Consider the power series $ f(x)=\sum_{n=2}^\infty \log (n) x^n $ Consider the power series $ f(x)=\sum_{n=2}^\infty \log (n) x^n$. The radius of convergence is
(a) $ \ 0 $ , 
(b) $ \infty $
(c) $ 1 $
Answer: Let $ a_n=\log (n).$ Then $a_{n+1}=\log(n+1).$  Now I need help .
Then $ \frac{a_n}{a_{n+1}}=\frac{\log(n)}{\log(n+1)} <1 $ 
Then $R <1.$
 A: Use the ratio test. Find $x$ s.t.
$$ \lim_{n \to \infty} \left| \frac{\log(n+1) x^{n+1}}{\log n \,x^n}\right| < 1 $$
$$ |x| \lim_{n \to \infty} \left| \frac{\log(n+1)}{\log n}\right| < 1 $$
$$ |x|<1 $$
So your sum converges when $|x|<1$ and subsequently $R=1$.
A: An alternative approach: the geometric series is analytic with radius the convergence $1$, and 

Theorem: the derivative of an analytic function is also analytic with the same radius of convergence, and it power series representation is the term-by-term derivative of the power series representation of the original function

The above imply that the series $\sum_{k=1}^\infty kx^{k-1}$ (the derivative of the geometric series $\sum_{k=0}^\infty x^k$) have also radius of convergence $1$. And it is easy to see that
$$|\ln(k)x^k|<|kx^{k-1}|,\quad k\in\Bbb N_{>0}$$
when $x\in(-1,1)$. Thus the radius of convergence of your series is, at least, $1$. By last we can see that the series diverges when $x=1$, hence it radius of convergence is $1$.
A: $$\lim \frac{a_n}{a_{n+1}}=\lim \frac{\log n}{\log(n+1)}=\lim \frac{\frac{1}{n}}{\frac{1}{n+1}}=\lim \frac{n+1}{n}=1$$
Hence, radius of convergence is $1$.
A: For $x>1$ the series clearly diverges.
Let's $0<x<1$. Then for every $n>1$, we have $$x^{n+1}\log(n+1)/ x^{n}\log(n)=x.\log(n+1)/\log(n)$$ and the limit of that expression is $x<1$, so by the Alembert criteria the series converge.
Then the radius of convergence is $1$.
A: First when $x = 1$, it clearly diverges because the terms are increasing. This means that the radius of convergence is no more than 1. This eliminates b.
When $x = 1/4$, we see that our sequence is bounded above by $\sum_{n=2}^\infty (1/2)^n$ which we know converges. Hence the radius of convergence is at least 1/2. This eliminates a.
Hence the answer is c.
A: Ratio test and equivalents:
$$\frac{a_n}{a_{n+1}}=\frac{\log n}{\log(n+1)}\sim_\infty\frac nn=1,$$
hence the radius of convergence is $1$.
