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mathmari

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I want to show that series $$f(x)=\sum_{k=1}^{\infty}2^k\sin (3^{-k}x)$$ is continuously differentiable. We have that $|2^k\sin (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$, or not?

The sum $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges as a geometric series. So by comparison test the series $f(x)$ converges absolutely.

The derivative is equal to $$f'(x)=\sum_{k=1}^{\infty}2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)$$

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

The sum $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges as a geometric series. So by comparison test the series $f'(x)$ converges absolutely. So, we have tha tthe function and its derivative are absolutely convergent. Does this imply that the function $f(x)$ is continuousy differentiable? (Wondering)