<s>[INST] 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.
 [/INST] 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.
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