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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?

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$\lim_{n \rightarrow + \infty} f \left( \frac{1}{n^2} \right) = \lim_{x \rightarrow 0^+} f(x)$ is not true in general. For example take $f(x) = \sin (\pi / x)$ – Crostul Aug 7 '14 at 13:40
Consider the function $f(x)$ that is equal to zero if $x$ is $\frac{1}{n}$, and one otherwise. – Baby Dragon Aug 7 '14 at 13:40
up vote 4 down vote accepted

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 $\lim_{n \to \infty}f(\frac{1}{n}) = 0$

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What happens with $$f(x)=\sin\left(\frac{2\pi}{x}\right)\quad ?$$

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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!

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This does not work. $f(1/n) = n^2$ and the sum $\sum f(1/n)$ therefore diverges. I think you mean $f(x) = x^2$ right?! – Winther Aug 7 '14 at 14:01
yes, edited, right – Hamou Aug 7 '14 at 14:04

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