Prove the sequence $\sum_{k=0}^n \frac{1}{(n+k)^2}$ is convergent I'm having a hard time trying to prove that the sequence $\{a_n\}$ whose general term $a_n$ is
$$\sum_{k=0}^n \frac{1}{(n+k)^2}$$
is convergent. I'm trying to prove it by definition, that is to say, by finding a lower/upper bound and by proving that it is decreasing/increasing using induction.
By subtracting $a_n$ from $a_{n+1}$ we obtain the following for $n=m$ (if I'm not mistaken):
$$
\frac{1}{(2m+1)^2} + \frac{1}{(2m+2)^2} - \frac{1}{m^2}
$$
Which is less than $0$ for $n=1$ from which I have assumed the sequence is decreasing and therefore trying to prove that $a_{n+1} - a_n < 0$ using induction. But I am terribly stuck! Thanks in advance for any suggestion.
 A: It might be easier to bound the sum from above by a telescoping sum and then invoke the Squeeze Theorem:
$$0\le\sum_{k=0}^n{1\over(n+k)^2}\lt\sum_{k=0}^n{1\over(n+k-1)(n+k)}=\sum_{k=0}^n\left({1\over n+k-1}-{1\over n+k}\right)={1\over n-1}-{1\over2n}={n+1\over2n(n-1)}\to0$$
A: 
Hint:
\begin{align*}
\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{(n+k)^2}\leq\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}
\end{align*}

We also have $\sum_{k=1}^\infty\frac{1}{k^\alpha}$ is convergent iff $\alpha>1$.
A: To prove that this sequence converges, we will use the monotone convergence theorem which states

If $(a_n)$ is either
  
  
*
  
*increasing and bounded above
  
*decreasing and bounded below
  
  
  then it is a convergent sequence. 

In order to use this result, we first look at the difference $a_{n+1}-a_{n}$ to determine if your sequence is increasing or decreasing. We have
\begin{align}
a_{n+1}-a_n &= \sum_{k=0}^{n+1} \frac{1}{(n+1+k)^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\
&=\sum_{k=0}^{n+1} \frac{1}{(n+(k+1))^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\
&=\sum_{k=1}^{n+2} \frac{1}{(n+k)^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\
&=\left(\sum_{k=1}^{n} \frac{1}{(n+k)^2} \frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} +\right) - \left(\frac{1}{n^2}+\sum_{k=1}^n \frac{1}{(n+k)^2}\right)\\
&=\frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} - \frac{1}{n^2}\\
&\leq \frac{1}{(2n)^2} + \frac{1}{(2n)^2} -\frac{1}{n^2}\\
&= -\frac{1}{2n^2}\leq 0
\end{align}
So your sequence is decreasing. Furthermore, it is clear that $a_n\geq 0$ for any $n\in\mathbb{N}$. That is, your sequence is bounded below. By the monotone convergence theorem, it converges.
A: You don't need induction.  $\frac{1}{(2m+2)^2}+\frac{1}{(2m+1)^2}-\frac{1}{m^2}\lt \frac{2}{(2m)^2}-\frac{1}{m^2}=\frac{1}{2m^2}-\frac{1}{m^2}\lt 0$.
A: Just for your curiosity since you already received good answers.
$$S_n=\sum_{k=0}^n \frac{1}{(n+k)^2}=\psi ^{(1)}(n)-\psi ^{(1)}(2 n+1)$$ where appears the polygamma function (don't worry : you will learn about it sooner or later).
Using the asymptotics for large values of $n$
$$S_n=\frac{1}{2 n}+\frac{5}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ For example
$$S_{10}=\frac{3056206830982561}{54192375991353600}\approx 0.0563955$$ while the above expansion gives
$\frac{9}{160}=0.05625$
A: $\displaystyle\sum_{k = 0}^{n}{1 \over \left(n + k\right)^{2}} =
{1 \over n}\left[{1 \over n}\sum_{k = 0}^{n}{1 \over
\left(1 + k/n\right)^{2}}\right]
\,\,\,\stackrel{\color{red}{\mathrm{as}\ n\ \to\ \infty}}{\sim}\,\,\,
{1 \over n}\int_{0}^{1}{\mathrm{d}x  \over \left(1 + x\right)^{2}} =
{1 \over 2n}$
