For a series with $\sum u_n'(x)$ not uniformly convergent, and
If $f '(x) = \lim_{n\to\infty} f_n'(x) $
where $f(x)=\lim_{n\to\infty} f_n(x) $ and $ f_n(x) $ $=u_1+u_2+ . . . +u_n$
Then the series $\sum u_n(x)$ can be differentiated term by term, is this condition true for any series?
On
This is a stronger version (uniformly integrability condition) of what you need:
If a sequence of absolutely continuous functions {$f_n$} converges pointwise to some $f$ and if the sequence of derivatives {$f_n’$} converges almost everywhere to some $g$ and if {$f_n’$} is uniformly integrable then $\lim\limits_{n\mapsto \infty} f_n’ = g= f’$ almost everywhere. Where the derivative of $f$ is $f’$. If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.
Proof : by FTC $f_n(x) – f_n(a) = \int_a^x f_n’ dx$
By Vitali convergence theorem : $\lim\limits_{n\mapsto \infty}\int_a^x f_n’ dx = \int_a^x g dx$
Therefore $\lim\limits_{n\mapsto \infty}( f_n(x) – f_n(a))= \int_a^x g dx$
$f(x)-f(a) = \int_a^x g dx$
$f(x)’=g$ almost everywhere
If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.
If $\sum_{k}f_k(x)=f(x)$, even if it's not uniformly convergent on $]a,b[$, it will be uniformly convergent on $[c,d]$ for all $a<c<d<b$. Therefore, if $\sum_{k}f_k'(x)=f'(x)$ on $]a,b[$, and $x\notin\{a,b\}$, then you can always differentiate $\sum_{k}f_k(x)$ term by term.