TITLE: Prove it converges: $a_n=\frac{1}{1+1^2}+\frac{1}{1+2^2}+...+\frac{1}{1+n^2}$
QUESTION [0 upvotes]: $(a_n)_{n>=1}$ 
$a_n=\frac{1}{1+1^2}+\frac{1}{1+2^2}+...+\frac{1}{1+n^2}$ 
I need to prove the series converges. I am really confused by this kind of exercises, I was taught in high school that a series converges if it is monotone and bounded, nothing more, this was the only way to prove a series converges.  
Now my confusion comes from the fact that there's no place on google that mentions this, instead there seems to be a couple of 'tests' to help decide whether a series converges or diverges.  
How do I apply the 'highschool method' to solve this exercise? More specifically how do I prove it is bounded, cause it is clearly increasing(right?).  
Do I just take the limit of $\frac{1}{1+n^2}$ and that is the upper bound while the lower bound would be the first term$(\frac{1}{2})$?

REPLY [1 votes]: As others have pointed out,
$$a_n=\sum_{k=1}^{n}\dfrac{1}{1+k^2}<\sum_{k=1}^{n}\dfrac{1}{k^2}$$
But does $\sum_{k=1}^{n}\dfrac{1}{k^2}$ converge?
Notice that $\forall{k\in{\mathbb{N}}},\;\text{we have}$
$$\frac{1}{k^2}\le\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}$$
Now,
$$\sum_{k=2}^{n}(\frac{1}{k-1}-\frac{1}{k})$$ is a telescoping series, and 
$$\sum_{k=2}^{n}(\frac{1}{k-1}-\frac{1}{k})=1-\frac{1}{n+1}\le1$$
So 
$$\sum_{k=2}^{n}\dfrac{1}{k^2}\le\sum_{k=2}^{n}(\dfrac{1}{k-1}-\dfrac{1}{k})=1-\frac{1}{n+1}\le1$$
Note the first summation starts from $k=2$ instead of $k=1$.
However, when $k=1,\; \dfrac{1}{k^2}=1$.
Therefore,
$$\sum_{k=1}^{n}\dfrac{1}{1+k^2}\lt\sum_{k=1}^{n}\dfrac{1}{k^2}\le 1+\sum_{k=2}^{n}(\frac{1}{k-1}-\dfrac{1}{k})=2-\dfrac{1}{n+1}\le 2$$
Since $\forall{k}\;\text{such that}\;1\le k\le n,\;\dfrac{1}{1+k^2}\gt 0$, the series is increasing, i.e. $\forall{N\gt n},\;$
$$\sum_{k=1}^{N}\dfrac{1}{1+k^2}\gt\sum_{k=1}^{n}\dfrac{1}{1+k^2}$$
Recalling that the series $\sum_{k=1}^{n}\dfrac{1}{1+k^2}$ is bounded above, it then follows that $$\lim_{n\to \infty}\sum_{k=1}^{n}\dfrac{1}{1+k^2}$$ converges. Similarly, $$\lim_{n\to \infty}\sum_{k=1}^{n}\dfrac{1}{k^2}$$
also converges.
{"set_name": "stack_exchange", "score": 0, "question_id": 3417630}