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We can calculate the partial sum $\sum_{n=0}^{20} \frac{\sin(n\pi /40)}{2^n}$ directly, but for large value of n, is there an easy way to calculate a series of this type?

bbw
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  • Write $\sin{(n \pi/40)}$ as the imaginary part of $e^{i n \pi/40}$. The result will be the imaginary part of a geometric series. – Ron Gordon Jan 12 '18 at 05:31
  • Notice that $f(x):=2^{-x}\cdot \sin(\displaystyle {x\pi\over 40})\to 0(x\to +\infty,)$ then we can use $\displaystyle\int_{n}^m f(x)dx$ to approximate $\displaystyle\sum_{i=n}^m f(i)$ when $n,m$ are large integers. Hope this can help. – painday Jan 12 '18 at 06:09
  • See also : $$#76$$ of https://archive.org/details/treatiseonplanet00hobs – lab bhattacharjee Jan 12 '18 at 06:17

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$$\sum_{r=a}^b\dfrac{\sin(nx)}{2^n}=$$ imaginary part of

$$\sum_{r=a}^b\dfrac{e^{inx}}{2^n}=\sum_{r=a}^b\left(\dfrac{e^{ix}}2\right)^n$$ which is a finite Geometric Series

lab bhattacharjee
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  • See also: https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi – lab bhattacharjee Jan 12 '18 at 05:38
  • oh wait, but i end up with $Im(\sum_{n=0}^{20} {\frac{e^{i\theta}}{2}}^n$. And the partial sum of this geometric series is $\frac{2^{21}-{e^{i\theta}}^{21}}{2^{21}-2^{21}e^{i\theta}}$ . But in my question $\theta=\frac{\pi}{40}$, to calculate $e^{i\theta}$, i have to calculate $\cos(\frac{\pi}{40})$ and $\sin(\frac{\pi}{40})$ . The problem becomes what are the values of the trig functions at $\theta=\frac{\pi}{40}$..... – bbw Jan 12 '18 at 06:21
  • @bbw, We can safely keep $$\cos\dfrac{\pi}{40}$$ or $$\sin\dfrac{\pi}{40}$$ in the answer. All we need to ensure is there should not be any $i$ in the final answer – lab bhattacharjee Jan 12 '18 at 06:35