Problem involving summing exponential series:

Question

I can show the first part (i) (a), but the second part (b) i think it should be $S=\infty$ since the denominator is zero with that value of $\theta$. However, this is not the answer, any ideas?

Thanks.

Answer 1

For any $n$, $$1+1+1+1+1+1+1+1+1+1=10.$$ Alternatively, denoting $z=e^{i\theta}$, $$S=\frac{z^{\frac12}(z^{10}-1)}{z^{\frac12}-z^{-\frac12}}=\frac{z(z^{10}-1)}{z-1}.$$ By L'Hospital, $$\lim_{z\to1}\frac{z^{10}-1}{z-1}=\lim_{z\to1}10z^{9}.$$ Or, if you want to do it the hard way, $$\lim_{\theta\to2n\pi}\frac{e^{i\frac\theta2}(e^{i10\theta}-1)}{2i\sin\frac\theta2} =\lim_{\theta\to2n\pi}\frac{\frac i2e^{i\frac\theta2}(e^{i10\theta}-1)+i10e^{\frac i2\theta}e^{i10\theta}}{i\cos\frac\theta2}=\frac{\frac i2(-1)^n(1-1)+i10(-1)^n1}{i(-1)^n}=10.$$

Answer 2

If $\theta=2n\pi$, $e^{ki\theta}=e^{2\pi(nk)i}=1$ if $k\in\Bbb Z$. Thus we have $$S=10.$$