Today's Putnam Problem of the Day

Quote:

Originally Posted by simplependulum

Evaluate the sum

$<br /> S = \frac{1}{ \cos{0^o} \cos{1^o} } + \frac{1}{ \cos{1^o} \cos{2^o} } + ... + \frac{1}{ \cos{88^o} \cos{89^o} }<br />$

Source : Harvard Mathematics Department : Putnam Competition

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