Inspired by Rahil Sehgal

Number Theory Level 3

Which of the following is greater?

$\large 100^{300} \quad \text{or} \quad 300!$

Cannot be determined

300! \lt 100^{300}

300! \gt 100^{300}

300! = 100^{300}

2 solutions

Brian Charlesworth
May 15, 2017

Comparing the natural logs of each of these quantities, we have that $\ln(100^{300}) = 300\ln(100) \approx 1381.55$ , and by Stirling's approximation $\ln(300!) \approx 300\ln(300) - 300 \approx 1411.13$ , so $\boxed{300! \gt 100^{100}}$ .

Note that Stirling's approximation for large $N$ is very accurate, well within 1% for $N = 300$ .

Thanks a lot sir for making us aware of the stirling's approximation! I enjoyed learning it. As N increases, error % decreases. So, can we say that error % becomes 0 as N approaches infinity?

Kaushik Chandra - 4 years ago

Yes, that's correct. A more accurate bounding of $ln(n!)$ is

$\dfrac{1}{2}\ln(2\pi) + \left(n + \dfrac{1}{2}\right)\ln(n) - n + \dfrac{1}{12n + 1} \lt \ln(n!) \lt \dfrac{1}{2}\ln(2\pi) + \left(n + \dfrac{1}{2}\right)\ln(n) - n + \dfrac{1}{12n}$

For $n = 300$ this bounds $\ln(300!)$ to an interval of width $\approx 0.000000077$ , with a value of $1414.90585$ . Thus the approximation I used in my solution is accurate to $0.27$ %. The percentage error in the approximation $\ln(n!) = n\ln(n) - n$ is of the order

$100 \times \dfrac{\dfrac{1}{2}\ln(n)}{\left(n + \dfrac{1}{2}\right)\ln(n) - 1}$ %,

or roughly $\dfrac{50}{n}$ % for very large $n$ , which $\to 0$ as $n \to \infty$ .

Brian Charlesworth - 4 years ago

Wonderful problem as well as solution as always. (+1)

Rahil Sehgal - 4 years ago

@Rahil Sehgal – Thanks! And thanks also for the inspiration. :)

Brian Charlesworth - 4 years ago

I appreciate your answer a lot. Mathematically, your answer sounds great but here is something strange.

Error % = error / original no. N × 100 % As the number N tends to infinity, error also tends to infinity. So, isn't it an indeterminate form ?

Thank you!

Kaushik Chandra - 4 years ago

@Kaushik Chandra – You're right that the error $\dfrac{1}{2}\ln(n)$ tends to infinity as $n$ tends to infinity, so the error calculation is of the indeterminate form $\infty / \infty$ . However, for very large $n$ the percentage is $\frac{50}{n}$ %, which does go to $0$ as $n$ goes to infinity. So while the absolute error value goes to infinity, the percentage error goes to $0$ .

Brian Charlesworth - 4 years ago

@Brian Charlesworth – Again astounded by the magic of mathematics. Thanks a lot sir Wizard for unfolding the magic!

Kaushik Chandra - 4 years ago

Tarit Goswami
May 25, 2017

By Stirling's approximation $300!>(\frac{300}{e})^{300} >(100)^{300}$

Inspired by Rahil Sehgal

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