Question

Infinite sum of the sequence (2^(1/n)-1)
Prove Divergence or Convergence

Answer 1

check whether the common ratio is |r| < 1
if < 1, it's convergent. Otherwise, it's divergent

Answer 2

See, that's the thing, you can't do that with 2^(1/n), ratio test doesn't solve anything.

Answer 3

limit comparison with 1/n

Answer 4

\[\lim_{n \rightarrow \infty} n2^{1/n} - n\] isn't that an indeterminate form

Answer 5

\[\lim_{n\to\infty}\frac{2^{1/n}-1}{1/n}\]

use L'Hospitals rule

Answer 6

\[-\frac{ 2^{1/n} \log 2 }{ n ^{2} } / n ^{-2}\] like so?

Answer 7

\[\lim_{n\to\infty}\frac{2^{1/n}-1}{1/n}=\lim_{n\to\infty}\frac{2^{1/n}\left(\frac{-1}{n^2}\right)\ln(2)}{\frac{-1}{n^2}}=\lim_{n\to\infty}2^{1/n}\ln(2)\]

Answer 8

That goes to ln(2), which is less than 1, yes? Which is a constant... meaning that it converges because of definition of LCT?

Answer 9

no

Answer 10

ln(2) is a positive finite number so \[\sum_{n=1}^\infty(2^{1/n}-1)\]

does what \[\sum_{n=1}^\infty\frac{1}{n}\] does

Answer 11

Yes, I understand. So lim of An/Bn goes to a constant, and An does what Bn does. Bn diverges, so An must diverge, yes?

Answer 12

yes

Answer 13

Thanks for the help!

Answer 14

np