# Thread: Does the sequence converge?

1. ## Does the sequence converge?

Does the sequence {[sigma] from n=1 to k ( 1/(k^2+n)^(1/2) )} k=1 to infinity.

I'm not sure how to do the symbols yet but its saying sigma from n=1 to k of 1 divided by the square root of k squared plus n and the whole thing goes from k=1 to infinity. this question stumped the math tutor so i hope someone can figure it out.

oh, and this was a hint given by the prof: C_k= [sigma] from n=1 to k ( 1/(k^2+n)^(1/2) ).............(k^2+1)^(1/2) <= (k^2+n)^(1/2) <= (k^2+k)^(1/2)......a_k<=c_k<=b_k

thanks!

2. Originally Posted by tn11631

Does the sequence {[sigma] from n=1 to k ( 1/(k^2+n)^(1/2) )} k=1 to infinity.

I'm not sure how to do the symbols yet but its saying sigma from n=1 to k of 1 divided by the square root of k squared plus n and the whole thing goes from k=1 to infinity. this question stumped the math tutor so i hope someone can figure it out.

oh, and this was a hint given by the prof: C_k= [sigma] from n=1 to k ( 1/(k^2+n)^(1/2) ).............(k^2+1)^(1/2) <= (k^2+n)^(1/2) <= (k^2+k)^(1/2)......a_k<=c_k<=b_k

thanks!
I have no idea what this says. Maybe, $\lim_{k\to\infty}\sum_{n=1}^{k}\frac{1}{\sqrt{k^2+ n}}$

3. Originally Posted by Drexel28
I have no idea what this says. Maybe, $\lim_{k\to\infty}\sum_{n=1}^{k}\frac{1}{\sqrt{k^2+ n}}$
Yes that is exactly what it means, only on my sheet there were brackets around the whole thing like this { } (sigma and all) and then on the right side (outside the brackets) its n=1 on the bottom and infinity on the top..more of like how a sequence would look.

4. Originally Posted by tn11631
Yes that is exactly what it means, only on my sheet there were brackets around the whole thing like this { } (sigma and all) and then on the right side (outside the brackets) its n=1 on the bottom and infinity on the top..more of like how a sequence would look.
Oh and the brackets were there in place of the lim stuff..hope that clears it up

5. Originally Posted by tn11631
Yes that is exactly what it means, only on my sheet there were brackets around the whole thing like this { } (sigma and all) and then on the right side (outside the brackets) its n=1 on the bottom and infinity on the top..more of like how a sequence would look.
It's probably a Riemann sum and you can find the actual value. But, since we only care about whether it converges, prove by induction that it's increasing and then note that $\sum_{n=1}^{k}\frac{1}{k^2+n}\leqslant\sum_{n=1}^k \frac{1}{k}=1$. It follows from the fact that every monotonically increasing sequence bounded sequence is convergent.

6. oh I forgot to read the bottom of the question (my bad) it says..."If it converges, find its limit" would you then have to go and find the actual sum?

7. Originally Posted by tn11631
oh I forgot to read the bottom of the question (my bad) it says..."If it converges, find its limit" would you then have to go and find the actual sum?
Take my earlier advice and look for a Riemann sum.

8. Yikes lol ok..Thanks a bunch