# Calculate the radius of convergence of $\sum \frac{\ln(1+n)}{1+n} (x-2)^n$

Calculate the radius of convergence of the following: $$\sum \frac{\ln(1+n)}{1+n} (x-2)^n$$

Will you please help me figure out how to calculate: $$\lim_{n\to \infty} \frac{\ln(2+n)}{2+n} \frac{1+n}{\ln(1+n)}$$ which is required for the solution?

HINT. It is clearly that $$\frac{1+n}{2+n}\to 1$$ On the logarithmic terms it is possible operate in this way $$\frac{\ln(2+n)}{\ln(1+n)}=\frac{\ln n +\overbrace{\ln(1+2/n)}^{\to 0}}{\ln n +\underbrace{\ln(1+1/n)}_{\to 0}}$$ Therefore the limit tends to 1.

$$\lim_{n\rightarrow+\infty}\frac{1+n}{2+n}=1$$

The other bit needs more care

$$\frac {\log(2+n)}{\log(1+n)}=\frac{\log{n}+\log(1+\frac{2}{n})}{\log{n}+\log(1+\frac{1}{n})}$$

The limit is $1$ and the radius of convergence is $1$.

You can use this other characterisation of the radius of convergence: $$R=\sup\bigl\{r\mid r\ge 0,\enspace a_nr^n \xrightarrow[n\to\infty]{}0 \bigr\}$$

Here this gives: $$R=\sup\Bigl\{r \mathrel{\Big\vert} r\ge 0,\enspace \frac{\ln(1+n)}{1+n}\,r^n \xrightarrow[n\to\infty]{}0 \Bigr\} =1.$$