# Is this series convergent or divergent?

• Nov 1st 2009, 04:53 PM
uberbandgeek6
Is this series convergent or divergent?
Does this series converge or diverge?

( nthRoot(2) - 1 )

I wanted to break it up into 2 series, but I don't think I can because I believe you must know that both parts are convergent to split it. Note that I am only allowed to use the integral test, test for divergence, comparison test, ratio test, alternating series test, and root test to find the answer, since we haven't learned anything else in class yet. If someone could just give me a hint as to what test to use or what to compare it to, that'd be great. Thank you!
• Nov 1st 2009, 05:20 PM
umgambit
In order to solve this problem, you need fo know that putting f(x)=2^x, f'(x)=(ln2)2^x.

That being said, note that
$\displaystyle f'(0)=ln(2)=\lim_{x\to 0}\frac {f(x)-f(0)}{x},$
hence
$\displaystyle ln(2)=\lim_{x\to 0}\frac {2^x-1}{x},$
Putting x=1/n, you get
$\displaystyle ln(2)=\lim_{n\to +\infty}\frac {2^{1/n}-1}{1/n},$
Since ln(2)>0,
$\displaystyle \left(\sum 1/n\right) and \left(\sum 2^{1/n}-1\right)$
have the same nature.
Since the first diverges (the harmonic series), your series diverges also.
• Nov 3rd 2009, 05:29 AM
umgambit
I just realized you asked for a hint and not a complete solution. I'm sorry if I spoiled your exercise.