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?

Help please!


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 $\sum \frac{ \cos nx }{n^2}$.

  • $\begingroup$ Where did that theorem state series is differentiable? $\endgroup$ – Gitika Jun 2 '20 at 12:17
  • $\begingroup$ 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. $\endgroup$ – mathcounterexamples.net Jun 2 '20 at 12:19
  • $\begingroup$ From there,we get that the uniform limit of the series is differentiable. $\endgroup$ – Gitika Jun 2 '20 at 12:25
  • $\begingroup$ Sum of series is differentiable..so the series is differentiable..is it? $\endgroup$ – Gitika Jun 2 '20 at 12:28
  • $\begingroup$ Sum of series is just what is called series. $\endgroup$ – 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.


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