convergence of the series through subsets Let $A_n \subset \mathbb{N}$ such that $A_{n}\subset A_{n+1}$ and $\lim A_n=\mathbb{N}$. I am interested in knowing if the next sequence converges
$$\sum_{j\in A_n}\frac{1}{j^2}$$
The candidate when $n$ increases to infinity is $\sum_{j\geq 1}\frac{1}{j^2}$. but I have no idea how to prove it , I would appreciate any suggestions .
EDIT:     $\lim A_n=\mathbb{N}$ it means $\bigcup A_n=\mathbb{N}$.
 A: Your limit exists, because $\{\sum_{j\in A_n}\frac1{j^2}\}$ is a bounded increasing sequence, so convergent. 
For any $k$, there exists $n$ such that $\{1,2,\ldots,k\}\subset A_n$. Then
$$
\sum_{j=1}^k\frac1{j^2}\leq\sum_{j\in A_n}\frac1{j^2}\leq\sum_{j=1}^\infty\frac1{j^2}.
$$
So $\limsup_{n}\sum_{j\in A_n}\frac1{j^2}=\sum_{j=1}^\infty\frac1{j^2}$. But the sequence $\{\sum_{j\in A_n}1/j^2\}$ is monotone increasing, and then
$$
\lim_{n}\sum_{j\in A_n}\frac1{j^2}=\sum_{j=1}^\infty\frac1{j^2}.
$$
A: That 
$$\sum_{j\in A_n}\frac{1}{j^2}\le \sum_{j=1}^\infty \frac{1}{j^2}$$ 
is obvious. Let $\epsilon >0$, then there is $N_\epsilon$ so that 
$$\sum_{j=1}^{N_\epsilon} \frac{1}{j^2}\ge \sum_{j=1}^\infty \frac{1}{j^2}-\epsilon.$$
Let $N_2$ be large so that $\{1,2, \cdots, N_\epsilon\} \subset A_{N_2}$ so 
$$\sum_{j\in A_{N_2}} \frac{1}{j^2}\ge \sum_{j=1}^\infty  \frac{1}{j^2}-\epsilon.$$
as $A_n \subset A_{n+1}$, for all $n\ge N_2$, 
$$\sum_{j\in A_{n}} \frac{1}{j^2}\ge \sum_{j=1}^\infty  \frac{1}{j^2}-\epsilon.$$
Then 
$$\sum_{j\in A_n} \frac{1}{j^2} \to \sum_{j=1}^\infty \frac{1}{j^2}.$$
