S shakalandro New member Joined Nov 29, 2008 Messages 36 Apr 7, 2009 #1 If Sn = 1/sqrt(n^2 + 1) + 1/sqrt(n^2 + 2) + ... + 1/sqrt(n^2 + n), show that (1 + 1/n)^-.5 < Sn < 1. I cannot find a way to show this and I need help.
If Sn = 1/sqrt(n^2 + 1) + 1/sqrt(n^2 + 2) + ... + 1/sqrt(n^2 + n), show that (1 + 1/n)^-.5 < Sn < 1. I cannot find a way to show this and I need help.
D DrMike Full Member Joined Mar 31, 2009 Messages 251 Apr 7, 2009 #2 shakalandro said: If Sn = 1/sqrt(n^2 + 1) + 1/sqrt(n^2 + 2) + ... + 1/sqrt(n^2 + n), show that (1 + 1/n)^-.5 < Sn < 1. I cannot find a way to show this and I need help. Click to expand... What happens if you replaces all the terms in Sn by the biggest term? You'll get something bigger than Sn, but what? And what happens if you replace all the terms by the smallest term? Now you have some bounds on Sn. What are they? Do they imply (1 + 1/n)^-.5 < Sn < 1 ? I haven't tried doing this, but it is the first thing I would try.
shakalandro said: If Sn = 1/sqrt(n^2 + 1) + 1/sqrt(n^2 + 2) + ... + 1/sqrt(n^2 + n), show that (1 + 1/n)^-.5 < Sn < 1. I cannot find a way to show this and I need help. Click to expand... What happens if you replaces all the terms in Sn by the biggest term? You'll get something bigger than Sn, but what? And what happens if you replace all the terms by the smallest term? Now you have some bounds on Sn. What are they? Do they imply (1 + 1/n)^-.5 < Sn < 1 ? I haven't tried doing this, but it is the first thing I would try.