# Thread: Limit of product of sequences

1. ## Limit of product of sequences

Evaluate lim sqrt(n)*[sqrt(n+1)-sqrt(n)]

I know that limit sqrt(n) = Infinity and that limit (sqrt(n+1)-sqrt(n)) = 0. And I know that sqrt(n)*(sqrt(n+1)-sqrt(n)) = sqrt(n)/(sqrt(n+1)+sqrt(n)).

I believe the limit of the product of these sequences is 1/2, but I am not sure how to get there. Thanks for your help.

2. This might be very simplistic, but could you use l'Hospital's Rule? It would certainly work for the real variable case, and would give you, as you intimated, 1/2.

3. Originally Posted by tarheelborn
Evaluate lim sqrt(n)*[sqrt(n+1)-sqrt(n)]
I know that sqrt(n)*(sqrt(n+1)-sqrt(n)) = sqrt(n)/(sqrt(n+1)+sqrt(n)).
I believe the limit of the product of these sequences is 1/2, but I am not sure how to get there. Thanks for your help.
$\displaystyle \frac{{\sqrt n }}{{\sqrt {n + 1} + \sqrt n }} = \frac{1} {{\sqrt {1 + \frac{1}{n}} + 1}}$

4. Thank you for the simplification. Now I need to find N and I am not sure how to combine sqrt[(1/n)+1]+sqrt(n). Or would I be able to ignore the +1 part of it and somehow combine sqrt(1/n) and sqrt(n)?

5. why do u need to find N? you're asked to find the limit. Just find it! u know $\displaystyle \lim_{n->\infty}\frac{1}{n}=0$ then you're done be applying this to Plato's note

6. I have to prove the limit, so I definitely do need to find N. Thanks.

7. In that case, go back to Plato's formula: and try to find N such that for a given $\displaystyle \epsilon$ and if $\displaystyle n>N$, then, $\displaystyle |\frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2}|<\epsilon$
i.e. $\displaystyle \frac{\sqrt{1+\frac{1}{n}}-1}{2\sqrt{1+\frac{1}{n}}+2}<\epsilon$ (note that we remove the absolute value signs - this quantity is positive!).
Now, this means: $\displaystyle \frac{\sqrt{1+\frac{1}{n}}-1}{2\sqrt{1+\frac{1}{n}}+2}<\frac{\sqrt{1+\frac{1} {n}}-1}{4}<\epsilon$ for large $\displaystyle n$ (check this!)
Therefore: $\displaystyle \sqrt{1+\frac{1}{n}}-1<4\epsilon$
Hence: $\displaystyle \sqrt{1+\frac{1}{n}}<1+4\epsilon$
$\displaystyle 1+\frac{1}{n}<1+16\epsilon^2+8\epsilon$
$\displaystyle n>\frac{1}{16\epsilon^2+8\epsilon}$ You found your N!
Note that you don't have to find the best value of N. Any value that works if fine to prove the limit.

8. Thank you so much!

9. Just one more thing to not: N is an integer! (obviously)...So, N is equal to the smallest integer larger than $\displaystyle \frac{1}{16\epsilon^2+8\epsilon}$.