Find the domain of convergence of the series $\sum ^{\infty}_{n=1}\frac{n}{n+1}\left(\frac{2x+1}{x}\right)^n$

Question

Find the domain of convergence of the series

$$\sum ^{\infty}_{n=1}\frac{n}{n+1}\left(\frac{2x+1}{x}\right)^n$$

My idea

suppose if choose $y=\frac{2x+1}{x}$ then $\sum ^{\infty}_{n=1}\frac{n}{n+1}\left(\frac{2x+1}{x}\right)^n=\sum ^{\infty}_{n=1}\frac{n}{n+1}y^n$ and it has radius $1$ then how we processed for further

Answer 1

Your idea is fine. Your series converges if and only if $x\in\left(-1,-\frac13\right)$. That's because$$\left\{x\in\mathbb R\,\middle|\,\left\lvert\frac{2x+1}x\right\rvert<1\right\}=\left(-1,-\frac13\right)$$and because$$\sum_{n=1}^\infty y^n\text{ converges}\iff\sum_{n=1}^\infty\frac n{n+1}y^n\text{ converges,}$$by Abel's test.

Answer 2

Note that: $$\lim_{n\to\infty}\frac{n}{n+1}=1$$ so we can say that: $$\lim_{n\to\infty}\frac{n}{n+1}\left(\frac{2x+1}{x}\right)^n=\lim_{n\to\infty}\left(\frac{2x+1}{x}\right)^n$$ and we know that for a series to converge of this form we require: $$\left|\frac{2x+1}{x}\right|<1$$ using algebra we can work out that the boundaries of this are $-1$ and $-\frac 13$. Now reapply this to the equation and we get that: $$x\in\left(-1,-\frac 13\right)$$