# Really hard infinite series test

## 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.

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

AKG
Homework Helper
Do you know the comparison test?

Gib Z
Homework Helper
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
Staff Emeritus
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!

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
Homework Helper
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.

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

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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
Homework Helper
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
Homework Helper
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 an > 0, bn > 0, and the limit of an/bn is greater than 0, then if $\sum b_n$ diverges, then so does $\sum a_n$. Compare an = 21/n-1 to bn = n-1, since you know $\sum n^{-1}$ diverges.