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Evaluate ∑n=1∞(sin⁡n2n)\sum_{n=1}^{\infty}\left(\frac{\sin n}{2^{n}}\right)

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Full solution

Q. Evaluate ∑n=1∞(sin⁡n2n)\sum_{n=1}^{\infty}\left(\frac{\sin n}{2^{n}}\right)
  1. Given Series: We are given the infinite series ∑n=1∞sin⁡n2n\sum_{n=1}^{\infty}\frac{\sin n}{2^n}. To evaluate this, we will use the concept of power series and complex numbers. We know that sin⁡(n)\sin(n) can be expressed in terms of complex exponentials using Euler's formula: sin⁡(n)=ein−e−in2i\sin(n) = \frac{e^{in} - e^{-in}}{2i}. Let's substitute this into our series.
  2. Substitute Euler's Formula: Substituting sin⁡(n)\sin(n) into the series, we get ∑n=1∞ein−e−in2i⋅2n\sum_{n=1}^{\infty}\frac{e^{in} - e^{-in}}{2i \cdot 2^n}. This simplifies to ∑n=1∞ein−e−in2n+1i\sum_{n=1}^{\infty}\frac{e^{in} - e^{-in}}{2^{n+1}i}. Now we have two separate series: one for ein2n+1i\frac{e^{in}}{2^{n+1}i} and another for −e−in2n+1i-\frac{e^{-in}}{2^{n+1}i}.
  3. First Geometric Series: Let's consider the first series ∑n=1∞ein2n+1i\sum_{n=1}^{\infty}\frac{e^{in}}{2^{n+1}i}. This is a geometric series with a common ratio of ei2\frac{e^{i}}{2}. The sum of a geometric series a1−r\frac{a}{1-r} can be used if |r| < 1. Here, aa is the first term of the series, and rr is the common ratio. The first term when n=1n=1 is ei22i=ei4i\frac{e^{i}}{2^2i} = \frac{e^{i}}{4i}.
  4. Second Geometric Series: The common ratio rr is ei/2e^{i}/2. Since the magnitude of eie^{i} is 11 and the magnitude of 22 is 22, the magnitude of the common ratio ∣ei/2∣|e^{i}/2| is 1/21/2, which is less than 11. Therefore, we can use the geometric series sum formula. The sum of the first series is a/(1−r)=(ei/(4i))/(1−(ei/2))a/(1-r) = (e^{i}/(4i))/(1 - (e^{i}/2)).
  5. Combine Series Sums: Now let's consider the second series ∑n=1∞−e−in2n+1i\sum_{n=1}^{\infty}\frac{-e^{-in}}{2^{n+1}i}. This is also a geometric series with a common ratio of −e−i/2-e^{-i}/2. The first term when n=1n=1 is −e−i/(4i)-e^{-i}/(4i). The common ratio rr is −e−i/2-e^{-i}/2, and its magnitude is also 1/21/2, which is less than 11.
  6. Find Common Denominator: The sum of the second series is a1−r=−e−i4i\frac{a}{1-r} = \frac{-e^{-i}}{4i} / 1−−e−i21 - \frac{-e^{-i}}{2}. Now we have the sums of both series, and we can add them together to find the total sum of the original series.
  7. Simplify Numerator: Adding the two sums together, we get (ei4i)/(1−(ei2))+(−e−i4i)/(1−(−e−i2))(\frac{e^{i}}{4i})/(1 - (\frac{e^{i}}{2})) + (\frac{-e^{-i}}{4i})/(1 - (\frac{-e^{-i}}{2})). To simplify this expression, we need to find a common denominator and combine the terms.
  8. Expand Denominator: The common denominator for the two fractions is (1−(ei/2))(1−(−e−i/2))(1 - (e^{i}/2))(1 - (-e^{-i}/2)). Multiplying the numerators by the necessary factors to get this common denominator, we can combine the fractions.
  9. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[(1−(ei2))(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 - (-\frac{e^{-i}}{2})\right)]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)].
  10. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[(1−(ei2))(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 - (-\frac{e^{-i}}{2})\right)]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]. Simplifying the numerator, we notice that ei4i(e−i2)\frac{e^{i}}{4i}(\frac{e^{-i}}{2}) and e−i4i(ei2)\frac{e^{-i}}{4i}(\frac{e^{i}}{2}) are complex conjugates and will cancel each other out. The numerator then becomes ei4i−e−i4i\frac{e^{i}}{4i} - \frac{e^{-i}}{4i}.
