$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-5)^n}{n5^n}$ derivatives and intervals of convergence? My math professor is knee-deep in questions from me and many students as he said, it might have to do with him not writing our homework questions (no Taylor series or Maclaurin yet):
I have to find the intervals of convergence for several derivatives of: $$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-5)^n}{n5^n}$$ first off, I can see that $r=x-5$ so $\lvert x-5\rvert\lt1$ giving the interval $(4,6)$ so all I have to do is check the endpoints from here.
I started getting derivatives, I find that $$f^{'}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}n(x-5)^n}{n5^n}$$
= $$f^{'}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-5)^n}{5^n}$$ notice my lower limit for the sum, it's still $1$, nor did I decrease the power on $x$ because it seems to make sense. I know that if the lower limit was $n=0$, then we can bump it up to $1$, and differentiate as if there wasn't a sum involved (as in I would decrease the power on $x$)
Mostly I'm confused on how derivatives of series work, especially with respect to what $n$ will equal
Assuming my $f^{'}$ is correct, I test the endpoints of my hypothetical interval of convergence and clearly find that both when $x=4$ and $x=6$, the series converges, so I get the interval: $[4,6]$ for $f^{'}$ which happens to be wrong:
When $x=4$:  $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-1)^n}{5^n}$$
Ignoring the alternating parts (the $(-1)^n$'s), we get that $b_n=\frac{1}{5^n}$ converges to zero, and $b_n$ also decreases, so it converges by the alternating series test, the same goes for if:
When $x=6$: $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(1)^n}{5^n}$$
I don't know if it has to do with me differentiating or something else, but the answer is not $[4,6]$ for $f^{'}$?
 A: Your derivative is incorrect, and I think it's because you're skipping steps. It's not uncommon, when differentiating power series, to re-index the sum so that an $x^n$ appears instead of an $x^{n-1}$, but I think you should write these steps more explicitly.
We have
$$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-5)^n}{n5^n}.$$
Taking the derivative, we take the derivative of each term:
\begin{align*}
f'(x) &= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}n(x-5)^{n-1}}{n5^n} \\
&= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-5)^{n-1}}{5^n}.
\end{align*}
Making the substitution $m = n - 1$, or $n = m + 1$, we get
\begin{align*}
f'(x) &= \sum_{m=0}^{\infty}\frac{(-1)^{m+2}(x-5)^m}{5^{m+1}} \\
&= \frac{1}{5}\sum_{m=0}^{\infty}\frac{(-1)^m(x-5)^m}{5^m}.
\end{align*}
You've also made an error, somehow, in your calculation of the radius of convergence: it should be $5$, not $1$. Note that, if $a_n$ is the $n$th term of the sum, then
\begin{align*}
\frac{a_{n+1}}{a_n} &= \frac{(-1)^{n+2}(x-5)^{n+1}}{(n+1)5^{n+1}} \div \frac{(-1)^{n+1}(x-5)^n}{n5^n} \\
&= \frac{(-1)^{n+2}(x-5)^{n+1} \times n 5^n}{(n+1)5^{n+1} (-1)^{n+1}(x-5)^n} \\
&= \frac{-n(x - 5)}{5(n + 1)} = -\frac{1}{5} \cdot \frac{n}{n+1} \cdot (x - 5).
\end{align*}
So, $|a_{n+1} / a_n| \to \frac{1}{5}|x - 5|$, which is less than $1$ if and only if $|x - 5| < 5$.
A similar calculation reveals $f'(x)$ has the same radius of convergence (indeed, the radius of convergence of the derivative is always the same as the original function). Give it another try with this in mind.
