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I've been trying for several days now to find the limit of $\dfrac{\log(n!)}{\log(n+1)}$ when $n$ tends to $\infty$.

I have no idea how to continue... I've tried to ask in other forums but I always get answers that only hint to the solution. I would really appreciate to get an explicit solution if possible since I'm stuck...

(The main goal is to find the radius of convergence of the complex power series where $a_{n}=\log(n!)$)

Thank you!

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    hint: $\frac{\log(n!)}{\log(n+1)}=\frac{\log(n)}{\log(n+1)}+\frac{\log(n-1)}{ \log(n+1)}+....=$ – 2017-01-18
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    By Stirling's approximation, $\log(n!)\approx n\log n-n$. Case closed. – 2017-01-18
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    Or, by Cesaro-Stoltz, $$\lim_{n\to +\infty}\frac{\log(n!)}{\log(n+1)}=\lim_{n\to +\infty}\frac{\log n}{\log\left(1+\frac{1}{n}\right)} = +\infty.$$ – 2017-01-18
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    If Stirlings approximation is too high tech then you approximate $\log(n!)$ by the integral $\int_1^n \log(x){\rm d}x$ to get the same result. – 2017-01-18
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    And the radius of convergence of $$\sum_{n\geq 1}\log(n!) x^n$$ is just one, by bounding $\log(n!)$ between $n$ and $n^2$, for instance. – 2017-01-18

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[Just to offer a more elementary answer than Jack D'Auricio's comments:]

Let $k$ be a fixed natural number. The estimation $$\frac{\log n!}{\log(n+1)}\geq\sum_{j=1}^k\frac{\log(n+1-j)}{\log(n+1)},$$ for $n\geq k$, gives us that $$\lim_{n\to\infty}\frac{\log n!}{\log(n+1)}\geq k.$$ As $k$ was arbitrary it follows that $$\lim_{n\to\infty}\frac{\log n!}{\log(n+1)}=+\infty. $$

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    Hm, not as rigorous as I'd like. Perhaps let $k$ be a function of $n$. – 2017-01-18
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    @SimpleArt I don't see any lack of rigor. I've edited the answer to make it clearer. – 2017-01-18
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    Ok, whatever then. But I don't think you should say that $k$ is arbitrary. If you could have it depend on $n$, it'd work out quite nicely. – 2017-01-18
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    @SimpleArt I don't see the depending-on-$n$ alternative you're talking about. What I am showing is that the limit is bigger than any natural number so it must be $\infty$. The reasoning is quite like when you prove that a certain nonnegative quantity is zero by showing that it is smaller than every $\varepsilon>0$. – 2017-01-18
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    Yes, it is :P Just ignore me. – 2017-01-18
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With some log properties and Riemann sums:

$$\log(n!)=\sum_{k=1}^n\log(k)>\int_1^n\log(x)\ dx=n\log(n)-n+1$$

Thus,

$$\frac{\log(n!)}{\log(n+1)}>\frac{n\log(n)-n+1}{\log(n+1)}$$

whereupon we find divergence as $n\to\infty$ quite easily with L'Hospital's rule, some basic expansion, etc.

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For $k\ge2$ we have the following:

$$1-{1\over k}\le{1\over2}+{1\over3}+\cdots+{1\over k}\le\int_1^k{dx\over x}=\ln k$$

(The first inequality is easily checked for $k=2$ and $3$, and holds for $k\ge4$ since ${1\over2}+{1\over3}+{1\over4}\gt1$.) Thus

$$\begin{align} \ln(n!) &=\ln2+\ln3+\cdots+\ln n\\ &\ge\left(1-{1\over2}\right)+\left(1-{1\over3}\right)+\cdots+\left(1-{1\over n}\right)\\ &=n-1-\left({1\over2}+{1\over3}+\cdots+{1\over n}\right)\\ &\ge n-1-\ln n \end{align}$$

so that

$${\ln(n!)\over\ln(n+1)}\ge{n-1\over\ln(n+1)}-1\to\infty\quad\text{as }n\to\infty$$

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Since $n! > (n/e)^n$ (proved by induction from $(1+1/n)^n < e < (1+1/n)^{n+1}$), $\log n! > n(\log n - 1)$ so, an even stronger result holds: $\frac{\log(n!)}{n} \gt \log n - 1 \to \infty $.