# interval of convergence for power series

I've been working through a few of these and was hoping to get some help on this problem. I'm thinking the interval of convergence is just 1, since the numerator grows far faster for values greater, but it seems like a much solution involving one of the comparison/convergence tests is required.

Find all values of $x$ for which the series converges $$\sum_{n=1}^{\infty}\dfrac{x^n}{ln(n+1)}$$

• I know. nobody responded there so I wanted to see a solution to that and the other question the person asked – john fowles Aug 4 '17 at 14:30

hint

$$\ln (n+1)=\ln (n)(1+\frac {\ln (1+\frac {1}{n})}{\ln (n)})$$ $$\sim \ln (n)$$

use ratio test.

• ah. so this gives me $\dfrac{x\times\ln(n)}{\ln(n+1)}$ and since i already approximated that $\ln(n+1)$ is ~ $ln(n)$ this cancels out leaving $x$. so the interval of convergence is $-1\geq x \leq 1$. is this on the right track? – john fowles Aug 4 '17 at 14:24
• @johnfowles Yes ${}{}{}{}{}$ the radius is one. – hamam_Abdallah Aug 4 '17 at 14:29

For $n\ge1$ you have $$|x|^n \ge \frac{|x|^n}{\ln(n+1)} \ge \frac{|x|^n}{n}$$

One gives you $R \le 1$ and the other $R \ge 1$

Hint: You can divide the exercise into three different parts:

1. The case when $x>0$ (we have that,$\forall x>1$, $\lim_{n->\infty} \frac{x^n}{\log (n+1)} = \infty$ )

2. The case when $x=0$ (this one is just trivial)

3. The case when $x<0$ (you could rewrite it as $(-1)^n \frac{|x|^n}{\log(n+1)}$ and use the Leibniz Test)

You should be able to get that the series converges for $-1 \le x<1$