Testing for convergence

httpete

Im doing an example from Stewart's Calculus book and I cant see how to go about it.

Its the series, n=1 to infinity, 1/(n^1/2) - 1/((n+1)^1/2)

I got some partial sums from wolfram alpha and they converge so I know the series has a sum. And I know I can split the above series into two separate series, one minus the other. This is question 7 on section 11.2, page 694. This is before any tests like ratio or comparision tests are shown so I am looking for a way to do it without those methods.

TheBody

httpete wrote: »

Im doing an example from Stewart's Calculus book and I cant see how to go about it.

Its the series, n=1 to infinity, 1/(n^1/2) - 1/((n+1)^1/2)

I got some partial sums from wolfram alpha and they converge so I know the series has a sum. And I know I can split the above series into two separate series, one minus the other. This is question 7 on section 11.2, page 694. This is before any tests like ratio or comparision tests are shown so I am looking for a way to do it without those methods.

Are you sure it converges? Looks like the difference of two p-series with a power [latex]p=\frac{1}{2}[/latex] for both bits. That would imply that they diverge.

Michael Collins

This is a telescoping series, check out the wiki page for the solution method (of differences):

http://en.wikipedia.org/wiki/Telescoping_series

httpete

Michael Collins wrote: »

This is a telescoping series, check out the wiki page for the solution method (of differences):

http://en.wikipedia.org/wiki/Telescoping_series

OK, I approached it as a telescoping series and after cancelling the terms I was left with
[latex]\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{n-1}}[/latex]

[latex]1 - \frac{1}{\sqrt{n-1}}[/latex]

Then I find the limit of that
[latex]\lim_{n \to \infty}(1 - \frac{1}{\sqrt{n-1}})[/latex]

[latex]1 - \lim_{n \to \infty}(\frac{1}{\sqrt{n-1}})[/latex]

As n approaches infinity, sqrt(n-1) also approaches infinity so left with

[latex]1 - 0[/latex]

= 1

Is that correct?

httpete

I am also having problems with some other ones from the same set of exercises in 11.2.

****** Question 9 -

Let [latex]\{a_n\} = \frac{2n}{3n+1}[/latex]

a. Determing whether [latex]$a_n[/latex] is convergent.
b. Determine whether [latex]\sum_{n=1}^\infty \frac{2n}{3n+1}[/latex] is convergent.


a. I just take the limit as n goes to infinity and that gives me 2/3 so the sequence is convergent with a limit of 2/3. Correct?

b. I basically do the same thing again, take the limit of 2n/(3n+1), which is equal to 2/3, and as this limit is not equal to zero the series is divergent. Correct?




****** Question 18 -
Determine whether the geometric series is divergent or convergent
[latex]\sum_{n=0}^\infty \frac{1}{(\sqrt{2})^n}[/latex]

Im not really sure what to do here as n=0 so I can't just apply the geometric series formula [latex]s_n = \frac{a}{1-r}[/latex]

Can I specify the first term explicity and then say that
[latex]\sum_{n=0}^\infty \frac{1}{(\sqrt{2})^n} = 1 + \sum_{n=1}^\infty \frac{1}{(\sqrt{2})^n}[/latex]

Then I have to get the 'n' into the form of 'n-1' so I can use the s = a/(1-r) formula
[latex]\sum_{n=0}^\infty \frac{1}{(\sqrt{2})^n} = 1 + \sum_{n=1}^\infty \frac{1}{(\sqrt{2})(\sqrt{2})^(n-1)}[/latex]

Am I on the right track here or am I getting it wrong?

Michael Collins

httpete wrote: »

OK, I approached it as a telescoping series and after cancelling the terms I was left with
[latex]\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{n-1}}[/latex]

[latex]1 - \frac{1}{\sqrt{n-1}}[/latex]

Then I find the limit of that
[latex]\lim_{n \to \infty}(1 - \frac{1}{\sqrt{n-1}})[/latex]

[latex]1 - \lim_{n \to \infty}(\frac{1}{\sqrt{n-1}})[/latex]

As n approaches infinity, sqrt(n-1) also approaches infinity so left with

[latex]1 - 0[/latex]

= 1

Is that correct?

Yes, this is the correct answer. It should be [latex]\frac{1}{\sqrt{n+1}}[/latex] above though.

httpete wrote: »

I am also having problems with some other ones from the same set of exercises in 11.2.

****** Question 9 -

Let [latex]\{a_n\} = \frac{2n}{3n+1}[/latex]

a. Determing whether [latex]$a_n[/latex] is convergent.
b. Determine whether [latex]\sum_{n=1}^\infty \frac{2n}{3n+1}[/latex] is convergent.


a. I just take the limit as n goes to infinity and that gives me 2/3 so the sequence is convergent with a limit of 2/3. Correct?

