Finding $\lim\limits_{n \to \infty} \frac{\ln(n+1)}{\ln(n)} \frac{\ln(n!)}{\ln((n+1)!)}$ I am trying to find the limit as $n \to \infty$ of the function below:
$$f(n) = \frac{\ln(n+1)}{\ln(n)} \frac{\ln(n!)}{\ln((n+1)!)}$$
The textbook only gives me an answer but I don't know how it got to it.
I got confused with the factorials within logs.
Edit: I understand that this could be expanded to:
$$f(n) = \frac{\ln(n+1)}{\ln(n)} \frac{\ln(n) + \ln(n-1) + ...}{\ln(n+1) + \ln(n) + \ln(n-1) + ...}$$
I'm confused which terms get reduced to zero as $n \to \infty$ or how to group them.
 A: We have that
\begin{align*}
f(n)
&=\frac{\ln(n+1)}{\ln n} \frac{\ln(n!)}{\ln((n+1)!)}\\
&=\frac{\ln n+\ln(1+1/n)}{\ln n} \frac{\ln((n+1)!)-\ln(n+1)}{\ln((n+1)!)}\\
&=\biggl(1+\frac{\ln(1+1/n)}{\ln n}\biggr)\biggl(1-\frac{\ln(n+1)}{\ln((n+1)!)}\biggr).
\end{align*}
By Stirling's approximation,
$$
\ln n!=n\ln n-n+O(\ln n).
$$
Hence, we see that $\ln n!\sim n\ln n$, where $\sim$ means that the ratio of the two sides tends to $1$ as $n\to\infty$. We obtain
\begin{align*}
\frac{\ln(n+1)}{\ln((n+1)!)}
&=\frac{(n+1)\ln(n+1)}{\ln((n+1)!)}\cdot\frac{\ln(n+1)}{(n+1)\ln(n+1)}\\
&=\frac{(n+1)\ln(n+1)}{\ln((n+1)!)}\cdot\frac1{(n+1)}\to0
\end{align*}
as $n\to\infty$ since the first term converges to $1$ as $n\to\infty$. So the sequence $f(n)$ converges to $1$ as $n\to\infty$.
A: I'm not sure if this is what your textbook is going for, but I would use Stirling's approximation, which states that
$$\ln(n!) = n \ln n - n + O(\ln n)$$
check here for reference as to what that notation means. With this, we can evaluate
$$\frac{\ln(n!)}{\ln(n)} = n \frac{\ln(n) - 1}{\ln(n)} \approx n$$
$$\frac{\ln(n + 1)}{\ln((n + 1)!)} = \ln(n + 1) \frac{1}{(n + 1)[\ln(n + 1) - 1]} \approx \frac{1}{n + 1}$$
$$\implies \lim_{n \to \infty} f(n) \approx \lim_{n \to \infty} \frac{n}{n + 1} = 1$$
A: $$\frac{\ln(n!)}{\ln((n+1)!)}=\frac{\ln(n!)}{\ln(n!)+\ln(n+1)}.$$
As the factorial has $n/2$ factors larger than $n/2$, $\ln(n!)>n/2\ln(n/2)$ and the above ratio tends to$1$.
A: The limit of the first factor is clearly $1$ because $$\log(n + 1) = \log n + \log(1 + 1/n)$$ and hence $$\frac{\log(n + 1)}{\log n} = 1 + \frac{\log(1 + 1/n)}{\log n} \to 1 + 0 = 1$$ as $n \to \infty$.
The second factor is $a_{n}/b_{n}$ where $a_{n} = \log n!$ and $b_{n} = \log (n + 1)!$. We can see that $b_{n} \to \infty$ monotonically and hence by Cesaro-Stolz Theorem if $(a_{n + 1} - a_{n})/(b_{n + 1} - b_{n}) \to L$ then so does $a_{n}/b_{n}$.
Now $$\frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} = \frac{\log(n + 1)}{\log (n + 2)} \to 1$$ as shown for the first factor above. Hence the second factor $a_{n}/b_{n} \to 1$. And therefore the whole product tends to $1$ as $n \to \infty$.
