Let $f(1)= 4, f'(x)= \sum_{k=0}^∞ \frac{(-1)^{k} (x-1)^{k}}{(k+1)!}$ then find $f''(1)$. Let $f(1)= 4, f'(x)= \sum_{k=0}^∞ \frac{(-1)^{k} (x-1)^{k}}{(k+1)!}$ then find $f''(1)$.
I directly differentiated $f'(x)$ and put $x=1$, then I got $0$.
If I expand the series $f'(x)$ then then coefficient of $(x-1)$ is $f"(1)$, which is $-1/2$.
Answer is given $-1/2$.
What is the problem in direct differentiation$?$
And how do we find derivative at a point in such cases$?$
 A: Your sum is
$$f'(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k} (x-1)^{k}}{(k+1)!} \tag{1}\label{eq1A}$$
Differentiating this term by term gives
$$f''(x) = \sum_{k=0}^{\infty} \frac{k(-1)^{k} (x-1)^{k-1}}{(k+1)!} \tag{2}\label{eq2A}$$
The first term, i.e., for $k = 0$, is $0$. The second term, i.e., for $k = 1$, is $\frac{(-1)^{1}(x-1)^{1-1}}{(1+1)!} = \frac{-1(1)}{2!} = -\frac{1}{2}$ since $(x - 1)^{0} = 1$. All other terms have a factor of $x - 1$ to a power of $1$ or higher so, at $x = 1$, they would all be $0$, thus resulting in
$$f''(1) = -\frac{1}{2} \tag{3}\label{eq3A}$$
which matches the given answer. If I understood your question correctly, this confirms what you also got.
Note the original derivative series is basically a modified version of the Taylor series for the exponential function (and, thus, as explained in the Exponential function section of Wikipedia's "Uniform convergence" article, the result is uniformly convergent, with this being a requirement for term by term differentiation, as done in \eqref{eq2A}). To see this, you have
$$\begin{equation}\begin{aligned}
f'(x) & = \sum_{k=0}^{\infty} \frac{(-1)^{k} (x-1)^{k}}{(k+1)!} \\
& = \sum_{k=0}^{\infty} \frac{(1-x)^{k}}{(k+1)!} \\
& = \frac{1}{1-x}\left(\sum_{k=0}^{\infty} \frac{(1-x)^{k+1}}{(k+1)!}\right) \\
& = \frac{1}{1-x}\left(\sum_{k=1}^{\infty} \frac{(1-x)^{k}}{k!}\right) \\
& = \frac{1}{1-x}\left(\sum_{k=0}^{\infty} \frac{(1-x)^{k}}{k!} - 1\right) \\
& = \frac{1}{1-x}\left(e^{1-x} - 1\right) \\
& = \frac{e^{1-x} - 1}{1-x}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
I assume you used this closed form expression when you stated you directly differentiated $f'(x)$ and then substituted $x = 1$. Well, differentiating \eqref{eq4A} gives
$$\begin{equation}\begin{aligned}
f''(x) & = \frac{-e^{1-x}}{1-x} + \frac{(-1)(e^{1-x} - 1)}{(-1)(1-x)^2} \\
& = \frac{(-e^{1-x})(1-x) + (e^{1-x} - 1)}{(1-x)^2} \\
& = \frac{x\left(e^{1-x}\right) - 1}{(1-x)^2}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Since $f''(x)$ is indeterminate at $x = 1$, I'll use L'Hôpital's rule twice to get
$$\begin{equation}\begin{aligned}
f''(1) & = \lim_{x \to 1}\frac{x\left(e^{1-x}\right) - 1}{(1-x)^2} \\
& = \lim_{x \to 1}\frac{\left(1 - x\right)\left(e^{1-x}\right)}{2(1-x)(-1)} \\
& = \lim_{x \to 1}\frac{\left(x - 2\right)\left(e^{1-x}\right)}{2} \\
& = -\frac{1}{2}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
As you can see, the same value is obtained this way as well.
