Calculate the following sum for integers n≥2:∑k=0n−22ktan⁡(π2n−k)

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asserena3wx · Accepted Answer

We have this nice identitytan⁡(θ)=cot⁡(θ)−2cot⁡(2θ)Making use of this, and denoting ∑k=0m2ktan⁡(2kθ) as S, we get thatS=tan⁡(θ)+2tan⁡(2θ)+4tan⁡(4θ)+…+2mtan⁡(2mθ)cot⁡(θ)−2cot⁡(2θ)+2cot⁡(2θ)−…+2mcot⁡(2mθ)−2m+1cot⁡(2m+1cot⁡(2m+1θ)=cot⁡(θ)−2m+1cot⁡(2m+1θ)In your case, θ=π2n and m=n−2. Hence, we get the sum to beS=cot⁡(π2n)−2n−1cot⁡(2n−1⋅π2n)=cot⁡(π2n)−2n−1cot⁡(π2)=cot⁡(π2nProof for tan⁡(θ)=cot⁡(θ)−2cot⁡(2θ)cot⁡(θ)−tan⁡(θ)=cos⁡(θ)sin⁡(θ)−sin⁡(θ)cos⁡(θ)=cos2(θ)−sin2(θ)sin⁡(θ)cos⁡(θ)=2cos⁡(2θ)sin⁡(2θ)=2cot⁡(2θ)

Cheryl Stark · Accepted Answer

Consider∏k=0n−2cos⁡(2kθ)Multiplying numerator and denominator by 2sin⁡(θ) we get,2sin⁡(θ)cos⁡(θ)2sin⁡(θ)∏k=1n−2cos⁡(2kθ)=sin⁡(2θ)2sin⁡(θ)∏k=1n−2cos⁡(2kθ)Now, repeatedly multiplying and dividing by 2, we can reduce the above to,∏k=2n−2cos⁡(2kθ)=sin⁡(2n−1θ)2n−1sin⁡(θ)Take logs on both sides,∑k=0n−2ln⁡(cos⁡(2kθ))=ln⁡(sin⁡(2n−1θ))−ln⁡(2n−1)−ln⁡(sin⁡(θ))Differentiating both sides w.r.t. θ we get,−∑k=0n−22ktan⁡(2kθ)=2n−1cot⁡(2n−1θ)−cot⁡(θ)Substitute θ=π2n above to get,∑k=0n−22ktan⁡(π2n−k)=cot⁡(π2n)

Calculate the following sum for integers n\geq2: \sum_{k=0}^{n-2}2^k\tan(\frac{\pi}{2^{n-k}})

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