How do I find the radius and interval of convergence of $\sum_{n=1}^\infty {(-1)^n(x+2)^n \over n} $ $$\sum_{n=1}^\infty {(-1)^n(x+2)^n \over n} $$
I used the ratio test to test for absolute convergence, but I'm sort of stuck on:
$$n(x+2) \over n+1$$
 A: It will converge if and only if $$\lim_{n \to +\infty}\left|\frac{\frac{(-1)^{n+1}(x+2)^{n+1}}{n+1}}{\frac{(-1)^n(x+2)^n}{n}}\right| < 1,$$ that is, if and only if: $$\lim_{n \to + \infty}\left|\frac{n}{n+1}(x+2)\right| < 1 \iff |x+2|<1.$$
So the inteval of convergence is $]-3,-1]$ and the radius is $1$. The point $-1$ is included, because the series will converge by the alternating test (thanks for Pauly B for pointing this in the comments).
A: Hint, the ratio needs to be such that
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|<1 $$
So, what values of $x$ will satisfy this condition after the limit has been taken?
A: In this case applying root test is better
$$\lim_{n\rightarrow \infty}\sqrt[n]{\Bigg|\frac{(-1)^n(x+2)^n}{n}}\Bigg| = |x+2|$$
Since $\lim_{n \rightarrow \infty} \sqrt[n]{n}=1$
You want $|x+2| < 1$.
A: it should be
$\lim_{n\to\infty}\left|\frac{n(x+2)}{n+1}\right|=|x+2|\lim_{n\to\infty}\frac{n}{n+1}$
hopefully you can work out that limit.
A: we can use the series of $\log(1+x)$
$$\log(1+x)=\sum_{n=1}^{\infty }\frac{(-1)^{n+1}x^n}{n}=-\sum_{n=1}^{\infty }\frac{(-1)^{n}x^n}{n}$$
we know the condition of this series $|x|<1$
when the x=x+2 then condition becomes $$|x+2|<1$$
