# Sum of a series

1. May 9, 2012

### arl146

Find the sum of the series:

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

I already know that this is convergent. I don't see this as a geometric series so that means I have to use partial sums. And I know that the sum is equal to the limit of the sequence of partial sums. I just don't know how to don't out, there's not good enough examples in the book that go along with this

2. May 9, 2012

3. May 9, 2012

### micromass

Staff Emeritus
It is divergent as the term sequence converges to 1, and not 0.

4. May 9, 2012

### arl146

Oh yea I lied it is divergent. But I don't know why it is divergent either haha. How does it converge to 1? The terms are 2 + sqrt(2) + 2^(1/3) + 2^(1/4)+...

5. May 9, 2012

### Infinitum

Just a guess, here. I'm still trying to learn these type of series.

Since the term at n = infinity converges to 1, the sum would continue for ever, 1 being added at every successive ∞ + 1, ∞ +2, so on. Therefore you would never get a exact limiting sum, which makes it divergent.

I believe this is correct, though it would be great if someone confirmed it.

6. May 9, 2012

### arl146

Definitely makes sense, I didn't think of it that way!

7. May 9, 2012

### Ray Vickson

The series is divergent, and you do not need powerful software to tell you that. The nth term a_n= 2^(1/n) does not approach 0 as n approaches infinity.

RGV

8. May 9, 2012

### arl146

So when I look for convergence or divergence of a series, if it doesn't approach 0, it's automatically divergent? And then all other numbers are convergent? What are like the 'rules' to it

9. May 9, 2012

### Ray Vickson

If the nth term does not --> 0 the series is divergent; if the nth term does --> 0 the series may converge or it may diverge: additional work is needed to tell which occurs.

RGV

10. May 10, 2012

### arl146

what kind of additional work? also, this is the first section on series, meaning we havent been taught any of the convergence tests...

11. May 11, 2012

### Staff: Mentor

Using the tests that you'll be soon seeing.

12. May 11, 2012

### Staff: Mentor

What Ray is talking about here is something you'll see very soon - the n-th term test for divergence. It is pretty important, but deceptively simple, saying that if the limit of the n-th term is not 0, then the series it's part of diverges.

It doesn't say anything at all about series whose n-th term has a limit of zero.

13. May 11, 2012

### arl146

i feel lost with the book, it doesn't show how to determine if a series (without it being geometric) converges or diverges but then for every question they ask if the series converges or diverges. its confusing

14. May 11, 2012

### Staff: Mentor

I think you are missing something that is in your book. If they expect you to determine convergence/divergence, they must have given you some tools to work with. Take a closer look at any examples and theorems they present in that section.

15. May 11, 2012

### Whovian

The Ratio Test and the Integral Test are always nice.

16. May 11, 2012

### arl146

we didnt learn the ratio test or integral test yet. but from this post and another post of mine, i think i may have it