Limit of product of sequences

**tarheelborn** · June 8th 2010, 12:31 PM

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.

**Ackbeet** · June 8th 2010, 12:37 PM

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.

**Plato** · June 8th 2010, 12:38 PM

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.

$\frac{{\sqrt n }}{{\sqrt {n + 1} + \sqrt n }} = \frac{1}<br /> {{\sqrt {1 + \frac{1}{n}} + 1}}$

**tarheelborn** · June 8th 2010, 12:48 PM

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)?

**mohammadfawaz** · June 8th 2010, 02:02 PM

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

**tarheelborn** · June 9th 2010, 05:10 AM

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

**mohammadfawaz** · June 9th 2010, 05:34 AM

In that case, go back to Plato's formula: and try to find N such that for a given $\epsilon$ and if $n>N$ , then, $|\frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2}|<\epsilon$
i.e. $\frac{\sqrt{1+\frac{1}{n}}-1}{2\sqrt{1+\frac{1}{n}}+2}<\epsilon$ (note that we remove the absolute value signs - this quantity is positive!).
Now, this means: $\frac{\sqrt{1+\frac{1}{n}}-1}{2\sqrt{1+\frac{1}{n}}+2}<\frac{\sqrt{1+\frac{1} {n}}-1}{4}<\epsilon$ for large $n$ (check this!)
Therefore: $\sqrt{1+\frac{1}{n}}-1<4\epsilon$
Hence: $\sqrt{1+\frac{1}{n}}<1+4\epsilon$
$1+\frac{1}{n}<1+16\epsilon^2+8\epsilon$
$n>\frac{1}{16\epsilon^2+8\epsilon}$ 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.

**tarheelborn** · June 9th 2010, 06:30 AM

Thank you so much!

**mohammadfawaz** · June 9th 2010, 06:42 AM

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

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