Proving the result $294!<100^{300}<295!$

Question

Proving $294!<100^{300}<295!$

$\bf{My\; Try::}$ I have used Stirling Approximation $$n!\approx \left(\frac{n}{e}\right)^n\cdot \sqrt{2\pi n}$$

Put $n=294$ and $n=295$,

$$294!\approx \left(\frac{294}{e}\right)^{294}\cdot \sqrt{2\pi \cdot 294}$$

and $$295!\approx \left(\frac{295}{e}\right)^{295}\cdot \sqrt{2\pi \cdot 295}$$

Now i did not understand hoe can i solve it, Help required, Thanks

Switch to logarithms in order to avoid dealing with huge numbers. Use Stirling's formula as $$\log n! \approx n\log n-n+\frac{1}{2}\log(2\pi n)$$ — Jack D'Aurizio, Oct 20 '16 at 14:19
If you choose $\log_{10}$, you can easily very the double-inequality. — Peter, Oct 20 '16 at 14:22
@Peter If we ignore that part and prove the second half inequality, it'd be much much easier. — Simply Beautiful Art, Oct 20 '16 at 14:24
Strictly speaking, using the Stirling's approximation, we also need to estimate the error. An induction proof would be nice, not sure whether it will work. — Peter, Oct 20 '16 at 14:28
@Peter It'll underapproximate $295!$, so we don't really have to worry for that part. — Simply Beautiful Art, Oct 20 '16 at 14:29
@Peter: we also have an inequality: $$\log n!=n\log n-n+\frac{1}{2}\log(2\pi n)+E(n)$$ where $E(n)$ is bounded between $\frac{1}{12n+1}$ and $\frac{1}{12n}$, see http://math.stackexchange.com/a/1409131/44121 — Jack D'Aurizio, Oct 20 '16 at 14:29
SimpleArt and Jack D'Aurizio : I know, I only wanted to mention it. — Peter, Oct 20 '16 at 14:33
The logarithms of the involved terms are roughly $1381,1381.5$ and $1386.5$, so we do not need to worry about $E(n)$, that is way less than $\frac{1}{2}$. — Jack D'Aurizio, Oct 20 '16 at 14:35

lhf · Answer 1 · 2017-07-11T16:29:05.187

Wikipedia tells us that $$ n\ln \left({\frac {n}{e}}\right)+1\leq \ln n!\leq (n+1)\ln \left({\frac {n+1}{e}}\right)+1 $$ Dividing by $\ln 100$ gives $$ 299.005 < \log_{100} 294! < 301.25 \\ 300.24 < \log_{100} 295! < 302.48 $$ which is almost what is needed, except for the upper bound for $\log_{100} 294!$.

The actual values are $$ \log_{100} 294! \approx 299.82 \\ \log_{100} 295! \approx 301.06 $$ and so we do need tight bounds to solve the question.

Using $$ {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}<n!<{\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}e^{\frac {1}{12n}} $$ and taking logs as before we get $$ 299.82240 < \log_{100} 294! < 299.82247 $$ which does it.

Proving the result $294!<100^{300}<295!$

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