Today's Putnam Problem of the Day

NonCommAlg

MHF Hall of Honor
May 2008
2,295
1,663
Evaluate the sum

\(\displaystyle
S = \frac{1}{ \cos{0^o} \cos{1^o} } + \frac{1}{ \cos{1^o} \cos{2^o} } + ... + \frac{1}{ \cos{88^o} \cos{89^o} }
\)

Source : Harvard Mathematics Department : Putnam Competition
we have \(\displaystyle \frac{1}{\cos n \cos (n+1)} = \frac{1}{\sin 1} \left(\frac{\sin(n+1)}{\cos(n+1)} - \frac{\sin n}{\cos n} \right)\) and so \(\displaystyle (\sin 1)S\) is just a simple telescoping sum: \(\displaystyle (\sin 1)S= \sum_{n=0}^{88} \left(\frac{\sin(n+1)}{\cos(n+1)} - \frac{\sin n}{\cos n} \right)=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}.\) thus \(\displaystyle S=\frac{\cos 1}{\sin^21}.\)