How do I plug in endpoints into a power series?

Question

I don't understand how to plug in the endpoints into the original power series.

The original power series is $$ \sum_{n=0}^\infty {(-1)^n x^n\over{n+1}} $$ What I have so far is this:

I applied the ratio test

$$ a_n= {(-1)^n x^n\over{n+1}}$$ $$ \lim_{n\to\infty} |{a_{n+1}\over a_n}|= \lim_{n\to\infty} |{{(-1})^{n+1}x^{x+1} \over n+2} *{n+1 \over{(-1)^n}x ^n}| = \lim_{n\to \infty} |{x^{n+1}\over n+2}*{n+1 \over x^n}| $$ $$=\lim_{n \to \infty} |x*{n+1\over n+2}|= |x| \lim_{n\to \infty} {n+1\over n+2} $$ $$ |x|\lt1 \Rightarrow -1 \lt x\lt 1$$ Now, this is where I am stuck. I don't know how to plug it into the original series.

When $x=-1$ $$ \sum_{n=0}^\infty {(-1)^n (-1)^n\over{n+1}}$$ And when $x=1$ $$ \sum_{n=0}^\infty {(1)^n (-1)^n\over{n+1}}$$ and...now what? What do I do with the n's?

Answer 1

Note that $(-1)^n(-1)^n=(-1)^{2n}=1$ and you end up with the well-known (and divergent) harmonic series.

Answer 2

When $x=-1$:

$$\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n+1}=\sum_{n=0}^{\infty} \frac{(-1)^n (-1)^n}{n+1}=\sum_{n=0}^{\infty} \frac{(-1)^{2n} }{n+1} \\ =\sum_{n=0}^{\infty} \frac{1 }{n+1} \text{, it is the harmonic series,and we know that this diverges.}$$

When $x=1$:

$$\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n+1}=\sum_{n=0}^{\infty} \frac{(-1)^n 1^n}{n+1}=\sum_{n=0}^{\infty} \frac{(-1)^n }{n+1}, \text{ and,using the Dirichlet criterion,you can conclude that this series converges.}$$