Discover the Convergent Sum of Series: 2^(1/n) Using Partial Sums

[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

 

[Infinitum]

I find this strange. Wolfram-alpha yields it is divergent...

http://www.wolframalpha.com/input/?i=sum+2^1/n

Is there a proof as to how it is convergent?

 

[micromass]

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

 

[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)+...

 

[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.

 

[arl146]

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

 

[Ray Vickson]

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

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

 

[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

 

[Ray Vickson]

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

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

 

[arl146]

what kind of additional work? also, this is the first section on series, meaning we haven't been taught any of the convergence tests...

 

[Mark44]

what kind of additional work?

Using the tests that you'll be soon seeing.

also, this is the first section on series, meaning we haven't been taught any of the convergence tests...

 

[Mark44]

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.

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.

 

[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

 

[Mark44]

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.

 

[Whovian]

The Ratio Test and the Integral Test are always nice.

 

[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