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Is the following true or false for $N\geq 3$?

$\log N - \log e + c >\frac{\log N!}{N}> \log N - \log{e}$ for some $ c< \log e$ where the logarithm is w.r.t. base $10.$

Calculating some values suggest it should be true. But if it is true, is there any elementary way by which we can show this.

In fact if we take the log out, I am getting , $ 10^{c} \frac{N}{e} > (N!)^{\frac{1}{N}} > \frac{N}{e}$ but I can't see whether it is true for all $N\geq 3$ and for some $c< \log e$.

Any help would be appreciated. Thanks in advance.

Mathmo123
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math is fun
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2 Answers2

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You can use the fact that

$$e\left(\frac ne\right)^n \le n! \le en\left(\frac ne\right)^n$$

Which is equivalent to

$$e^{1/n} \frac ne \le n!^{1/n} \le (en)^{1/n} \frac ne$$

For the left part of the inequality; since $e^{1/n} > 1$, $$n!^{1/n} \ge e^{1/n} \frac ne > \frac ne$$

For the right part of the inequality, focus on the $(en)^{1/n}$. This function as a maximum at $n=1$, where it attains the value of $e$. After this point it is decreasing. Therefore, for $n > 3$, the function is lower than $e$, which means that you can find a $c$ close enough to $\log e$ such that $10^c >(en)^{1/n}$ for all $n > 3$

The following proof of the first inequality is lifted from the italian wikipedia page on the Stirling approximation.

Let $n$ be an integer. Then

$$\ln n! = \sum_{k = 1}^n \ln k = \sum_{k = 1}^n k\ln k - \sum_{k = 1}^n (k - 1)\ln k $$

$$= n \ln n - \sum_{k = 1}^{n - 1} k \left[\ln(k + 1) - \ln k \right] = n \ln n - \sum_{k = 1}^{n - 1} k \int_{k}^{k + 1} \frac{dt}{t}$$

$$= n \ln n - \sum_{k = 1}^{n - 1} \int_{k}^{k + 1} \frac{ \lfloor t \rfloor dt}{t} = n \ln n - \int_{1}^{n} \frac{ \lfloor t \rfloor dt}{t} = n \ln n - (n-1) + \int_{1}^{n} \frac{ \{ t \} dt}{t},$$

where $\lfloor x \rfloor$ and $\left \{ x \right \}$ are the integer part and the fractional part of $x$, respectively.

This implies $$n \ln n - (n-1) \leq \ln n! \leq n \ln n - (n-1) + \ln n$$

and after exponentiating you'll find $$ e \left(\frac{n}{e}\right)^n \leq n! \leq e n\left(\frac{n}{e}\right)^n $$

Gary
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Ant
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Following the way you started, using Riemann sums (here is another example) and considering that $\color{green}{f(x)=\ln{x}}$ is monotone increasing $$ 10^{c} \cdot\frac{N}{e} > (N!)^{\frac{1}{N}} > \frac{N}{e} \iff\\ c\ln{10}+\color{blue}{\ln{N}-1}> \color{red}{\frac{1}{N}\left(\sum\limits_{k=1}^N \ln{k}\right)}> \color{blue}{\ln{N}-1}$$

However (Riemann sums) $$\sum\limits_{k=1}^N \ln{k} = \sum\limits_{k=2}^N \ln{k} = \sum\limits_{k=2}^N \color{green}{f(k)}\cdot (k+1-k)\leq \int\limits_{2}^{N+1} \color{green}{f(x)}dx= x(\ln{x}-1)\Big |_{2}^{N+1}=\\ (N+1)(\ln{(N+1)}-1)-2(\ln{2}-1)$$ $$\sum\limits_{k=1}^N \ln{k} = \sum\limits_{k=2}^N \ln{k} = \sum\limits_{k=2}^N \color{green}{f(k)}\cdot (k-(k-1))\geq \int\limits_{1}^{N} \color{green}{f(x)}dx= x(\ln{x}-1)\Big |_{1}^{N}=\\ N(\ln{N}-1)+1$$

So $$\ln{N}-1+\frac{1}{N}\leq \color{red}{\frac{1}{N}\left(\sum\limits_{k=1}^N \ln{k}\right)} \leq\frac{N+1}{N}(\ln{(N+1)}-1)-\frac{2(\ln{2}-1)}{N}$$ One part becomes easy to see $$\color{blue}{\ln{N}-1}<\ln{N}-1+\frac{1}{N}\leq \color{red}{\frac{1}{N}\left(\sum\limits_{k=1}^N \ln{k}\right)}$$ The other $$\color{red}{\frac{1}{N}\left(\sum\limits_{k=1}^N \ln{k}\right)} \leq\frac{N+1}{N}(\ln{(N+1)}-1)-\frac{2(\ln{2}-1)}{N}=\\ \ln{(N+1)}-1+\frac{\ln{(N+1)}-1}{N}-\frac{2(\ln{2}-1)}{N}=\\ \color{blue}{\ln{N}-1}+\ln{\left(1+\frac{1}{N}\right)}+\frac{\ln{(N+1)}-1}{N}-\frac{2(\ln{2}-1)}{N}$$ What is left to show is $$\ln{\left(1+\frac{1}{N}\right)}+\frac{\ln{(N+1)}-1}{N}-\frac{2(\ln{2}-1)}{N}=\\ \frac{\ln{\left(1+\frac{1}{N}\right)^N}+\ln{(N+1)}-2\ln{2}+1}{N}<\\ \frac{1+\ln{(N+1)}-2\ln{2}+1}{N}< \frac{\ln{(N+1)}+0.62}{N}< c\ln{10}$$ Since $\ln{10}>2$, any $c$ satisfying $$\frac{\ln{(N+1)}+0.62}{2N}\leq c$$ will do. Function $\frac{\ln{(x+1)}+0.62}{2x}$ is decreasing (check the 1st derivative), thus we can take (for $x=1$) $$c= \frac{\ln{2}+0.62}{2}\approx 0.65657...<1=\ln{e}$$

rtybase
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