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
 A: For the series $\sum_{n = 1}^\infty a_n$, where there is an $N$ such that $a_n \neq 0$ for all $n \geq N$, the ratio test has you calculate
$$  L = \lim_{n \rightarrow \infty} \left|  \frac{a_{n+1}}{a_n}\right|  $$
If $L < 1$, the series converges absolutely.  If $L = 1$ or the limit fails to exist, the test is inconclusive.  If $L > 1$ then the series is divergent.
In both of your examples, you have lost/ignored the absolute value.
For your first example, as I type this, you don't have a sum, but I will guess you meant
$$  f(x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} (x-5)^n}{n 5^n}  \text{.}  $$
The numerator is only zero when $x = 5$, and that sum is easy to do, so we apply the ratio test to determine what happens when $x \neq 5$.  We compute \begin{align*}
L &= \lim_{n \rightarrow \infty} \left| \frac{\frac{(-1)^{(n+1)+1} (x-5)^{(n+1)}}{(n+1) 5^{(n+1)}}}{\frac{(-1)^{n+1} (x-5)^n}{n 5^n}} \right|  \\
    &= \lim_{n \rightarrow \infty} \left| \frac{(-1)^{(n+1)+1} (x-5)^{(n+1)}}{(n+1) 5^{(n+1)}} \cdot \frac{n 5^n}{(-1)^{n+1} (x-5)^n} \right|  \\
    &= \lim_{n \rightarrow \infty} \left| \frac{(-1)^{(n+1)}(-1) (x-5)^{n}(x-5)}{(n+1) 5^{n}5} \cdot \frac{n 5^n}{(-1)^{n+1} (x-5)^n} \right|  \\
    &= \lim_{n \rightarrow \infty} \left| \frac{(-1) (x-5)}{(n+1) 5} \cdot \frac{n}{1} \right|  \\
    &= |(-1) (x-5)| \lim_{n \rightarrow \infty} \left| \frac{n}{5(n+1)} \right|  \\
    &= \frac{1}{5} |x-5|  \text{.}
\end{align*}
The ratio test assures us the series converges if $L < 1$, so when \begin{align*}
\frac{1}{5} |x-5| &< 1  \\
|x-5| &< 5  \\
-5 < x-5 &< 5  \\
0 < x &< 10  \text{.}  
\end{align*}
The ratio test is inconclusive when $L = 1$, that is at $x = 0$ and $x = 10$, so we inspect those individually:


*

*$x = 0$:  \begin{align*}  
\sum_{n=1}^\infty \frac{(-1)^{n+1} (0-5)^n}{n 5^n}
    &= \sum_{n=1}^\infty \frac{(-1)^{n+1} (-1)^n 5^n}{n 5^n}  \\
    &= \sum_{n=1}^\infty \frac{(-1)^{2n+1}}{n}  \\
    &= \sum_{n=1}^\infty \frac{-1}{n}  \\
    &= - \sum_{n=1}^\infty \frac{1}{n}  \text{,}
\end{align*}
which is minus a diverging $p$-series (in particular, it is minus the harmonic series, which diverges).

*$x = 10$:  \begin{align*}  
\sum_{n=1}^\infty \frac{(-1)^{n+1} (10-5)^n}{n 5^n}
    &= \sum_{n=1}^\infty \frac{(-1)^{n+1} 5^n}{n 5^n}  \\
    &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}  \text{,}
\end{align*}
which is an alternating series, whose terms monotonically decrease with limit zero.  By the alternating series test, this series converges.  (We could also recognize this as the alternating harmonic series, which converges.)


Therefore, the interval of convergence of $f(x)$ is $(0,10]$.
You should be able to adapt the above to your other series.  If you have difficulties, ask in comments.
A: $$f(x)=\sum_{n=1}^\infty\underbrace{\frac{(-1)^{n+1}(x-5)^n}{n5^n}}_{a_n}$$
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
$$\int f(x)\,\mathrm dx=\sum_{n=1}^\infty\frac{(-1)^{n+1}(x-5)^{n+1}}{n(n+1)5^n}$$
