Consider the power series $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x+2)^n}{n2^n}$$
Find it's radius of convergence, interval of convergence. Also find for which $x$ the series is absolutely convergent and conditionally convergent.
My attempt :
Let, $$a_n = \frac{(-1)^{n+1}}{n2^n}$$
Then by ratio test we have the radius of convergence, $$\rho = \lim_{n\to \infty} |\frac{a_n}{a_{n+1}}| = \frac{1}{2}$$
Then the interval of convergence is $$ |x+2| \lt \frac{1}{2} \\
\implies x \in (-\frac{5}{2},-\frac{3}{2})$$
We know that in the interval of convergence the power series is absolutely convergent. So we now check the absolute convergence at the end points.
For $x= -\frac{5}{2}$, the series becomes,
$$\sum_{n=1}^{\infty} \frac{(-1)^{2n+1}}{n2^{2n}}$$ which is absolutely convergent and that can be tested using the comparison test with the series $\sum \frac{1}{4^n}$.
For $x = -\frac{3}{2}$,the series becomes,
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n2^{2n}}$$ which is also absolutely convergent and that can be tested using the comparison test with the series $\sum \frac{1}{4^n}$.
Is my answer correct? Can anyone please confirm?
Best Answer
Actually,$$\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}\frac{(n+1)2^{n+1}}{n2^n}=2,$$and therefore the radius of convergence is $2$, not $\frac12$. So, your series converges absolutely on $(-4,0)$ and diverges when $x\notin[-4,0]$.
If $x=0$, then your series is $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$, which converges conditionally. And if $x=-4$, then your series is $\displaystyle\sum_{n=1}^\infty-\frac1n$, which diverges.