  11. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[(1−(ei2))(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 - (-\frac{e^{-i}}{2})\right)]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]. Simplifying the numerator, we notice that ei4i(e−i2)\frac{e^{i}}{4i}(\frac{e^{-i}}{2}) and e−i4i(ei2)\frac{e^{-i}}{4i}(\frac{e^{i}}{2}) are complex conjugates and will cancel each other out. The numerator then becomes ei4i−e−i4i\frac{e^{i}}{4i} - \frac{e^{-i}}{4i}. The denominator can be expanded using the distributive property: (1−(ei2))(1+(e−i2))=1−(ei2)(e−i2)−(ei2)+(e−i2)(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2})) = 1 - (\frac{e^{i}}{2})(\frac{e^{-i}}{2}) - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}). Since eie−i=1e^{i}e^{-i} = 1, the denominator simplifies to 1−14−(ei2)+(e−i2)=34−(ei−e−i2)1 - \frac{1}{4} - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}) = \frac{3}{4} - (\frac{e^{i} - e^{-i}}{2}).
  12. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[1−(ei2)(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[1 - (\frac{e^{i}}{2})(1 - (-\frac{e^{-i}}{2}))]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]. Simplifying the numerator, we notice that ei4i(e−i2)\frac{e^{i}}{4i}(\frac{e^{-i}}{2}) and e−i4i(ei2)\frac{e^{-i}}{4i}(\frac{e^{i}}{2}) are complex conjugates and will cancel each other out. The numerator then becomes ei4i−e−i4i\frac{e^{i}}{4i} - \frac{e^{-i}}{4i}. The denominator can be expanded using the distributive property: (1−(ei2))(1+(e−i2))=1−(ei2)(e−i2)−(ei2)+(e−i2)(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2})) = 1 - (\frac{e^{i}}{2})(\frac{e^{-i}}{2}) - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}). Since eie−i=1e^{i}e^{-i} = 1, the denominator simplifies to 1−14−(ei2)+(e−i2)=34−(ei−e−i2)1 - \frac{1}{4} - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}) = \frac{3}{4} - (\frac{e^{i} - e^{-i}}{2}). The denominator can be further simplified by recognizing that (ei−e−i2)(\frac{e^{i} - e^{-i}}{2}) is actually sin⁡(1)\sin(1) using Euler's formula. So the denominator becomes [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]00.
  13. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[(1−(ei2))(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 - (-\frac{e^{-i}}{2})\right)]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]. Simplifying the numerator, we notice that ei4i(e−i2)\frac{e^{i}}{4i}(\frac{e^{-i}}{2}) and e−i4i(ei2)\frac{e^{-i}}{4i}(\frac{e^{i}}{2}) are complex conjugates and will cancel each other out. The numerator then becomes ei4i−e−i4i\frac{e^{i}}{4i} - \frac{e^{-i}}{4i}. The denominator can be expanded using the distributive property: (1−(ei2))(1+(e−i2))=1−(ei2)(e−i2)−(ei2)+(e−i2)(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2})) = 1 - (\frac{e^{i}}{2})(\frac{e^{-i}}{2}) - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}). Since eie−i=1e^{i}e^{-i} = 1, the denominator simplifies to 1−14−(ei2)+(e−i2)=34−(ei−e−i2)1 - \frac{1}{4} - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}) = \frac{3}{4} - (\frac{e^{i} - e^{-i}}{2}). The denominator can be further simplified by recognizing that (ei−e−i2)(\frac{e^{i} - e^{-i}}{2}) is actually sin⁡(1)\sin(1) using Euler's formula. So the denominator becomes [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]00. Now we have the total sum as [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]11. This can be simplified by multiplying the numerator and denominator by [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]22 to get rid of the complex denominator.
  14. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[(1−(ei2))(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 - (-\frac{e^{-i}}{2})\right)]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]. Simplifying the numerator, we notice that ei4i(e−i2)\frac{e^{i}}{4i}(\frac{e^{-i}}{2}) and e−i4i(ei2)\frac{e^{-i}}{4i}(\frac{e^{i}}{2}) are complex conjugates and will cancel each other out. The numerator then becomes ei4i−e−i4i\frac{e^{i}}{4i} - \frac{e^{-i}}{4i}. The denominator can be expanded using the distributive property: (1−(ei2))(1+(e−i2))=1−(ei2)(e−i2)−(ei2)+(e−i2)(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2})) = 1 - (\frac{e^{i}}{2})(\frac{e^{-i}}{2}) - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}). Since eie−i=1e^{i}e^{-i} = 1, the denominator simplifies to 1−14−(ei2)+(e−i2)=34−(ei−e−i2)1 - \frac{1}{4} - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}) = \frac{3}{4} - (\frac{e^{i} - e^{-i}}{2}). The denominator can be further simplified by recognizing that (ei−e−i2)(\frac{e^{i} - e^{-i}}{2}) is actually sin⁡(1)\sin(1) using Euler's formula. So the denominator becomes [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]00. Now we have the total sum as [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]11. This can be simplified by multiplying the numerator and denominator by [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]22 to get rid of the complex denominator. Multiplying by [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]22, we get [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]44. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[\left(1 - (\frac{e^{i}}{2})\right)\left(1 + (\frac{e^{-i}}{2})\right)]55.
