Weierstrass Function: Continuous and Bounded on $\mathbb{R}$

[mathmari]

We have that $$\lim\limits_{x\to x_0} f(x)=\lim\limits_{x\to x_0}\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\lim\limits_{x\to x_0}\phi \left (4^nx\right )$$

Is it correct that we can take the limit into the sum? :unsure:
Can we do that because the series converges?

 

[I like Serena]

Is it correct that we can take the limit into the sum?
Can we do that because the series converges?

It's because Tannery's theorem says:

Let $ S_n = \sum\limits_{k=0}^\infty a_k(n) $ and suppose that $ \lim\limits_{n\rightarrow\infty} a_k(n) = b_k $. If $ |a_k(n)| \le M_k $ and $ \sum\limits_{k=0}^\infty M_k < \infty $, then $ \lim\limits_{n\rightarrow\infty} S_n = \sum_{k=0}^{\infty} b_k $.

Can we find such $M_k$?

 

[mathmari]

It's because Tannery's theorem says:

Let $ S_n = \sum\limits_{k=0}^\infty a_k(n) $ and suppose that $ \lim\limits_{n\rightarrow\infty} a_k(n) = b_k $. If $ |a_k(n)| \le M_k $ and $ \sum\limits_{k=0}^\infty M_k < \infty $, then $ \lim\limits_{n\rightarrow\infty} S_n = \sum_{k=0}^{\infty} b_k $.

Can we find such $M_k$?

So we want to find an upper bound of $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |$, right?
We have that $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |\leq \left |\left (\frac{3}{4}\right )^n\right |\cdot \left |\phi (4^nx)\right |\leq 1\cdot 1=1$ or not?
But this $M_k$ does not satisfy the condition $ \sum\limits_{k=0}^\infty M_k < \infty $.

:unsure:

 

[I like Serena]

So we want to find an upper bound of $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |$, right?
We have that $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |\leq \left |\left (\frac{3}{4}\right )^n\right |\cdot \left |\phi (4^nx)\right |\leq 1\cdot 1=1$ or not?
But this $M_k$ does not satisfy the condition $ \sum\limits_{k=0}^\infty M_k < \infty $.

How about $M_k=\left (\frac{3}{4}\right )^k$?