No, it is divergent. You get this right in the next question below! If the limit of the general term doesn't go to zero as the index tends to infinity, then the series definately diverges.

Yes, you were quite right here, I was thinking you were working with a series, as MathsManiac says the sequence is convergent. Sorry about that, I hope I didn't confuse you. You seem to have a good understanding anyway.

httpete wrote: »

b. I basically do the same thing again, take the limit of 2n/(3n+1), which is equal to 2/3, and as this limit is not equal to zero the series is divergent. Correct?

Yes, as far as I can see this is exactly the same question, just asked very slightly differently. So the answer is the same as per the last question just above.

httpete wrote: »

****** Question 18 -
Determine whether the geometric series is divergent or convergent
[latex]\sum_{n=0}^\infty \frac{1}{(\sqrt{2})^n}[/latex]

Im not really sure what to do here as n=0 so I can't just apply the geometric series formula [latex]s_n = \frac{a}{1-r}[/latex]

Can I specify the first term explicity and then say that
[latex]\sum_{n=0}^\infty \frac{1}{(\sqrt{2})^n} = 1 + \sum_{n=1}^\infty \frac{1}{(\sqrt{2})^n}[/latex]

Then I have to get the 'n' into the form of 'n-1' so I can use the s = a/(1-r) formula
[latex]\sum_{n=0}^\infty \frac{1}{(\sqrt{2})^n} = 1 + \sum_{n=1}^\infty \frac{1}{(\sqrt{2})(\sqrt{2})^(n-1)}[/latex]

Am I on the right track here or am I getting it wrong?

You can't just jump straight in and apply the sum to infinity. That formula already assumes the series is convergent.

Take the geometric series with first term a = 1 and common ratio r = 2:

1 + 2 + 4 +8 + ...

This obviously diverges. Yet

[latex] \displaystyle S_{\infty} = \frac{a}{1-r} = \frac{1}{1-2} = -1 [/latex]

which is clearly not correct.

For geometric series you should first check the limit of the general term as the index goes infinity - this limit should go to zero, as you know. If the series passes that, then next step should probably be the Ratio Test.

http://en.wikipedia.org/wiki/Ratio_test

TheBody wrote: »

Are you sure it converges? Looks like the difference of two p-series with a power [latex]p=\frac{1}{2}[/latex] for both bits. That would imply that they diverge.

I think the minus makes a big difference though. It's the terms of this series that are made up of one p-series minus the same series one index advanced, and this actually helps the convergence. I don't think it is quite the same as two p-series combined. In that case (p = 0.5), even if one was subtracted from the other, you wouldn't get absolutely convergence, only perhaps conditional.

MathsManiac

Michael Collins wrote: »

No, it is divergent. You get this right in the next question below! If the limit of the general term doesn't go to zero as the index tends to infinity, then the series definately diverges.



Yes, as far as I can see this is exactly the same question, just asked very slightly differently. So the answer is the same as per the last question just above.

I think actually that the OP is correct with this one and that you have misunderstood the question. Part (a) is about whether the sequence is convergent, (which it is,) and part (b) is about whether the series is convergent (which it's not).

MathsManiac

For the geometric series, you can use the formula for the sum of a finite geometric series first, (for which convergence is not relevant).

Sn = a(1 - r^n)/(1 - r)

This gives you an expression for the nth partial sum.

Then take the limit of that sequence of partial sums as n goes to infinity. If the limit exists, then the series is convergent and you have found its sum.

Michael Collins

MathsManiac wrote: »

I think actually that the OP is correct with this one and that you have misunderstood the question. Part (a) is about whether the sequence is convergent, (which it is,) and part (b) is about whether the series is convergent (which it's not).

Ah yes, you are quite right. I figured something weird was happening there, I forgot that notation. Thanks.

CJC86

httpete wrote: »

****** Question 18 -
Determine whether the geometric series is divergent or convergent
[latex]\sum_{n=0}^\infty \frac{1}{(\sqrt{2})^n}[/latex]

The question here is just to determine whether the series is divergent or convergent, not calculate the limit, so just use the ratio test.

If you want to calculate the limit, you can just use the geometric series formula you were using as [latex]| {\frac{1}{\sqrt{2}} | < 1[/latex], with [latex]a=1[/latex] and [latex]r = \frac{1}{\sqrt{2}}[/latex]

MathsManiac

CJC86 wrote: »

The question here is just to determine whether the series is divergent or convergent, not calculate the limit, so just use the ratio test.

If you want to calculate the limit, you can just use the geometric series formula you were using as [latex]| {\frac{1}{\sqrt{2}} | < 1[/latex], with [latex]a=1[/latex] and [latex]r = \frac{1}{\sqrt{2}}[/latex]

But the OP said that the ratio and comparison were off limits (so to speak!)

So I would assume the intention of the question is to answer the question from first principles: by definition, an infinite series is convergent iff the sequence of partial sums has a limit.