Converge or Diverge? sum, n = 1 to infinity, of 2^(1/n)

[monomoco]

If the series converges, find the sum:

Summation from n=1 to infinity of 2 ^ (1/n)

Is this really just 2^0 = 1? Seems like I'm missing something...

 

[stapel]

Summation from n=1 to infinity of 2 ^ (1/n)

Is this really just 2^0 = 1? Seems like I'm missing something...

The first term is 2[sup:2sbc5j0h]1[/sup:2sbc5j0h] = 1; the second term is 2[sup:2sbc5j0h]1/2[/sup:2sbc5j0h], or about 1.414. The sum of the first two terms is about 2.414.

How are you getting that the sum of infinitely-many terms will be less than the sum of just the first two terms? :shock:

Eliz.

 

[monomoco]

Re: Converge or Diverge?

I was taking the limit of n -> infinity. Is this wrong?

 

[pka]

Re: Converge or Diverge?

\(\displaystyle \left( {\sqrt[n]{2}} \right) \to 1\)
Thus the series fails the first test so it diverges.

 

[monomoco]

Re: Converge or Diverge?

Thanks for the answers.
Since it's not 0 it diverges, right?

 

[stapel]

I was taking the limit of n -> infinity. Is this wrong?

The way you posted the exercise, this appears to be a summation, so you should be trying to find the limit of the SUM, not of each term of the sum.

Did you mean this actually to be a sequence, rather than a series...?

Eliz.