One way to prove if an infinite sum is to use the following theorem
Theorem Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a, b]$ and such that $f_n(x_0)$ converges for some point $x_0$ on $[a,b]$. If $f_n^\prime$ converges uniformly on $[a,b]$, then $f_n$ converges uniformly on $[a, b]$, to a function $f$, and $$ f^\prime(x) = \lim_{n\to\infty} f_n^\prime(x) \qquad (a \leq x \leq b) $$
Does this theorem have a converse?
An example of what I mean:
In Elementary Classical Analysis (Marsden) there is the following exercise (p. 316, #52):
Can the series $$x = \sum_{k=1}^\infty \frac{x^k}{k} - \frac{x^{k+1}}{k+1} \quad 0 \leq x \leq 1 $$ be differentiated term by term?
Letting $S_n = \sum_{k=1}^n \frac{x^k}{k} - \frac{x^{k+1}}{k+1} = x-\frac{x^{n+1}}{n+1} $, then $S_n' = 1 - x^{n+1}$. This converges uniformly to $1$ on $[0,r], r < 1$.
So, on $[0,r], r < 1$, we can differentiate term by term by the theorem. (term-by-term is an odd way of expressing the point of this exercise I feel. I think it is phrased this way because $S_n$ are differentiated term by term.)
In this case, we know that the sum is equal to $x$, so the derivative is equal to $1$. The term-wise differentiation yields $0$ at $x=1$, so term-wise differentiation is not possible at $x=1$.
But is there a general method to disproving this? (i.e., if the series was not equal to something that is easy to differentiate, can we conclude anything about it's differentiability from this theorem?
The problem is you could have a sequence $f_n$ that converges to $0$ everywhere on $[0,1]$, but with a very large bump that gets closer and closer to $0$. A classical example is something like $f_n(x) = nxe^{-nx^2}$.
The problem is to also obtain $f_n'(x) \to 0$ everywhere on $[0,1]$ which the preceding example does not satisfy. A little experimentation with desmos led me to $$f_n(x) = nx^2e^{-nx^3}.$$
It isn't hard to show that $f_n(x) \to 0$ and $f_n'(x) \to 0$ for all $x \in [0,1]$, but the convergence is highly non-uniform.