  15. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[1−(ei2)(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[1 - (\frac{e^{i}}{2})(1 - (-\frac{e^{-i}}{2}))]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]. Simplifying the numerator, we notice that (ei4i)(e−i2)(\frac{e^{i}}{4i})(\frac{e^{-i}}{2}) and (e−i4i)(ei2)(\frac{e^{-i}}{4i})(\frac{e^{i}}{2}) are complex conjugates and will cancel each other out. The numerator then becomes (ei4i)−(e−i4i)(\frac{e^{i}}{4i}) - (\frac{e^{-i}}{4i}). The denominator can be expanded using the distributive property: (1−(ei2))(1+(e−i2))=1−(ei2)(e−i2)−(ei2)+(e−i2)(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2})) = 1 - (\frac{e^{i}}{2})(\frac{e^{-i}}{2}) - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}). Since eie−i=1e^{i}e^{-i} = 1, the denominator simplifies to 1−14−(ei2)+(e−i2)=34−(ei−e−i2)1 - \frac{1}{4} - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}) = \frac{3}{4} - (\frac{e^{i} - e^{-i}}{2}). The denominator can be further simplified by recognizing that (ei−e−i2)(\frac{e^{i} - e^{-i}}{2}) is actually sin⁡(1)\sin(1) using Euler's formula. So the denominator becomes [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]00. Now we have the total sum as [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]11. This can be simplified by multiplying the numerator and denominator by [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]22 to get rid of the complex denominator. Multiplying by [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]22, we get [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]44. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]55. Using Euler's formula again, we know that [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]66. Substituting this into our expression, we get [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]77.
  16. Final Simplification: After combining the fractions, we get [ei4i(1−(−e−i2))−e−i4i(1−(ei2))]/[1−(ei2)(1−(−e−i2))][\frac{e^{i}}{4i}(1 - (-\frac{e^{-i}}{2})) - \frac{e^{-i}}{4i}(1 - (\frac{e^{i}}{2}))]/[1 - (\frac{e^{i}}{2})(1 - (-\frac{e^{-i}}{2}))]. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]. Simplifying the numerator, we notice that ei4i(e−i2)\frac{e^{i}}{4i}(\frac{e^{-i}}{2}) and e−i4i(ei2)\frac{e^{-i}}{4i}(\frac{e^{i}}{2}) are complex conjugates and will cancel each other out. The numerator then becomes ei4i−e−i4i\frac{e^{i}}{4i} - \frac{e^{-i}}{4i}. The denominator can be expanded using the distributive property: (1−(ei2))(1+(e−i2))=1−(ei2)(e−i2)−(ei2)+(e−i2)(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2})) = 1 - (\frac{e^{i}}{2})(\frac{e^{-i}}{2}) - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}). Since eie−i=1e^{i}e^{-i} = 1, the denominator simplifies to 1−14−(ei2)+(e−i2)=34−(ei−e−i2)1 - \frac{1}{4} - (\frac{e^{i}}{2}) + (\frac{e^{-i}}{2}) = \frac{3}{4} - (\frac{e^{i} - e^{-i}}{2}). The denominator can be further simplified by recognizing that (ei−e−i2)(\frac{e^{i} - e^{-i}}{2}) is actually sin⁡(1)\sin(1) using Euler's formula. So the denominator becomes [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]00. Now we have the total sum as [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]11. This can be simplified by multiplying the numerator and denominator by [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]22 to get rid of the complex denominator. Multiplying by [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]22, we get [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]44. This simplifies to [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]55. Using Euler's formula again, we know that [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]66. Substituting this into our expression, we get [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]77. Simplifying the expression, we can cancel out the [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]88's and we get [ei4i−ei4i(e−i2)−e−i4i+e−i4i(ei2)]/[(1−(ei2))(1+(e−i2))][\frac{e^{i}}{4i} - \frac{e^{i}}{4i}(\frac{e^{-i}}{2}) - \frac{e^{-i}}{4i} + \frac{e^{-i}}{4i}(\frac{e^{i}}{2})]/[(1 - (\frac{e^{i}}{2}))(1 + (\frac{e^{-i}}{2}))]99. This is the sum of the original series.

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