Comparing large numbers

Taking logarithm (base 10) of both sides, we have

$2.n > \sum_{t=2}^{n} log (t)$

we can have an estimation for $\sum_{t=2}^{n} log (t)$ , using Abel's lemma (take the sequence $a_n=1$ ) and the function $f(x)=logx$ .

$\sum_{t=2}^{n} log (t) \approx n.log n -(n-1)$

with this estimation, for $n \approx 1000$ the inequality starts to change sign. However, as proven practically, it is far from the answer (268). what one can do is to use the logarithmic version to facilitate the numerical evaluation of the inequality.

To solve this, we need the Stirling approximation, which is correct asymptotically.

$n! \approx \sqrt{2\pi n}(n/e)^n \rightarrow \frac{1}{2}\log_{10}({2\pi n})+n\log_{10}{n}-n\log_{10}{e}$

Now, this is manageable, and if you just make a plot against $2n$ you'll see that the two functions cross between $n = 268$ and $n = 269$ .

Key idea: Trying to compute the factorial directly is a recipe for disaster.