# MATHS BITE: e^π > π^e

Consider the Taylor Series for :

If x>0, then

Letting x = π/e – 1 gives us

Multiplying by e, we find that

Raising everything to the power of e gives us the final, desired result:

M x

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## One comment

1. I really like this inequality. If we take log both sides, we get $\frac{\pi}{\log \pi} > e$. And this doesn’t appears to be obvious.

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