Determine the values of $x$ for which the power series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n+1} \cdot (\frac{2x+1}{x})^{n } $ converges .
Try:
I've found the radius of convergence for the above power series is '$1$'. So we've to now solve the inequality $\frac{|2x+1|}{|x|} < 1$ to get the interval of convergence . After solving the inequality I've got the interval of convergence as $(-1,\frac{-1}{2}]$ .
Question:
I'm not sure about my answer . Please help!
EDIT:
The correct interval of convergence for the given power series is $(-1,\frac{-1}{3})$ .
$x > 0$ implies $-x < 2x+1 < x$. This gives $x < -1$ and $x > \frac{-1}{3}$ which is not possible simultaneously.
For $x < 0$ we have $-x > 2x+1 > x$. This gives $-1 < x < \frac{-1}{3}$. So the required interval is $\left(-1,\frac{-1}{3}\right)$.