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