1. ## Prove by induction?

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

2. Originally Posted by tn11631
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
There is nothing to prove about that expression.
Please post the exact question as you have.

3. Originally Posted by Plato
There is nothing to prove about that expression.
Please post the exact question as you have.
Sorry I was told to prove by induction to see if the sequence converges or diverges and im still so confused..here is the original question:

does this sequence {sigma from n=1 to k (1/(k^2+n)^(1/2)) } k=1 to infinity ..Sorry about that..thanks

4. Originally Posted by tn11631
Sorry I was told to prove by induction to see if the sequence converges or diverges and im still so confused..here is the original question:
does this sequence {sigma from n=1 to k (1/(k^2+n)^(1/2)) } k=1 to infinity ..Sorry about that..thanks
Still not quite clear.
Is it $S_k = \sum\limits_{n = 1}^k {\frac{1}{{\sqrt {k^2 + n} }}}$?
So that we are asked if the $S_k$ sequence converges?
If so is the sequence increasing and bounded above by 1?

5. Originally Posted by Plato
Still not quite clear.
Is it $S_k = \sum\limits_{n = 1}^k {\frac{1}{{\sqrt {k^2 + n} }}}$?
So that we are asked if the $S_k$ sequence converges?
If so is the sequence increasing and bounded above by 1?
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.

6. Any monotone bounded sequence of real numbers converges.

7. Originally Posted by Plato
Any monotone bounded sequence of real numbers converges.
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?

8. 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 $1\le n\le k$ we have that $\frac{1}{\sqrt{k^{2}+k}}\le \frac{1}{\sqrt{k^{2}+n}}\le \frac{1}{\sqrt{k^{2}+1}},$ then $\frac{k}{\sqrt{k^{2}+k}}\le \sum\limits_{n=1}^{k}{\frac{1}{\sqrt{k^{2}+n}}}\le \frac{k}{\sqrt{k^{2}+1}},$ then as $k\to\infty$ we get $S_k\to1.$

9. Thanks a bunch! and thanks to every one who helped!!
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