$\lim_{n \to \infty}\frac{((n+1)!)^{\frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}$ help Could someone please help me solve $$\lim_{n \to \infty}\frac{((n+1)!)^{\frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}.$$
I am familiar with the ratio and $n$th square root rule, and have tried doing this problem with it, but without success. 
 A: I'll be using the well known fact that $n^{1/n}\to 1.$ The expression equals
$$(n+1)^{1/(n+1)}\frac{(n!)^{1/(n+1)}}{(n!)^{1/n}}.$$
Since $(n+1)^{1/(n+1)}\to 1,$ we can forget about that term. We are then looking at
$$\frac{1}{(n!)^{1/n-1/(n+1)}} = \frac{1}{(n!)^{1/[n(n+1)]}}.$$
But note $1\le (n!)^{1/[n(n+1)]}\le (n!)^{1/n^2}\le (n^n)^{1/n^2} = n^{1/n} \to 1.$ Thus $(n!)^{1/n(n+1)}\to 1,$ and hence the original limit is $1.$
A: Just adding to the answer pool.  Let $a_n = \frac{n^n}{n!}$.  Like so, you can show $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = e$.  Then make use of the fact that the root test is stronger than the ratio test to yield $\lim_{n\to\infty} \sqrt[n]{a_n} = e$.  With that, we have
$$
\lim_{n\to\infty} \frac{(n+1)!^{\frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}
= \lim_{n\to\infty} \frac{n+1}{n}\cdot \lim_{n\to\infty}\frac{n}{n+1}\cdot \frac{(n+1)!^{\frac{1}{n+1}}}{n!^{\frac{1}{n}}}
= 1 ~.
$$
A: Hint
$$\ln \left(\frac{((n+1)!)^{\frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}\right)=\frac{1}{n+1}\ln((n+1)!)-\frac{1}{n}\ln(n!)\\=\frac{n \ln((n+1)!)-n \ln(n!)-\ln(n!)}{n(n+1)}\\
=\frac{n \ln(n+1)-\ln(n!)}{n(n+1)}$$
Now,
$$\lim_n \frac{n \ln(n+1)}{n(n+1)}=\frac{\ln(n+1)}{n+1}=0$$
And 
$$\ln(n!) \leq \ln(n^n) =n \ln(n)$$
and hence 
$$0 \leq \frac{\ln(n!)}{n(n+1)} \leq \frac{\ln(n)}{n+1}$$
A: $ l= \lim_{n\to\infty}  \frac{\sqrt[n+1] {(n+1)!}}{\sqrt[n] n!}$ =
$ \lim_{n\to\infty}   \frac{\sqrt[n+1] {(n+1)!}}{\sqrt[n+1] n!}\cdot   \frac{\sqrt[n+1] n!}{\sqrt[n] n!} $
$\lim_{n\to\infty} \sqrt[n+1] \frac{(n+1)!}{n!} \cdot (n!)^{\frac{-1}{(n+1)n}} $
So, $l=   \lim_{n\to\infty} \sqrt[n+1] {(n+1)} \cdot (n!)^{\frac{-1}{(n+1)n}} $
Observation :$ \lim_{n\to\infty} \sqrt[n+1] {(n+1)} = 1$
Continuing :  $l=\lim_{n\to\infty} (n!)^{\frac{-1}{(n+1)n}}  $
=  $\frac{1}{\lim_{n\to\infty} (n!)^{\frac{1}{(n+1)n}} }$
Let $L = \lim_{n\to\infty} (n!)^{\frac{1}{(n+1)n}}$ . Now we take the natural logarithm on both sides $\rightarrow$ $ln(L) = \lim_{n\to\infty}  \frac{ ln(1)+ln(2)+\cdots+ln(n)}{n^2+n}$
Observation : $ b_n= n^2+n $ is a sequence which is increasing  and is  unbounded . 
Now, we apply the Stolz-Cesaro Theorem : $b_n>0 , b_n<b_{n+1}$ and $b_n$ unbounded so that  $\lim_{n\to\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$
So, we have : $   \lim_{n\to\infty}  \frac{ln(n+1)}{2(n+1)} $
Note that  : Since $\ln(n+1)$ grows asymptotically slower than the polynomial $n+1$ as n approaches $\infty$ the limit will be equal to 0.
So, $   \lim_{n\to\infty}  \frac{ln(n+1)}{2(n+1)} =0$
We conclude that : $ ln(L)=0 \rightarrow L= 1$ and $l=\frac{1}{L} \rightarrow l= 1$
So,  $ l= \lim_{n\to\infty}  \frac{\sqrt[n+1] {(n+1)!}}{\sqrt[n] n!} =1 $
A: Looking for more tahn the limit itself
$$a_n=\frac{((n+1)!)^{\frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}\implies \log(a_n)=\log \left(\frac{((n+1)!)^{\frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}\right)=\frac{1}{n+1}\ln((n+1)!)-\frac{1}{n}\ln(n!)$$
Using twice Stirling approximation and continuing with Taylor series for large values of $n$, then 
$$\log(a_n)=\frac{1}{n}-\frac{\log (2 \pi  n)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=1+\frac{1}{n}-\frac{\log (2 \pi  n)-1}{2n^2}+O\left(\frac{1}{n^3}\right)$$
