How would I prove this by induction...1/(k^2+n)^(1/2)...basically in words, 1 divided by the square root of k squared plus n
um, in my books its stated really weird but i think setting the whole thing equal to the sequence S_k takes away the original brackets so i think that is correct. And yes the sequence is increasing and it looks like its approching one so I'm going to say that it is bounded above by 1. I just don't know how to prove its converging and what its limit is.
Oh duh! so just by showing that it is increasing shows that it in monotone sequnce and that its approaching 1 therefore its bounded above by 1..Thanks I can't believe I missed that..however, since it converges, what would be its limit would it be 1? or is that just its least upper bound?
well the limit is indeed 1, but not because it does have a bound.
actually, this is solved by using the squeeze theorem, since for $\displaystyle 1\le n\le k$ we have that $\displaystyle \frac{1}{\sqrt{k^{2}+k}}\le \frac{1}{\sqrt{k^{2}+n}}\le \frac{1}{\sqrt{k^{2}+1}},$ then $\displaystyle \frac{k}{\sqrt{k^{2}+k}}\le \sum\limits_{n=1}^{k}{\frac{1}{\sqrt{k^{2}+n}}}\le \frac{k}{\sqrt{k^{2}+1}},$ then as $\displaystyle k\to\infty$ we get $\displaystyle S_k\to1.$