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As the title suggests, I would like to find the sum of the following series:

$S(x)=1+ 2\sin x+ \dfrac{2^{2}\sin 2x}{2!}+ \dfrac{2^{3}\sin 3x}{3!}+ \dots$. I have attempted the question and I am unsure how to proceed. Please kindly correct me if I have made any mistakes along the way.

Sol: Let $C(x)=1+ 2\cos x+ \dfrac{2^{2}\cos 2x}{2!}+ \dfrac{2^{3}\cos 3x}{3!}+ \dots$ and consider $iS(x)= i (1+ \sum_0^\infty \frac{2^{n} \sin nx}{n!})$.

Then, summing both series up, we have $C(x)+iS(x)=(1+ \sum_0^\infty \frac{2^{n} \cos nx}{n!})+i (1+ \sum_0^\infty \frac{2^{n} \sin nx}{n!}) =1+i +\sum_0^\infty \frac{2^{n}}{n!}(\cos nx + i\sin nx) = 1+i + \sum_0^\infty e^{2}.e^{inx}= 1+i+ \sum_0^\infty e^{2+inx}$

since $e^{x}= \sum_o^\infty \frac{x^{n}}{n!}$ (in this case, $x=2$) and using De Moivre's Theorem, $\cos nx + i\sin nx=e^{inx}$.

Now, it remains to evaluate the sum $T(x)= 1+i+ \sum_0^\infty e^{2+inx}$. Since the second term is a geometric series with first term $e^{2}$ and common ratio $e^{ix}$, we have $T(x)=1+i+ \frac{e^{2}}{1-e^{ix}}=1+\frac{e^{2}}{1-e^{-x}(1+ \cos x)}+i(1+ \frac{e^{2}}{1-e^{x}}\sin x)$.

Lastly, comparing the imaginary part, $S(x)= 1+ \frac{e^{2} \sin x}{1-e^{-x}}$.

My concern is whether I can conclude that $\sum_o^\infty \frac{2^{n}}{n!}=e^{2}$ and whether the general idea is correct. Any criticisms are welcomed!

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  • $\begingroup$ Easier approach would be to see that $\sin (nx) = \frac{e^{inx}-e^{-inx}}{2} $, and then see it's a geometric series. $\endgroup$ – Jakobian Jun 16 '18 at 10:59
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\begin{align} S(x) &=1+ 2\sin x+ \dfrac{2^{2}\sin 2x}{2!}+ \dfrac{2^{3}\sin 3x}{3!}+ \dots \\ &=1+{\bf Im}\sum_{k=1}^\infty\dfrac{2^ke^{ikx}}{k!} \\ &={\bf Im}\left(1+\sum_{k=1}^\infty\dfrac{(2e^{ix})^k}{k!}\right) \\ &={\bf Im}e^{2e^{ix}} \\ &=e^{2\cos x}\sin(2\sin x) \end{align}

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