Which one is bigger $100^{300}$ or $300!$?

Question

How to find which one is bigger $100^{300}$ or $300!$ without using a calculator? I have tried it for whole 2 years but could not find it yet.

Answer 1

Let $n\in\Bbb N$. We have $e<3$ and hence $$3^{3n}>e^{3n}=\sum_{k=0}^\infty\frac{(3n)^k}{k!}.$$ As all summands are positive, this implies $$ 3^{3n}>\frac{(3n)^k}{k!}=\frac{3^kn^k}{k!}$$ for all $k$. In particular, for $k=3n$, this becomes $3^{3n}>\frac{3^{3n}n^{3n}}{(3n)!}$, or $$(3n)!>n^{3n}.$$ For $n=100$, this gives us $$300! > 100^{300}. $$

Answer 2

Easy general method needing no cleverness, useful for many problems, See the diagram below:

For a function such as (natural) logarithm with $f(x) > 0$ and $f'(x) > 0,$ we get $$ \int_{a-1}^b \; f(x) dx < \sum_{k=a}^b \; f(k) < \int_{a}^{b+1} \; f(x) dx $$

Here $f$ is log base e, take $a=2$ and $b=n,$ later we will take $n=300$ $$ \int_{1}^n \; \log x \; dx < \sum_{k=2}^n \; \log k < \int_{2}^{n+1} \; \log x \; dx $$

An antiderivative of $\log x$ is $x \log x - x.$

$$ n \log n - n + 1 < \log n! < (n+1) \log (n+1) - n - 1 - 2 \log 2 + 2 $$

$$ 300 \log 300 - 299 < \log 300! < 301 \log 301 - 299 - \log 4 $$

$$ 300 ( \log 100 + \log 3)- 299 < \log 300! $$

As $e \approx 2.71828 < 3,$ we find $1 < \log 3$ and $300 < 300 \log 3,$ $$ 300 \log 100 < 300 \log 100 + 300 - 299 < \log 300! $$ $$ 100^{300} < 300! $$

Answer 3

More generally, for positive integer $n$, we have

$$(3n)!>n^{3n},$$

and taking logs of both sides and rearranging, the inequality is equivalent to

$$\frac{1}{n}\sum_{k=1}^{3n}\log (k/n)>0.$$

This can be shown by noting the LHS is just a right Riemann sum for $\int_0^3 \log xdx$ and since $\log x$ is monotonically increasing, the right Riemann sum will be an overestimate of the integral so

$$\frac{1}{n}\sum_{k=1}^{3n}\log (k/n)>\int_0^3 \log xdx=3\log 3-3>3\log e-3=0.$$