Really hard infinite series test

[Doom of Doom]

Homework Statement

Test to see whether the following series converges
$$\sum_{n=1}^\infty \sqrt[n]{2}-1$$

Homework Equations

All we've done so far is integral test, ratio test, and root tests.

The Attempt at a Solution

As n approaches infinity, the term apporaches 0, so it may or may not converge.

A ratio test reveals r=1, so it is inconclusive

A root test also gives us 1, so it is inconclusive

I have no idea how i would take the integral of that.


I do know that it diverges, but I have no idea how! I don't know why the prof would give such a hard homework problem.

 

[tim_lou]

use limit test and compare it to 1/n (use L'Hopital's rule)

 

[AKG]

Do you know the comparison test?

 

[Gib Z]

I don't understand you notation :(

But the gist of it is, we find a series that is term by term smaller than the series we wish to test. If the smaller series converges, the test is inconclusive but you can try another smaller series. But if the smaller series diverges, the larger one will as well. Is 1/n term by term smaller than your series?

 

[cristo]

It's this, Gib: $$\sum_{n=1}^\infty \sqrt[n]{2}-1$$. At least that's what the latex looked like it was trying to show!

 

[Doom of Doom]

1/n is bigger that this series term by term, so the comparison test does not work either!

Unless there is another series that is smaller than this series term-by-term that still diverges

 

[StatusX]

1/n is bigger that this series term by term, so the comparison test does not work either!

How'd you get this? I think it's wrong.

 

[Doom of Doom]

StatusX,
when n=1, the terms are equal. When n=2:
1/2= 0.5
[2^(1/2)-1] =0.41421

note that .5>.414.

When n=3:
1/3=.33333
[2^(1/2)-1] =0.25992

 

[Doom of Doom]

The notation should be
$$\sum_{n=1}^\infty (\sqrt[n]{2}-1\ ) )$$

sorry for any confusion. I can't seem to get LaTeX to close the parenthesis on the outer end (outside the -1)

 

[Dick]

1/n is bigger that this series term by term, so the comparison test does not work either!

Unless there is another series that is smaller than this series term-by-term that still diverges

1/n may be larger than the series, but what about ln(2)/n?

 

[StatusX]

Sorry, you're right. But the terms do tend to 1/n (from below), so for any c<1, c/n will eventually be dominated the series.

 

[AKG]

(Part of) The limit comparison test: if a_n > 0, b_n > 0, and the limit of a_n/b_n is greater than 0, then if $\sum b_n$ diverges, then so does $\sum a_n$. Compare a_n = 2^1/n-1 to b_n = n^-1, since you know $\sum n^{-1}$ diverges.