Complicated Series

**jules027** · May 16th 2008, 07:01 PM

Consider the Series: ∑ 1/[2^(k+(-1)^k)] from k=0 to ∞.

a. Use the ratio test to show the limit does not exist.

b. Use the root test to show the series converges.

**Mathstud28** · May 16th 2008, 07:18 PM

Originally Posted by jules027

Consider the Series: ∑ 1/[2^(k+(-1)^k)] from k=0 to ∞.

a. Use the ratio test to show the limit does not exist.

b. Use the root test to show the series converges.

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

Setting up the ratio test we get

$\lim_{n\to\infty}\bigg|\frac{1}{2^{(n+1)(-1)^{n+1}}}\cdot\frac{2^{n+(-1)^n}}{1}\bigg|=\lim_{n\to\infty}|2^{-((n+1)+(-1)^{n+1})}\cdot{2^{n+(-1)^n}|}=\lim_{n\to\infty}2^{2(-1)^n-1}=$ $\lim_{n\to\infty}\frac{4^{(-1)^n}}{2}$

Which is an oscilating function...to realize this also notice that $(-1)^n=\cos(\pi{n})$

In this case only because $n\in\mathbb{N}$

Next if we do the root test we get

$\lim_{n\to\infty}\bigg(2^{-(n+(-1)^n)}\bigg)^{\frac{1}{n}}=\lim_{n\to\infty}2^{-1-\frac{(-1)^n}{n}}=\lim_{n\to\infty}\frac{1}{2}\cdot{2^{\fr ac{(-1)^{n+1}}{n}}}=\frac{1}{2}\cdot{2^{0}}=\frac{1}{2} <1$

Thus since this is less than one the series converges

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