Show that $\sum_{n=1} ^{\infty} \frac {1}{(x+n)^2} \leq \frac{2}{x} $ 
Show that $\displaystyle \sum_{n=1} ^{\infty} \frac {1}{(x+n)^2} \leq \frac{2}{x} $

For any real number x $\geq 1$, I want to show the above
Very rusty on my analysis, I think I need to do a comparison test to show the series converges, but then not sure what after that to get the inequality. 
 A: Let $n\ge 1$, for $x> 0$ we have $$\frac{1}{(x+n)^2}\le\frac{1}{(x+t)^2}\le\frac{1}{(x+n-1)^2}\qquad\text{for }n-1\le t\le n$$
Then
\begin{align}
\int_{n-1}^{n}\frac{1}{(x+n)^2}\,dt&\le \int_{n-1}^{n}\frac{1}{(x+t)^2}\,dt\\[6pt]
\frac{1}{(x+n)^2}&\le\frac{1}{x+n-1}-\frac{1}{x+n}\\[6pt]
\sum_{n=1}^N\frac{1}{(x+n)^2}&\le\frac{1}{x}-\frac{1}{x+N}
\end{align}
Then, as $N\to\infty$ we get 
$$\sum_{n=1}^{\infty}\frac{1}{(x+n)^2}\le\frac{1}{x}$$ 
A: Use for $n\geq 1$
$$\frac{1}{(n+x)^2}\leq \frac{1}{(x+n-1)(x+n)}=\frac{1}{x+n-1}-\frac{1}{x+n}$$
A: Too long for a comment: 

As an aside, the series evaluates to $S(x)=\psi_1(x)-\dfrac1{x^2}:$ see trigamma function. For $x=0$, we have $S(0)=\dfrac{\pi^2}6:$ see Riemann $\zeta$ function for more information. And if the iterator would have started at $-\infty$ instead of $1$, its value would have been $\bigg[\dfrac\pi{\sin(\pi x)}\bigg]^2$, which can be shown by twice differentiating the natural logarithm of Euler's reflection formula for the $\Gamma$ function. 


$($When I studied analysis in college, no one told us the “story” or the meaning behind the various series presented during class hours, or mentioned in the exercise section. I hope that you will find this information interesting, and that it will enrich your learning experience$)$.
