1. ## Help in "Limits"...

1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
I know we have to use the rationalizing the denominator but I keep getting this is divergent.

2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

3) when limit n -> infinity, find n! / 2^n.
I know we have to use the ratio test in here, but I am stuck at (n+1)/2.

2. Originally Posted by Vedicmaths

1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
I know we have to use the rationalizing the denominator but I keep getting this is divergent.

2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

3) when limit n -> infinity, find n! / 2^n.
I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
For the first one multiply by the conjugate to get

$\displaystyle \lim_{n\to\infty}\bigg[\sqrt{n^2+n}-\sqrt{n^2+1}\bigg]\cdot\frac{\sqrt{n^2+n}+\sqrt{n^2+1}}{\sqrt{n^2+n} +\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{n^2+n-n^2+1}{\sqrt{n^2+n}+\sqrt{n^2+1}}$

Can you go from there?
For the second one try the same trick

For the third one

use the fact taht

$\displaystyle n!\sim{\sqrt{2\pi{n}}n^ne^{-n}}$

to get

$\displaystyle \lim_{n\to\infty}\frac{n!}{2^n}=\lim_{n\to\infty}\ frac{\sqrt{2\pi{n}}n^ne^{-n}}{2^n}=\infty$

But that wouldnt be the Ratio test I dont think?

If you meant to find the convergence of that series, we just showed by the n-th term test that it is divergent, no need for ratio test

3. Originally Posted by Vedicmaths
3) when limit n -> infinity, find n! / 2^n.
I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
Instead look at $\displaystyle \frac{2^n}{n!}$.
$\displaystyle \frac{2^{n+1}}{(n+1)!}\cdot \frac{n!}{2^n} = \frac{2}{n+1} \to 0$.
Thus, $\displaystyle \frac{2^n}{n!}\to 0 \implies \frac{n!}{2^n}\to \infty$

4. Originally Posted by Vedicmaths

1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
I know we have to use the rationalizing the denominator but I keep getting this is divergent.

2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

3) when limit n -> infinity, find n! / 2^n.
I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
3) The ratio test requires you to consider $\displaystyle \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|$.

$\displaystyle a_n = \frac{n!}{2^n}$

$\displaystyle a_{n+1} = \frac{(n+1)!}{2^{n+1}} = \frac{(n+1)n!}{(2) \, 2^n}$

Therefore $\displaystyle \left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{2}$.

Therefore $\displaystyle \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| = \lim_{n \rightarrow \infty} \frac{n+1}{2} > 1$.

Therefore .......

5. Dear Mathstud 28!

Thanks for the prompt response. Sir, for the second one, I myself got until (n-1) / [ sqrt(n^2 + n) + sqrt(n^2 + 1)] but then after that I am stuck and this is the same case with the second one too.
Do we have to split that in two forms like (a-b)/c is a/c - b/c and then try solving it??

Thanks again!

6. Originally Posted by Vedicmaths

1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
I know we have to use the rationalizing the denominator but I keep getting this is divergent.

2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

3) when limit n -> infinity, find n! / 2^n.
I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
Originally Posted by Vedicmaths
Dear Mathstud 28!

Thanks for the prompt response. Sir, for the second one, I myself got until (n-1) / [ sqrt(n^2 + n) + sqrt(n^2 + 1)] but then after that I am stuck and this is the same case with the second one too.
Do we have to split that in two forms like (a-b)/c is a/c - b/c and then try solving it??

Thanks again!
Consider the fact that as $\displaystyle n\to\infty$ $\displaystyle \sqrt{n^2+n}+\sqrt{n^2+1}\to{2n}$

7. Originally Posted by Vedicmaths
[snip]
2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]
[snip]
$\displaystyle \frac{\sqrt{n} \, (\sqrt{n+1} - \sqrt{n})}{1} \times \frac{(\sqrt{n+1} + \sqrt{n})}{(\sqrt{n+1} + \sqrt{n})} = \frac{\sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} ......$

8. Originally Posted by Vedicmaths
Dear Mathstud 28!

Thanks for the prompt response. Sir, for the second one, I myself got until (n-1) / [ sqrt(n^2 + n) + sqrt(n^2 + 1)] but then after that I am stuck and this is the same case with the second one too.
Do we have to split that in two forms like (a-b)/c is a/c - b/c and then try solving it??

Thanks again!
$\displaystyle \lim_{n\to\infty}\bigg[\sqrt{n^2+n}-\sqrt{n^2+1}\bigg]\cdot\frac{\sqrt{n^2+n}+\sqrt{n^2+1}}{\sqrt{n^2+n} +\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{n^2+n-n^2+1}{\sqrt{n^2+n}+\sqrt{n^2+1}}$

$\displaystyle = \lim_{n\to\infty}\frac{n+1}{\sqrt{n^2+n}+\sqrt{n^2 +1}} = \lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1 + \frac{1}{n}} +\sqrt{1 + \frac{1}{n^2}}}$

9. Thanks a lot guys!

That helps a lot!