Determining if the sequence convergers or diverges(IV)

[shamieh]

Determining if the sequence converges or diverges, if it converges find the limit

\(\displaystyle \sqrt{n^2 + n} - n\)

Wouldn't this just diverge if n--> infinity ?

I'm not sure what to do here? I can;t use lopitals...Also how would this converge to 1/2 is this a telescoping series?

 

[Prove It]

Determining if the sequence converges or diverges, if it converges find the limit

\(\displaystyle \sqrt{n^2 + n} - n\)

Wouldn't this just diverge if n--> infinity ?

I'm not sure what to do here? I can;t use lopitals...Also how would this converge to 1/2 is this a telescoping series?

No, $\displaystyle \begin{align*} \infty - \infty \end{align*}$ is an indeterminate form.

You would need to try to rationalise the numerator...

$\displaystyle \begin{align*} \sqrt{n^2 + n} - n &= \frac{\left( \sqrt{n^2 + n} - n \right) \left( \sqrt{n^2 + n} + n \right) }{\sqrt{n^2 + n} + n} \\ &= \frac{n^2 + n - n^2}{\sqrt{n^2 + n} + n} \\ &= \frac{n}{\sqrt{n^2 + n} + n} \\ &= \frac{n}{\sqrt{n^2 \left( 1 + \frac{1}{n} \right) } + n } \\ &= \frac{n}{n\sqrt{1 + \frac{1}{n}} + n} \\ &= \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \\ &\to \frac{1}{\sqrt{1 + 0} + 1} \textrm{ as } n \to \infty \\ &= \frac{1}{2} \end{align*}$

So the sequence converges to 1/2.

Also this is not a series, so again, don't try to see if this non-existent series is telescopic...

 

[shamieh]

Hey thank you so much, but on the 5th step I think I'm confused. Where did the n^2 go? did you pull it out infront of the square root because it is a constant multiplier and if so if you pulled it out n/n^2 would be 1/n not 1/1 right?? Then you would take lim as n -> infty and get 0?

 

[Prove It]

Hey thank you so much, but on the 5th step I think I'm confused. Where did the n^2 go? did you pull it out infront of the square root because it is a constant multiplier and if so if you pulled it out n/n^2 would be 1/n not 1/1 right?? Then you would take lim as n -> infty and get 0?

Surely you can see that $\displaystyle \begin{align*} \sqrt{n^2} = n \end{align*}$ (if n is nonnegative as it is since we are making it go to infinity)...

 

[shamieh]

Oh I see.