How to prove that ∑n=1∞14ncos2x2n=1sin2x−1x2

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How to prove that  ∑n=1∞14ncos2x2n=1sin2x−1x2

gekraamdbk · Accepted Answer

Often when sin⁡(x) and x appear in a formula such as this, it is based on the fact that  lima→0sin⁡(ax)a=x (1)  Since the summands are 4−kcos2(2−kx) , it appears that an identity involving a double angle formula is being exploited. Therefore, with an eye toward a telescoping sum, I started with  4sin2(2x)−1sin2(x)=4−4cos2(x)4sin2(x)cos2(x)  =4sin2(x)4sin2(x)cos2(x)  =1cos2(x)  Substitute x↦2−k and multiply by 4−k and sum (and telescope)  ∑k=1n4−kcos2(2−kx)=∑k=1n(4−k+1sin2(2−k+1x)−4−ksin2(2−kx))  =1sin2(x)−4−nsin2(2−nx}  →1sin2(x)−1x2  since, by (1)  limn→∞4−nsin2(2−nx)=1x2

porekalahr · Accepted Answer

Since  1=sin2θ+cos2θ  dividing by sin2θcos2θ, we get that  4csc2(2θ)=csc2(θ)+sec2(θ)  Replace θ as x2n we get that  sec2(x2n+1)=4csc2(x2n)−csc2(x2n+1)  Dividing by 4n+1, we get that  sec2(x2n+1)4n+1=csc2(x2n)4n−csc2(x2n+1)4n+1  Now telescopic summantion gives us  ∑n=0Nsec2(x2n+1)4n+1=csc2(x)−csc2(x2N+1)4N+1  Now letting N→∞, we get what you want.

How to prove that \sum_{n=1}^\infty\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac{1}{\sin^2x}-\frac{1}{x^2}

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