{v+n}-\sum_{n=0}^k \frac{1}{v+n+1}=\frac{1}{v}+ \left( \sum_{n=1}^k \frac{1}{v+n} \right)  -\left( \sum_{n=0}^{k-1} \frac{1}{v+n+1} \right) - \frac{1}{v+k+1}=\frac{1}{v}-\frac{1}{v+k+1}$$
And so it's easy to see that for $k \to \infty$ we obtain
$$\sum_{n=0}^{\infty} \frac{1}{(v+n)^2}>\frac{1}{v}$$
as desired.
A: It suffices to consider the case where $v > 0$. Use the fact that
$${1 \over (v + n)^2} > {1 \over (v+ n)(v + n + 1)}$$
$$= {1 \over v + n} -{1 \over v + n + 1}$$
Then add this over $n$. You get a telescoping sum adding up to ${\displaystyle {1 \over v}}$.
