N-sigma-tan-limit

Calculus Level 5

lim ⁡ n → ∞ ∑ r = 1 n 1 2 r tan ⁡ ( x 2 r ) \large \lim_{n\to\infty} \sum_{r=1}^n \dfrac1{2^r} \tan \left( \dfrac x{2^r} \right)

Let a , b , x a,b,x be all constants such that the limit above evaluates to a x b − cot ⁡ x ax^b - \cot x .

Find the value of 10 a + b 10a + b .


The answer is 9.

Aaron Jerry Ninan by Aaron Jerry Ninan , 20, India

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1 solution

Chew-Seong Cheong
Jan 27, 2017

L = lim ⁡ n → ∞ ∑ r = 1 n 1 2 r tan ⁡ ( x 2 r ) = lim ⁡ n → ∞ ∑ r = 1 n tan ⁡ 2 ( x 2 r ) 2 r tan ⁡ ( x 2 r ) = lim ⁡ n → ∞ ∑ r = 1 n 1 − 1 + tan ⁡ 2 ( x 2 r ) 2 r tan ⁡ ( x 2 r ) = lim ⁡ n → ∞ ∑ r = 1 n ( 1 2 r tan ⁡ ( x 2 r ) − 1 − tan ⁡ 2 ( x 2 r ) 2 r tan ⁡ ( x 2 r ) ) = lim ⁡ n → ∞ ∑ r = 1 n ( 1 2 r tan ⁡ ( x 2 r ) − 1 2 r − 1 tan ⁡ ( x 2 r − 1 ) ) = lim ⁡ n → ∞ ∑ r = 1 n ( cot ⁡ ( x 2 r ) 2 r − cot ⁡ ( x 2 r − 1 ) 2 r − 1 ) = lim ⁡ n → ∞ ( cot ⁡ ( x 2 n ) 2 n − cot ⁡ x ) = lim ⁡ n → ∞ ( cos ⁡ ( x 2 n ) 2 n sin ⁡ ( x 2 n ) − cot ⁡ x ) = lim ⁡ n → ∞ ( cos ⁡ ( x 2 n ) x sin ⁡ ( x 2 n ) x 2 r − cot ⁡ x ) = 1 x − cot ⁡ x = x − 1 − cot ⁡ x \begin{aligned} L & = \lim_{n \to \infty} \sum_{r=1}^n \frac 1{2^r} \tan \left(\frac x{2^r}\right) \\ & = \lim_{n \to \infty} \sum_{r=1}^n \frac {\tan^2 \left(\frac x{2^r}\right)}{2^r \tan \left(\frac x{2^r}\right)} \\ & = \lim_{n \to \infty} \sum_{r=1}^n \frac {1-1+\tan^2 \left(\frac x{2^r}\right)}{2^r \tan \left(\frac x{2^r}\right)} \\ & = \lim_{n \to \infty} \sum_{r=1}^n \left( \frac 1{2^r \tan \left(\frac x{2^r}\right)} - \frac {1-\tan^2 \left(\frac x{2^r}\right)}{2^r \tan \left(\frac x{2^r}\right)} \right) \\ & = \lim_{n \to \infty} \sum_{r=1}^n \left( \frac 1{2^r \tan \left(\frac x{2^r}\right)} - \frac 1{2^{r-1} \tan \left(\frac x{2^{r-1}}\right)} \right) \\ & = \lim_{n \to \infty} \sum_{r=1}^n \left(\frac {\cot \left(\frac x{2^r}\right)}{2^r} - \frac {\cot \left(\frac x{2^{r-1}}\right)}{2^{r-1}} \right) \\ & = \lim_{n \to \infty} \left(\frac {\cot \left(\frac x{2^n}\right)}{2^n} - \cot x \right) \\ & = \lim_{n \to \infty} \left(\frac {\cos \left(\frac x{2^n}\right)}{2^n \sin \left(\frac x{2^n}\right)} - \cot x \right) \\ & = \lim_{n \to \infty} \left(\frac {\cos \left(\frac x{2^n}\right)}{\frac {x \sin \left(\frac x{2^n}\right)}{\frac x{2^r}}} - \cot x \right) \\ & = \frac 1x - \cot x = x^{-1} - \cot x \end{aligned}

⟹ 10 a + b = 10 − 1 = 9 \implies 10a+b = 10-1= \boxed{9}

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