(Theoretical) Series of function and the function's limit Let $f(x)$ defined in $[0,1]$ and
$$\sum_{n=1}^\infty f\left(\frac 1n\right)$$
converges.
Prove or disprove:
1) $lim_{x \to 0^{+}} f(x)= 0 $ .
2) let $$a_n = f\left(\frac 1{n^2}\right)$$
   than $a_n \to 0.$
My try:
The first one i think i managed to solve:
1) If the given series converges than by the limit of the summand theorem 
$$\lim_{n \to \infty} f\left(\frac 1n\right)= 0$$
With simple assignment $x_n = \frac 1n$
$$\lim_{n \to \infty} f\left(\frac 1n\right)= \lim_{x\to 0^+}f(x) = 0$$
Is this way acceptable?
2) I wanted to use the same trick here:
Let $x_n = \frac 1{n^2}$
$$\lim_{n \to \infty} f\left(\frac 1{n^2}\right)= \lim_{x\to 0^+}f(x)$$
The problem is i can't find a way to show that
$$\sum_{n=1}^\infty f\left(\frac 1{n^2}\right)$$
converges.
Maybe the claim is false and there's a simple counter example?
 A: Take $f(x) = \sin(\frac{2\pi}{x}), x\neq0$ then $f(\frac{1}{n}) =0,\forall n = 1, 2,\cdots$, thus $\sum_{n=1}^{\infty}f(\frac{1}{n})$ converges.
But $\lim_{x \to 0^+}f(x)$ doesn't exsit, since if we define $x_n = \frac{1}{n+\frac{1}{4}}$ and $y_n = \frac{1}{n -\frac{1}{4}}$, then both $(x_n)_n$ and $(y_n)_n$ converges to $0$ but $f(x_n) = 1$ while $f(y_n) = -1$.
We do have $a_n = f(\frac{1}{n^2})$ converges to $0$, since that $\sum_{n=0}^{\infty} f(\frac{1}{n})$ convergs implies $\lim_{n\to \infty}f(\frac{1}{n}) = 0$. Then $f(\frac{1}{n^2})$ as a subsequence of $f(\frac{1}{n})$ converges to $0$ too.
Remark that if $f$ is supposed to be right-continuous at $0$, then we do have $\lim_{x \to 0^+}f(x) =0$, since the limit is supposed to exist and it has to be the same as $\lim_{n \to \infty}f(\frac{1}{n}) = 0$
A: Let $f$ defined by $f(x)=x^2$ if $x\in A$ and $f(x)=1$ otherwise, where $A=\{\dfrac{1}{n}, n\in \mathbb{N}^*\}$. (1) is not working!  
A: What happens with
$$f(x)=\sin\left(\frac{2\pi}{x}\right)\quad ?$$
