# Alternate series converges

• Jun 16th 2010, 11:49 AM
tarheelborn
Alternate series converges
Prove that (1-2) - (1-2^(1/2)) + (1 - 2^(1/3)) - (1 - 2^(1/4)) + ... converges.
• Jun 16th 2010, 12:09 PM
chiph588@
Quote:

Originally Posted by tarheelborn
Prove that (1-2) - (1-2^(1/2)) + (1 - 2^(1/3)) - (1 - 2^(1/4)) + ... converges.

Just show $\displaystyle \sqrt[n]{2}\to1$ monotonically, then just apply the alternating series test.
• Jun 16th 2010, 12:25 PM
tarheelborn
Ah, so since (1-2) - (1- 2^(1/2))+... = (-1)(2-1) + (1)(2^(1/2) -1)-... = \sum (-1)^n(2^(1/n)-1). Lim 2^(1/n) = 1 ==> Lim 2^(1/n)-1 = 0. Also, 2^(1/n)>=2^1/(n+1)>=2^(1/n+2)... so {2^(1/n)} is nonincreasing. Thus this series is convergent by alternate series test. That will work, right? Thanks!
• Jun 16th 2010, 12:40 PM
chiph588@
Quote:

Originally Posted by tarheelborn
Ah, so since (1-2) - (1- 2^(1/2))+... = (-1)(2-1) + (1)(2^(1/2) -1)-... = \sum (-1)^n(2^(1/n)-1). Lim 2^(1/n) = 1 ==> Lim 2^(1/n)-1 = 0. Also, 2^(1/n)>=2^1/(n+1)>=2^(1/n+2)... so {2^(1/n)} is nonincreasing. Thus this series is convergent by alternate series test. That will work, right? Thanks!

Correct. You might want to say why $\displaystyle 2^{\frac1n}>2^{\frac{1}{n+1}}$.