3

Prove that $\log(n!) \leq (n-1)\log(n)$ directly and by induction.

I'm having a hard time with this one. I tried:

$$\log(n!) + \log(n+1) \leq (n-1)\log(n+1) + \log(n+1)$$ $$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1)$$ $$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1) \leq n\log(n+1) - \log(n) + \log(n+1)$$ $$\log((n+1)!) \leq n\log(n+1) - \log(n) + \log(n+1) \leq n\log(n+1) + \log(n+1)$$ $$\log((n+1)!) \leq (n+1)\log(n+1)$$

But that's not what is asked, and I can't find a way of getting there.

Thomas Andrews
  • 186,215
bgly
  • 311

4 Answers4

4

Hint: You can use $n!\leqslant n^{n-1}$

zkutch
  • 14,313
0

By induction you have,

$$\log(n!)+\log(n+1)\le (n-1)\log(n)+\log(n+1)\to$$ $$\log[(n+1)!]\le (n-1)\log(n)+\log(n+1)$$

Remmember that $\log()$ is a strictly increasing function, so $\log(n)< \log(n+1)$. It means,

$$(n-1)\log(n)\le (n-1)\log(n+1)\to$$ $$(n-1)\log(n)+\log(n+1)\le (n-1)\log(n+1)+\log(n+1)=n\log(n+1).$$

Arnaldo
  • 21,758
0

\begin{align} \log(n!) &\leq (n-1)\log(n) \\ \iff \log(n) + \log(n!) &\leq \log(n^n) \\ \iff \log(nn!) &\leq \log(n^n) \end{align}

Since $\log$ is an increasing function, and $nn!\leq n^n$, the third inequality holds true for all $n \in \mathbb{N}$. I would try proving that $nn! \leq n^n$ myself, and then you can compare with my solution:

\begin{align} nn! &\leq n^n \iff n! \leq n^{n-1} \\[6pt] n! &= \underbrace{n \cdot (n-1) \cdot (n-2) \cdot (n-3) \cdot \ldots \cdot 2}_{n-1 \text{ terms}} \\[6pt] n^{n-1}&=\underbrace{n \cdot n \cdot n \cdot n \cdot \ldots \cdot n}_{n-1 \text{ terms}} \end{align}

The case $n=1$ is a mild annoyance, but can be dealt with easily.

Joe
  • 22,603
0

$\log(n!)$

$=\sum_{m=1}^{n}\log(m)$

$=\log(1)+\sum_{m=2}^{n}\log(m)$

$=\sum_{m=2}^{n}\log(m)$

$\log(m)$ is an increasing function

So

$\sum_{m=2}^{n}\log(m)$

$\le \sum_{m=2}^{n}\log(n)$

$ = (n-2+1)\log(n)$

$ = (n-1)\log(n)$

For induction proof:

Base case: $n=1$

$\log(1!) = 0$.

$(1-1)\log(1) = 0$.

So $\log(n!) \le (n-1)\log(n)$ is true for $n=1$.

Assume true for $n=k$

$\log(k!) \le (k-1)\log(k)$

Now prove true for $n=k+1$

$\log((k+1)!)$

$=\log(k+1) + \log(k!)$

$\le \log(k+1) + (k-1)\log(k)$ (by inductive assumption)

$\le \log(k+1) + (k-1)\log(k+1)$ (again since log is an increasing function)

$= ((k+1)-1)\log(k+1)$

So it's true for $n=k+1$

Ameet Sharma
  • 2,997