How does one (without the use of Calculator) determine that $5/6$ is less than $(35/36)^6$? How is this done mentally?
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6Bernoulli inequality, $(1+x)^n \geqslant 1 + nx$ for $x \geqslant -1$. – Daniel Fischer Aug 01 '13 at 20:22
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If you look at your second number, it can be written as
$$( 1 - \frac{1}{36} )^6 \ = \ 1 \ - \ 6 \cdot 1^5 \cdot \frac{1}{36} \ + \ \binom 62 \cdot 1^4 \cdot (\frac{1}{36})^2 \ - \ ... $$
with the remaining (unwritten) terms being very small (the first two terms equal $ \ \frac{5}{6} \ $ ) . So $ \ \frac{5}{6} \ $ ends up being slightly smaller than $ ( 1 - \frac{1}{36} )^6 \ $ (by a little less than $ \ 15 \cdot \frac{1}{36^2} \ = \ \frac{5}{432} \ \approx \ 0.0116 \ $ ) .
[This is what Daniel Fischer is briefly describing.]
colormegone
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