Not an usual Trigonometric Summation!

Calculus Level 5

L = lim n k = 1 n 1 2 k tan ( π 3 2 k + 1 ) \large{L = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{3\cdot2^{k+1}}\right)}

If L L can be represented as:

A π B C \large{\dfrac{A}{\pi^B} - \sqrt{C}}

where A , B , C A,B,C are positive integers, find the value of A + B + C A+B+C ?


Here is a similar problem : Limited Trigonometry .


The answer is 10.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Satyajit Mohanty
Jul 29, 2015

We have (By-multiplying Numerator and Denominator by sin ( x 2 n ) \sin\left(\dfrac{x}{2^n}\right) ):

lim n cos ( x 2 ) cos ( x 2 2 ) cos ( x 2 3 ) cos ( x 2 n ) = sin ( x ) x \lim_{n \to \infty} \cos \left(\dfrac{x}{2} \right)\cos \left(\dfrac{x}{2^2} \right)\cos\left(\dfrac{x}{2^3} \right) \ldots \cos \left(\dfrac{x}{2^n} \right) = \dfrac{\sin(x)}{x}

Differentiating above, we obtain:

n = 1 1 2 n tan ( x 2 n ) = 1 x cot ( x ) \sum_{n=1}^{\infty} \dfrac{1}{2^n} \tan\left(\dfrac{x}{2^n}\right)=\dfrac1x -\cot(x)

We'll apply the above result in a step below. Now,

L = lim n k = 1 n 1 2 k tan ( π 3 2 k + 1 ) = lim n k = 1 n 1 2 k tan ( π 6 1 2 k ) L = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{3\cdot2^{k+1}}\right) = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{6} \cdot \dfrac{1}{2^k}\right)

= lim n n = 1 1 2 n tan ( x 2 n ) = \lim_{n \to \infty} \sum_{n=1}^{\infty} \dfrac{1}{2^n} \tan\left(\dfrac{x}{2^n}\right) where x = π 6 x=\dfrac{\pi}{6} .

= 1 x cot ( x ) = 6 π cot ( π 6 ) =\dfrac1x - \cot(x) = \dfrac{6}{\pi} - \cot\left(\dfrac{\pi}{6}\right)

= 6 π 3 =\dfrac{6}{\pi} - \sqrt{3}

So, A = 6 , B = 1 , C = 3 , A + B + C = 10 A=6, B=1, C=3, A+B+C=\boxed{10} .

WOW! This is awesome! =D

Pi Han Goh - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...