Need help with calculus II series I am working on a problem that my professor isn't really explaining well, so i decided to ask here.
The following is the question
$ f(x) =\sum_{n=1}^\infty \frac{\mathrm{(-1)}^{n+1}\mathrm{(x-5)}^{n}}{(n\mathrm{5}^{n})} $
I am asked to find the interval of convergence of the following
$f(x)$
$f'(x)$
$f''(x)$
$\int f(x) $
So for $ f(x) $ i used the ratio test $ (\frac{a_n+1}{a_n} )$ and got that -> $ 1 - \frac{x}{5} < 1 $ Is this the interval of convergence? I am totally clueless.
My main question was for $ f'(x) $ I have the first derivative as follows
$ f'(x) = \frac{\mathrm{(-1)}^{n+1}\mathrm{(x-5)}^{n-1}}{\mathrm{5}^{n}} $
Using the ratio $ (\frac{a_n+1}{a_n} )$ test on this one got me -> $ - \frac{x-5}{5} $ -> $ \frac{x}{5} - 1 $
But this will never be greater than 1. 
I am not sure what I am doing wrong here.
Any help is appreciated

$$f(x)=\sum_{n=1}^\infty\underbrace{\frac{(-1)^{n+1}(x-5)^n}{n5^n}}_{a_n}$$
By the ratio test, the series converges if the following holds:
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+2}(x-5)^{n+1}}{(n+1)5^{n+1}}}{\frac{(-1)^{n+1}(x-5)^n}{n5^n}}\right|=\frac{|x-5|}5\lim_{n\to\infty}\frac n{n+1}=\frac{|x-5|}5<1$$
Solve the inequality for $x$ to get the interval of convergence, $0<x<10$. Don't forget to check for convergence at the endpoints:
$$x=0\implies f(0)=\sum_{n=1}^\infty\frac{(-1)^{n+1}(-5)^n}{n5^n}=-\sum_{n=1}^\infty\frac{5^n}{n5^n}=-\sum_{n=1}^\infty\frac1n$$
$$x=10\implies f(10)=\sum_{n=1}^\infty\frac{(-1)^{n+1}5^n}{n5^n}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$$
Do these series converge or diverge?

For the derivatives of $f$, you can differentiate the series as many times as needed. For instance,
$$f'(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}n(x-5)^{n-1}}{n5^n}=\sum_{n=1}^\infty\frac{(-1)^{n+1}(x-5)^{n-1}}{5^n}$$
Then you can apply the same process as above to determine the interval where the series converges.
For the integral of $f$, first notice that $f(5)=0$, since this makes $(x-5)^n=0$ for all $n\ge1$. Then integrating $f$ gives the series
$$\int f(x)\,\mathrm dx=\sum_{n=1}^\infty\frac{(-1)^{n+1}(x-5)^{n+1}}{n(n+1)5^n}$$
