# Series convergence question

• Apr 8th 2009, 04:05 PM
virtuoso735
Series convergence question
I'm stuck on a series convergence/divergence problem. Does this coverge or diverge: the sum of [(the nth root of 2)] - 1 as n goes to infinity? I tried the ratio test, but I got 1 as the limit, so that doesn't tell me anything. What should I try next?

Thanks.
• Apr 8th 2009, 04:31 PM
Jester
Quote:

Originally Posted by virtuoso735
I'm stuck on a series convergence/divergence problem. Does this coverge or diverge: the sum of [(the nth root of 2)] - 1 as n goes to infinity? I tried the ratio test, but I got 1 as the limit, so that doesn't tell me anything. What should I try next?

Thanks.

Try a limit comparison with $\sum_{n=1}^\infty \frac{1}{n}$ i.e. consider $\lim_{n \to \infty} \frac{2^{1/n} - 1}{\frac{1}{n}}$
• Apr 8th 2009, 04:58 PM
virtuoso735
Hmm, I tried using the limit comparison test above and it came out to be 0, which is inconclusive is it not?
• Apr 8th 2009, 05:22 PM
Jester
If we consider $\lim_{n \to \infty} \frac{2^{1/n} - 1}{\frac{1}{n}}$, if we let $n = \frac{1}{x}$ then $\lim_{x \to 0} \frac{2^{x} - 1}{x} = \lim_{x \to 0} 2^{x} \ln 2 = \ln 2$ (by L'Hopital's rule). Since $\sum_{n=1}^\infty \frac{1}{n}$ diverges, by the LCT, the original series diverges.
• Apr 8th 2009, 11:56 PM
matheagle
You mean 0+

Quote:

Originally Posted by danny arrigo
If we consider $\lim_{n \to \infty} \frac{2^{1/n} - 1}{\frac{1}{n}}$, if we let $n = \frac{1}{x}$ then $\lim_{x \to 0^+} \frac{2^{x} - 1}{x} = \lim_{x \to 0^+} 2^{x} \ln 2 = \ln 2$ (by L'Hopital's rule). Since $\sum_{n=1}^\infty \frac{1}{n}$ diverges, by the LCT, the original series diverges.

• Apr 9th 2009, 06:12 AM
Jester
Quote:

Originally Posted by matheagle
You mean 0+

Of course - thanks for pointing that for the OP.