# Thread: Convergence of a Series

1. ## Convergence of a Series

Hey I've been having a bit of trouble find the interval of convergence for:

$\sum_{n = 0}^{\infty} \frac{(-1)^n (2x + 1)^n}{n 2^{2n}}$

I did the ratio test:

$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1) |2x + 1| n}{2^2 (n+1)} \right| = \frac{1}{4} \left| 2x + 1 \right| < 1$

So $\frac{-5}{2} < x < \frac{3}{2}$

Now test x = -5/2:

$\sum_{n = 0}^{\infty} \frac{(-1)^n (-4)^n}{n 2^{2n}} = \sum_{n = 0}^{\infty} \frac{1}{n} \ \ \mbox{DV}$

Now test x = 3/2:

$\sum_{n = 0}^{\infty} \frac{(-1)^n (4)^n}{n 2^{2n}} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{n} \ \ \mbox{CV}$

So the interval of convergence is: $\boxed{\frac{-5}{2} < x \leq \frac{3}{2}}$

Does that seem right???

2. Originally Posted by TrevorP
Hey I've been having a bit of trouble find the interval of convergence for:

$\sum_{n = 0}^{\infty} \frac{(-1)^n (2x + 1)^n}{n 2^{2n}}$
You can kill this with the root test instead.

$\left| (-1)^n \frac{(2x+1)^n}{n2^{2n}} \right|^{1/n} = \frac{|2x+1|}{n^{1/n} 4} \to \frac{|2x+1|}{4}$

We want, $|2x+1| < 4 \implies -4 < 2x + 1 < 4 \implies -\frac{5}{2} < x < \frac{3}{2}$.

If $x=3/2$ then $\sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{n4^n}$ which converges.

If $x=-5/2$ then $\sum_{n=0}^{\infty} \frac{(-1)^n (-1)^n 4^n}{n4^n}$ which diverges.

3. Thanks, I'm not familiar with the root test but thanks for confirming my answer.