# Show that $\sum_{1}^\infty\frac{\sin(nx)}{n^3}$ is differentiable everywhere

I have recently been trying out some questions on series of functions.In one of the questions, I was given a series $$\sum_{1}^\infty\frac{\sin(nx)}{n^3}$$ and now I am supposed to show that the above series is differentiable at every real number and I need to find its derivative.

I was wondering when a series of the form $$\sum_{1}^\infty f_n(x)$$ said to be differentiable at x?..is it when each $$f_n(x)$$ is differentiable?.. and if that is so,then I assume that I am done with first half of the question.

In the second half,should I show that the given series is uniformly convergent by using the Weierstrass' M-Test and hence differentiate the series term by term?

The main theorem regarding differentiation of series of functions is provided at this link.

Here $$f_n(x) = \frac{\sin nx }{n^3}$$ is such that $$\sum f_n^\prime(x) = \sum \frac{ \cos nx }{n^2}$$ is normally convergent, therefore uniformly convergent by Weierstrass M-test as $$\sum 1/n^2$$ converges. As the series is also convergent at $$0$$, the theorem states that the series of functions is differentiable and has for derivative $$\sum \frac{ \cos nx }{n^2}$$.

• Where did that theorem state series is differentiable? – Gitika Jun 2 '20 at 12:17
• Last line $...f^\prime(x) = g(x)...$. Have also a look at the proof which is instructional and refers to Differentiation and Uniformly Convergent Sequences of Functions. – mathcounterexamples.net Jun 2 '20 at 12:19
• From there,we get that the uniform limit of the series is differentiable. – Gitika Jun 2 '20 at 12:25
• Sum of series is differentiable..so the series is differentiable..is it? – Gitika Jun 2 '20 at 12:28
• Sum of series is just what is called series. – mathcounterexamples.net Jun 2 '20 at 12:33

For functions from $$\Bbb R$$ to $$\Bbb R:$$

Suppose (i) $$f_n(0)$$ converges to a value $$f(0)$$ and (ii) Each $$f'_n$$ is continuous and $$f'_n$$ converges uniformly to $$g$$ on any bounded subset of $$\Bbb R.$$ Then $$g$$ is continuous and $$\lim_{n\to \infty} f_n(x)=\lim_{n\to \infty}f_n(0)+\int_0^xf'_n(t)dt=f(0)+\int_0^x g(t)dt.$$ Define the above limit as $$f(x).$$ By the Fundamental Theorem of Calculus, since $$g$$ is continuous we have $$f'(x)=(d/dx)[\,f(0)+\int_0^x g(t)dt\,]=g(x)=\lim_{n\to \infty}f'_n(x).$$

Let $$f_n(x)=\sum_{j=1}^n (\sin jx)/j^3.$$ Then $$f'_n$$ converges uniformly on all of $$\Bbb R,$$ and obviously $$f_n(0)$$ converges.

Every continuous $$h:\Bbb R \to \Bbb R$$ is the uniform limit of a sequence $$(h_n)_n$$ of differentiable functions but $$h$$ may be non-differentiable at some (or every) point. The sequence $$(h'_n(x))_n$$ may fail to converge at some $$x,$$ or the sequence $$(h'_n)_n$$ may fail to converge uniformly on some interval of positive length.