Determine the values such that series converges $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{2n+1}\left(\frac{1-x}{1+x}\right)^n$ Determine for what values of $x \in \Bbb R$ the series
$$\sum_{n = 1}^\infty \frac{(-1)^n}{2n+1}\left(\frac{1-x}{1+x}\right)^n$$
converges.
I have tried the alternating series test but I don't think I am doing it correctly because I keep getting infinity. Does that mean it converges for all values? Thank you.
 A: Applying the ratio test, one finds the limiting ratio:
$$
\left|\frac{1-x}{1+x}\right|.
$$
The series therefore converges if that ratio is $<1$ and diverges if it is $>1$.  What happens when it is equal to $1$ must be looked at separately; the ratio test doesn't help there.  So we have
$$
-1<\frac{1-x}{1+x}<1.
$$
We cannot multiply all three members by $1+x$ because that is sometimes positive and sometimes negative, depending on $x$.  So use a common denominator:
$$
\frac{-1-x}{1+x} < \frac{1-x}{1+x} < \frac{1+x}{1+x}
$$
The first inequality becomes
$$
0 < \frac{2}{1+x}
$$
and the second becomes
$$
\frac{2x}{1+x}>0.
$$
The first is satisfied if $x>-1$ and the second if either $x>0$ or $x<-1$.  You need both, so the solution is $x>0$.
A: Since $(1 - x)/(1 + x)$ is defined only for $x \neq -1$, we can rule out $x = -1$. Let $a_n(x)$ be the $n$th term of the series. Then 
$$\lim_{n\to \infty} \left|\frac{a_{n+1}(x)}{a_n(x)}\right| = \lim_{n\to \infty} \frac{2n+1}{2n+3}\left|\frac{1-x}{1+x}\right| = \left|\frac{1 - x}{1 + x}\right|.$$
Now $|(1 - x)/(1 + x)| < 1 \iff|1 - x| < |1 + x| \iff|1 - x|^2 < |1 + x|^2$. By expanding both sides of the equality, we get the equivalent inequality $x > 0$. By the ratio test, the series $\sum a_n(x)$ converges when $x > 0$ and diverges when $x < 0$. When $x = 0$, the alternating series test shows $\sum a_n(x)$ converges. Thus $\sum a_n(x)$ converges if and only if $x \ge 0$.
