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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?

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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.

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

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 $1$.

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

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