Prove using induction $\ln(n!)\leqslant n\ln(n)$ for $n\geqslant 1$.

Question

Base case true

$$\ln((n+1)!) = \ln(n+1)+\ln (n!)$$ Product rule But now I'm suck Idk how to prove that is less than or equal to $(n+1)\ln(n+1)$

Answer 1

Hint. $\ln$ is an increasing function.

Answer 2

Note $$\ln((n+1)!) = \ln(n+1)+\ln (n!) \le \ln (n+1)+n\ln n$$

From our inductive hypothesis. But since $\ln n$ is an increasing function, $$\ln((n+1)!) \le \ln (n+1)+n \ln n \le \ln (n+1)+n \ln (n+1)$$

This gives us that $$\ln (n+1)! \le (n+1)\ln n$$

Answer 3

"$\ln((n+1)!) = \ln(n+1)+\ln (n!)$"

That's 90% of it.

$\ln(n!) \le n\ln(n)$

so $\ln((n+1)!) = \ln(n+1) + \ln(n!) \le \ln(n+1) + n\ln(n)$.

We need to show $\ln(n+1) + n\ln(n) \le (n+1)\ln(n+1) = \ln(n+1) + n\ln(n+1)$

But $\ln(n+1) + n\ln(n) \le \ln(n+1) + n\ln(n+1)\iff \ln(n) \le \ln(n+1)$.

So if we can show $\ln (n) \le \ln(n+1)$ we are done. [As as $e > 1$, $\ln(n1) \le \ln(n+1) \iff e^{\ln(n)} = n \le e^{\ln(n+1)} = n+1$]