Summation with sine & cosine sequences

Geometry Level 3

Let a n = sin ( π 2 n π ) + cos n π a_n=\sin{\left(\dfrac{\pi}2-n\pi\right)}+\cos{n\pi} and b n = 10 cos ( 2 n π π 3 ) b_n=10\cos{\left(2n\pi-\dfrac{\pi}3\right)} . Find the value of n = 1 ( a n b n ) n \large\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{a_n}{b_n}\right)^n

2 3 \frac23 2 7 -\frac27 2 7 \frac27 2 3 -\frac23

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1 solution

Maggie Miller
Jul 21, 2015

Note for integer n n , cos n π = sin ( π 2 n π ) = ( 1 ) n \cos n\pi=\sin\left(\frac{\pi}{2}-n\pi\right)=(-1)^n , so a n = 2 ( 1 ) n a_n=2(-1)^n . Similarly, cos ( 2 n π π 3 ) = cos ( π 3 ) = 1 2 \cos\left(2n\pi-\frac{\pi}{3}\right)=\cos\left(\frac{\pi}{3}\right)=\frac{1}{2} , so b n = 5 b_n=5 .

Thus, n = 1 ( a n b n ) n = n = 1 ( 1 ) n 2 ( 2 5 ) n = n = 1 ( 1 ) n ( 2 5 ) n \displaystyle\sum_{n=1}^{\infty}\left(\frac{a_n}{b_n}\right)^n=\sum_{n=1}^{\infty}(-1)^{n^2}\left(\frac{2}{5}\right)^n=\sum_{n=1}^\infty(-1)^n\left(\frac{2}{5}\right)^n is a geoemtric series with ratio 2 5 -\frac{2}{5} , so

n = 1 ( a n b n ) n = 2 5 1 + 2 5 = 2 7 \displaystyle\sum_{n=1}^\infty\left(\frac{a_n}{b_n}\right)^n=\frac{-\frac{2}{5}}{1+\frac{2}{5}}=\boxed{-\frac{2}{7}} .

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