Given a sum find $a\in\mathbb R$ such that $R=2$

Question

Find $a\in\mathbb R$ such that the series power $$\sum_{n=1}^\infty{\frac{(1+a)^n}n(x-2)^n}$$ has convergence ratio equal to $2$.

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I use the D'Alembert criteria for the sum and came with $$\lim_{n\to+\infty}{\left|\frac{a_{n+1}}{a_n}\right|}<1\quad\Rightarrow\quad\cdots\quad\Rightarrow\quad|(1+a)(x-2)|<1,$$ but the statement says $R=2$, so $$|(1+a)(x-2)|<2\quad\Rightarrow\quad|1+a|<\frac 2{|x-2|}\quad\Rightarrow\quad\boxed{-\frac 2{|x-2|}-1<a<\frac 2{|x-2|}-1}.$$ Anyway the statement says $a\in\mathbb R$, and here $a$ depends on the value of $x$.

Am I doing something wrong? If so, what am I missing?

Thank you!

Answer 1

Your computations show that the radius of convergence is $$ R=\frac{1}{|1+a|} $$ Your task is to find $a$ such that $R=2$. But $$ R=2\iff |1+a|=\frac{1}{2}\iff a=\frac{-1}{2},\frac{-3}{2} $$

Answer 2

Hint

$$a_n = \frac{(1+a)^n}{n}$$

Answer 3

Perhaps a little easier: with Cauchy-Hadamard formula:

$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{\frac{|1+a|^n}n}=\limsup_{n\to\infty}\frac{|1+a|}{\sqrt[n]n}=|1+a|=$$

$$=\frac12\iff 1+a=\pm\frac12\iff a=\begin{cases}-\cfrac12\\{}\\-\cfrac32\end{cases}$$