Is this series convergent or divergent?

uberbandgeek6 · #1 $Old$ November 1st, 2009, 06:53 PM

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!

umgambit · #2 $Old$ November 1st, 2009, 07:20 PM

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
$f'(0)=ln(2)=\lim_{x\to 0}\frac {f(x)-f(0)}{x},$
hence
$ln(2)=\lim_{x\to 0}\frac {2^x-1}{x},$
Putting x=1/n, you get
$ln(2)=\lim_{n\to +\infty}\frac {2^{1/n}-1}{1/n},$
Since ln(2)>0,
$\left(\sum 1/n\right) and \left(\sum 2^{1/n}-1\right)$
have the same nature.
Since the first diverges (the harmonic series), your series diverges also.

umgambit · #3 $Old$ November 3rd, 2009, 07:29 AM

I just realized you asked for a hint and not a complete solution. I'm sorry if I spoiled your exercise.