# Thread: Proof of continuity of series

1. ## Proof of continuity of series

Let $k(x) = \sum_{n=1}^{\infty}\frac{\sin{(nx^2)}}{3^{n}}$. Prove $k$ is continuous on $\mathbb{R}$.

NOTE: for this proof, you're allowed to assume $\sin{(x)}$ is continuous on $\mathbb{R}$.

2. Originally Posted by Auxiliary
Let $k(x) = \sum_{n=1}^{\infty}\frac{\sin{(nx^2)}}{3^{n}}$. Prove $k$ is continuous on $\mathbb{R}$.

NOTE: for this proof, you're allowed to assume $\sin{(x)}$ is continuous on $\mathbb{R}$.

ideas:
1. the sum of continuous functions is continuous..
2. show that k(x) approaches on some function uniformly on R..

3. Originally Posted by kalagota
ideas:
1. the sum of continuous functions is continuous..
2. show that k(x) approaches on some function uniformly on R..
Wouldn't I just have to prove $3^n$ is continuous, as I can assume $\sin{nx^2}$ is?

4. Originally Posted by Auxiliary
Let $k(x) = \sum_{n=1}^{\infty}\frac{\sin{(nx^2)}}{3^{n}}$. Prove $k$ is continuous on $\mathbb{R}$.

NOTE: for this proof, you're allowed to assume $\sin{(x)}$ is continuous on $\mathbb{R}$.
Note that $\left| \frac{\sin (nx^2)}{3^n} \right| \leq \frac{1}{3^n}$ and the series $\sum_{n=1}^{\infty}3^{-n}$ converges. Thus, by Weierstrass test the series of functions $\sum_{n=1}^{\infty} \sin (nx^2) 3^{-n}$ converges uniformly to some function in $\mathbb{R}$. Thus, this function is continous because the uniform limit of continous functions is continous.

5. I think that this is a much more interesting question.
“Is f a differentiable function on $\Re$?”
It certainly is on any compact subset of $\Re$.
That follows from the M-Test.

6. Originally Posted by Plato
I think that this is a much more interesting question.
“Is f a differentiable function on $\Re$?”
It is a very interesting question. It looks very similar to the Weierstrass function (which is differenciable nowhere but continous).

Note: In complex analysis a beautiful consquence happens, namely say $f_n(z)$ are analytic on $\mathbb{C}$ converging uniformly to $f(z)$ then $f(z)$ is analytic on $\mathbb{C}$ (Weierstrass theorem, I think that is how it is called).