TITLE: How to prove $(1-\frac1{36})^{25}\lt\frac12$?
QUESTION [9 upvotes]: How to prove the inequality? 
$(1-\frac1{36})^{25}\lt\frac12$

I'm in trouble. 
Thank you very much for your help

REPLY [2 votes]: If you are willing to use that $\ln(1 - x) < -x$ (which is immediate from the power series for $\ln(1 - x)$) then it can be done quite quickly. Taking logs your inequality is equivalent to
$$25 \ln \bigg(1 - {1 \over 36}\bigg) < \ln (1/2)$$
Using $\ln(1 - x) < -x$ it suffices to show that
$$-{25 \over 36} < \ln (1/2)$$
Equivalently,
$$\ln 2 < {25 \over 36}$$
Note that $\ln 2 = .6931...$  and ${25 \over 36} = .6944..$
So the inequality does hold.