limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)

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- May 25th 2008, 07:28 PMszpengchaohow to prove this series
limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)

- May 25th 2008, 08:23 PMmr fantastic
- May 25th 2008, 08:31 PMReckoner
What are you trying to prove? I see no propositions in your post. Are you trying to show divergence? We have

This is a*p*-series, with , so you should know that it diverges. If you want to prove it without relying on the properties of*p*-series, one way is to use the Integral Test. Show that diverges (and note that is positive, decreasing, and continuous on ). - May 25th 2008, 09:03 PMszpengchaowrong
sorry, the question is wrong, my question is:

limit( sigma( k^(-1/2),k, 1, n)/sqrt(n),n goes to infinity)

i think this limit is 2

but cant find a way to prove it - May 25th 2008, 10:49 PMszpengchaowat
- May 26th 2008, 01:28 AMCaptainBlack
- May 26th 2008, 01:43 AMszpengchaowrong
mistake again, there is a sqrt of n there...

this is wat i got when finding integral of 1/sqrt(x) - May 26th 2008, 01:52 AMCaptainBlack
- May 26th 2008, 02:01 AMszpengchaoyes
yes...

- May 26th 2008, 04:10 AMflyingsquirrel
- May 26th 2008, 05:17 AMKrizalid

For each it's a Riemann sum for and the conclusion follows. - May 26th 2008, 06:10 AMszpengchaofollow
i can follow the last step...but how can we get that =2 then?

- May 26th 2008, 06:16 AMszpengchaoproof
i hav proved that 1/sqrt (x) is integrable. but i m trying to show that integral =2sqrt(a) , so i need to show riemanss sum tends to 2sqrt(a) then..that's what i m troubling with

- May 26th 2008, 11:15 AMKrizalid
Can't you do ?