1. ## Limits

Determine the following limits if they exists. If the limit exists, establish the convergence.

$(a) \lim _{n\rightarrow\infty} (\sqrt{(n+a)(n+b)} - n$ where $a,b$ $> 0.$

$(b) \lim _{n\rightarrow\infty} (n!)^{\frac{1}{n^2}}$

2. Originally Posted by yusukered07
Determine the following limits if they exists. If the limit exists, establish the convergence.

$(a) \lim _{n\rightarrow\infty} (\sqrt{(n+a)(n+b)} - n$ where $a,b$ $> 0.$

$(b) \lim _{n\rightarrow\infty} (n!)^{\frac{1}{n^2}}$
Let $L=\lim_{n\to\infty}\left(n!\right)^{\frac{1}{n^2}} =\lim_{n\to\infty}\left(\left(n!\right)^{\frac{1}{ n}}\right)^{\frac{1}{n}}$, applying the connection between the root and ratio tests we see that this is equal to $L=\lim_{n\to\infty}\frac{((n+1)!)^{\frac{1}{n+1}}} {(n!)^{\frac{1}{n}}}=\lim_{n\to\infty}\frac{(n!)^{ \frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}\cdot (n+1)^{\frac{1}{n+1}}=\lim_{n\to\infty}(n!)^{\frac {-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{ 1}{n^2}}}=\frac{1}{L}$. Thus, $L=\pm 1$ but since it is clearly positive we may conclude that $L=1$.

3. Is...

$\sqrt{(n+a)\cdot (n+b)} - n = \frac{(n+a)\cdot (n+b)}{\sqrt{(n+a)\cdot (n+b)} + n} =$

$= \frac{n\cdot (a+b)+ a b }{\sqrt{(n+a)\cdot (n+b)} + n} = \frac{a + b + \frac{a b}{n}}{1+ \sqrt{1 + \frac{a+b}{n} + \frac{a b}{n^{2}}}}$ (1)

... so that is...

$\lim_{n \rightarrow \infty} \sqrt{(n+a)\cdot (n+b)} - n = \frac{a+b}{2}$ (2)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by Drexel28
Let $L=\lim_{n\to\infty}\left(n!\right)^{\frac{1}{n^2}} =\lim_{n\to\infty}\left(\left(n!\right)^{\frac{1}{ n}}\right)^{\frac{1}{n}}$, applying the connection between the root and ratio tests we see that this is equal to $L=\lim_{n\to\infty}\frac{((n+1)!)^{\frac{1}{n+1}}} {(n!)^{\frac{1}{n}}}=\lim_{n\to\infty}\frac{(n!)^{ \frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}\cdot (n+1)^{\frac{1}{n+1}}=\lim_{n\to\infty}(n!)^{\frac {-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{ 1}{n^2}}}=\frac{1}{L}$. Thus, $L=\pm 1$ but since it is clearly positive we may conclude that $L=1$.
I am not sure this is fully rigorous: for instance, don't you need to prove first that the limit exists? (and a few steps would need justifications, like $\lim_{n\to\infty}(n!)^{\frac{-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{ 1}{n^2}}}$ or the previous one)

I would rather say: readily, $1\leq n!=n(n-1)\cdots 1\leq n^n$, hence $1\leq (n!)^{\frac{1}{n^2}}=\exp(\frac{\log (n!)}{n^2})\leq \exp(\frac{\log n}{n})$. Since $\frac{\log n}{n}\to 0$ (and $\exp$ is continuous at 0), the limit indeed exists and equals 1.

5. Originally Posted by Laurent
I am not sure this is fully rigorous: for instance, don't you need to prove first that the limit exists? (and a few steps would need justifications, like $\lim_{n\to\infty}(n!)^{\frac{-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{ 1}{n^2}}}$ or the previous one)

I would rather say: readily, $1\leq n!=n(n-1)\cdots 1\leq n^n$, hence $1\leq (n!)^{\frac{1}{n^2}}=\exp(\frac{\log (n!)}{n^2})\leq \exp(\frac{\log n}{n})$. Since $\frac{\log n}{n}\to 0$ (and $\exp$ is continuous at 0), the limit indeed exists and equals 1.
I don't think that the point of the exercise was to be fully rigorous. Besides, I don't think I made any extraordinary leaps.

6. Originally Posted by Drexel28
I don't think that the point of the exercise was to be fully rigorous. Besides, I don't think I made any extraordinary leaps.
I was actually a bit euphemistic in my answer : assuming the existence of the limit in the first place and playing with it is really really not a good habit to take (no harm intended!). Especially when, like here, we are asked "if the limit exists, establish the convergence"; but certainly not only in this case.
It is not because the only "possible" limit is 1 that the sequence converges to 1.
I'm just saying...

7. Originally Posted by Drexel28
I don't think that the point of the exercise was to be fully rigorous. Besides, I don't think I made any extraordinary leaps.
If you act this way when you are a mathematician, you'll be in some kind of trouble

8. Originally Posted by Moo
If you act this way when you are a mathematician, you'll be in some kind of trouble
Haha! I agree

That said...I struggle with complete rigor and not giving the OP the full answer. My response was meant to be a nudge for which they would fill in the details