# Prove or disprove that $\sum_{n=1}^{\infty} \frac{x^{3/2}\cos(nx)}{n^{5/2}}$ is differentiable on $(0, \infty)$

Let $$f(x) = \sum_{n=1}^{\infty} \frac{x^{3/2}\cos(nx)}{n^{5/2}}$$ on $$(0, \infty)$$.

(i) Is $$f(x)$$ differentiable on $$(0, \infty)$$?

(ii) Does the series uniformly converge to $$f$$ on $$(0, \infty)$$?

Any help or hints will be appreciated.

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Edit: What I know about the differentiability of a sequence of functions is that

If $$(f_n(x))$$ is defined on $$I = [a,b]$$ and

(i) Each $$f_n$$ is differentiable on $$I$$.

(ii) For some $$x_0 \in I$$, $$(f_n(x_0))$$ converges.

(iii) $$(f'_n)$$ converges uniformly.

Then $$(f_n)$$ converges uniformly, $$f$$ is differentiable on $$I$$ and

$$f'(x) = \lim_{n \rightarrow \infty} f'_n(x)$$ holds.

The above case seems not to satisfy the conditions of the above theorem.

At this point, I don't know how to proceed.

Hint for uniform convergence: If the series converged uniformly on $$(0,\infty),$$ then

$$\sup_{0

as $$n\to \infty.$$ Is this true?

• I see. if some series $\sum_{n=1}^{\infty} f_n (x)$ converges uniformly on $I$, then $sup_{x \in I} f_n(x) \rightarrow 0$, right? Jul 31, 2020 at 4:27
• Yes. If the series converges uniformly, then the general terms tend to $0$ uniformly. Jul 31, 2020 at 4:29
• @MarkViola Thanks. Jul 31, 2020 at 4:33
• It was my pleasure. Jul 31, 2020 at 4:34

HINT:

For any positive number $$L$$, we see that the series obtained by differentiation of the general terms, namely $$\displaystyle \sum_{n=1}^\infty \left(\frac32 \frac{x^{1/2}\cos(nx)}{n^{5/2}}-\frac{x^{3/2}\sin(nx)}{n^{3/2}} \right)$$, converges uniformly for $$x\in (0,L]$$.

• Ah! I have kept calculating the term with $sin(nx)$ wrongly, that's why I thought that the above series seems not to satisfy the hypotheses of the above theorem...Thanks. Jul 31, 2020 at 4:30
• You're welcome. My pleasure. Aug 21, 2020 at 16:05