# The series is continuously differentiable

I want to show that series $$f(x)=\sum_{k=1}^{\infty}2^k\sin (3^{-k}x)$$ is continuously differentiable.

We have that each $f_k=2^k\sin (3^{-k}x)$ is differentiable and its derivative is equal to $f_k'(x)=2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)$.

It holds that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}$$

To show that the series is continuously differentiable do we have to show that it converges uniformly?

• Observe that $2 <3$ Apr 7 '18 at 19:37
• Apr 7 '18 at 21:42
• So we have that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$$ right? Do we use the geometric series? @Salahamam_Fatima Apr 7 '18 at 22:27
• @MaryStar Yes {}{}{[{[{[{ Apr 7 '18 at 22:30
• We have that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$$ and the series $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges, since it is a geometric series. By the Comparison Test we get that the series $\sum_{k=1}^{\infty}\left |\frac{2^k}{3^k}\cos(x/3^k)\right |$ converges. So the series $\sum_{k=1}^{\infty}\frac{2^k}{3^k}\cos(x/3^k)$ is absolutely convergent. Does it hold that every absolutely convergent is also continuously differentiable? Or how do we continue from here? @Salahamam_Fatima Apr 7 '18 at 22:36

$$\sum_{k\geq 1}\frac{2^k}{3^k}\cos(x/3^k)$$ is an absolutely convergent series of continuous functions, hence a continuous function which can be termwise-integrated, leading to a continuously differentiable function, $f(x)$.
• We have that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$$ and the series $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges, since it is a geometric series. By the Comparison Test we get that the series $\sum_{k\geq 1}\left |\frac{2^k}{3^k}\cos(x/3^k)\right |$ converges. So the series $\sum_{k\geq 1}\frac{2^k}{3^k}\cos(x/3^k)$ is absolutely convergent. How could we continue from here? Could you explain it further to me? Apr 7 '18 at 22:34
• We have that $f(x)$ and $f'(x)$ converge absolutely, right? Does this imply that the series is continuously differentiable? Apr 8 '18 at 2:58
• @MaryStar: by the $3\varepsilon$-theorem, an absolutely convergent series of continuous functions is a continuous function. Apr 8 '18 at 3:28