Identity in Fourier summation Suppose the following series:
\begin{eqnarray}
\sum_{k'}k'f_{k'}
\end{eqnarray}
where $f_{k'}$ are some Fourier coefficients that result from a periodic function $f(t+T)=f(t)$:
\begin{eqnarray}
f_{k'}=\frac{1}{T}\int_{0}^{T}dt e^{ik'2\pi t/T}f(t).
\end{eqnarray}
Is it true then that:
\begin{eqnarray}
\sum_{k'}k'f_{k'}=\frac{1}{T}\int_{0}^{T}dt e^{ik'2\pi t/T}f(t)\left(\sum_{k'}k'e^{ik'2\pi t/T}\right)=0,
\end{eqnarray}
that is, given that:
\begin{eqnarray}
g(t)=\left(\sum_{k'}k'e^{ik'2\pi t/T}\right)=2i\sum_{k'=1}^{+\infty}k'\sin(2\pi k' t/T)=0
\end{eqnarray}
Is the above identity correct?
EDIT: Wolfram Alpha gives the wrong answer for the sum $g(t)$. So I would like to know how to determine $g(t)$ in closed form, if possible.
 A: If $f$ is smooth then modulo some constant $$\sum_{k'}k'f_{k'}=cf'(0).$$
So your main question amounts to asking whether $$f'(0)=0$$holds for every periodic smooth $f$.
Edit: So now we want to know what distribution the series $\sum ke^{ikt}$ converges to. This turns out to be very easy. I'm going to take advantage of the Littlewood Convention, to the effect that $2\pi=1$; you may want to clean up the following, inserting the missing constants:
$\newcommand\D{\mathcal D}$
$\newcommand\T{\Bbb T}$
$\newcommand{\ip}[2]{\langle #1,#2\rangle}$


Lemma The series $\sum_{k\in\Bbb Z}e^{ikt}$ converges to $\delta_0$ in the (weak) topology of $\D'(\T)$.


Proof. Say $s_n=\sum_{|k|\le n}$. If $\phi\in\D(\T)$ then $$\ip{s_n}\phi=\sum_{|k|\le n}\widehat \phi(k)\to \phi(0)=\ip{\delta_0}\phi,\quad(n\to\infty)$$which is exactly what it means to say $s_n\to\delta_0$ in $\D'(\T)$.
And now since differentiation is continuous on $\D'(\T)$ it follows that $$i\sum_{k\in\Bbb Z}ke^{ikt}=\delta_0',$$where by definition $$\ip{\delta_0'}\phi=-\phi'(0).$$
A: Amplifying a bit @DavidCUllrich's answer: ... by the way, using $k'$ as a summing index is a bit confusing (in the context of standard practice) when there's no $k$ without the prime... Especially dangerous if/when derivatives might be involved.
So, somewhat revising the notation, first, we have $f(x)=\sum_{k\in\mathbb Z} \widehat{f}(k)\,e^{2\pi ikx}$, under various hypotheses on periodic $f$.
In circumstances where it is legitimate to differentiate termwise (e.g., either for fairly smooth $f$, or interpreting the summating as converging in a Sobolev space...)
$$
f'(x) \;=\; 2\pi i\cdot \sum_k k\cdot \widehat{f}(k)\cdot e^{2\pi ikx}
$$
Then, if this had a pointwise-convergence sense,
$$
\sum_k k\cdot \widehat{f}(k) \;=\; {f'(0)\over 2\pi i}
$$
much as in @DavidC.Ullrich's answer. Again, the notational choice is considerably distracting... :)
EDIT: I can't speak for any software system's reasoning about why $\sum_k k\cdot e^{2\pi ikx}=0$ (nor in what sense this is claimed), but it is true that $\sum_k 1\cdot e^{2\pi ikx}$ is a Fourier series expansion of the periodic Dirac delta, the Dirac comb. Of course this does not converge pointwise anywhere, but it does converge (meaningfully and usefully) in a Sobolev space.
The distributional derivative is indeed $2\pi i\sum_k k\cdot e^{2\pi ikx}$, and this is absolutely correct in a Sobolev space, and the differentiation is likewise absolutely correct, if interpreted properly.
Since the (distributional) derivative of the Dirac comb is locally $0$ away from integers, we might decide to say that the corresponding Fourier series is $0$ away from integers... and since the (distributional) derivative is locally $0$ away from integers, we could also say that the Fourier series for the derivative is $0$ there (even though it does not converge pointwise at all, only in a Sobolev space).
