Summing Diminishing Cosines

Calculus Level 4

∑ n = 1 ∞ cos ⁡ ( 2 n ) 2 n = ? \large \sum_{n=1}^\infty \dfrac{\cos(2n)}{2^n} = \, ?

2 cos ⁡ ( 2 ) − 1 5 − 4 cos ⁡ ( 2 ) \frac{2\cos(2)-1}{5-4\cos(2)} cos ⁡ ( 2 ) − 1 5 − 4 cos ⁡ ( 2 ) \frac{\cos(2)-1}{5-4\cos(2)} 0 0 3 cos ⁡ ( 2 ) 5 + 4 cos ⁡ ( 2 ) \frac{3\cos(2)}{5+4\cos(2)} 1 1 2 cos ⁡ ( 2 ) − 1 5 − 3 cos ⁡ ( 2 ) \frac{2\cos(2)-1}{5-3\cos(2)}

Bar Hemo by Bar Hemo , 26, Israel

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2 solutions

Bar Hemo
Apr 14, 2016

∑ n = 1 ∞ c o s ( 2 n ) 2 n = ∑ n = 1 ∞ e i 2 n + e − i 2 n 2 1 2 n = ∑ n = 1 ∞ ( e i 2 2 ) n 1 2 + ( e − i 2 2 ) n 1 2 = 1 2 − e i 2 + 1 2 − e − i 2 − 1 = 2 c o s ( 2 ) − 1 5 − 4 c o s ( 2 ) \sum_{n=1}^\infty \frac{cos(2n)}{2^n} =\sum_{n=1}^\infty \frac{e^{i2n}+e^{-i2n}}{2}\frac{1}{2^n}=\sum_{n=1}^\infty (\frac{e^{i2}}{2})^n\frac{1}{2}+(\frac{e^{-i2}}{2})^n\frac{1}{2}=\frac{1}{2-e^{i2}}+\frac{1}{2-e^{-i2}}-1=\frac{2cos(2)-1}{5-4cos(2)}

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could you explain the step after summation .. im getting e^i2 extra in that

Dhruv Aggarwal - 5 years, 1 month ago

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Geometric progression

Bar Hemo - 5 years, 1 month ago
Harsh Poonia
Apr 18, 2020

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