series exercise asked in نادى الرياضيات فلسطين the by ربيع العامودي

Exercise:

Show that, $\sum\nolimits_{k=1}^{n}{\sin \left( \frac{\pi k}{n} \right)=\cot \left( \frac{\pi }{2n} \right)}$

Solution: we know that $\sin \theta =\frac{{{e}^{i\theta }}-{{e}^{-i\theta }}}{2i}=\frac{1}{2i}\left( {{e}^{i\theta }}-{{e}^{-i\theta }} \right)$

So $\sin \left( \frac{\pi k}{n} \right)=\frac{1}{2i}\left( {{e}^{i\frac{\pi k}{n}}}-{{e}^{-i\frac{\pi k}{n}}} \right)\Leftrightarrow \sum\nolimits_{k=1}^{n}{\sin \left( \frac{\pi k}{n} \right)=\frac{1}{2i}\sum\nolimits_{k=1}^{n}{\left( {{e}^{i\frac{\pi k}{n}}}-{{e}^{-i\frac{\pi k}{n}}} \right)}}$

Observe that $\sum\nolimits_{k=1}^{n}{{{e}^{i\frac{\pi k}{n}}}}=\sum\nolimits_{k=1}^{n}{{{\left( {{e}^{i\frac{\pi }{n}}} \right)}^{k}}}$ which is an Geometric series

Let $x={{e}^{i\frac{\pi }{n}}}\Leftrightarrow \sum\nolimits_{k=1}^{n}{{{x}^{k}}}=x+{{x}^{2}}+....+{{x}^{n}}=\frac{x\left( 1-{{x}^{n}} \right)}{1-x}$ hence $\sum\nolimits_{k=1}^{n}{{{e}^{i\frac{\pi k}{n}}}}=\frac{{{e}^{i\frac{\pi }{n}}}\left( 1-{{e}^{i\pi }} \right)}{1-{{e}^{i\frac{\pi }{n}}}}$

We can reach the above result by using factorization process which left for the reader .

Also $\sum\nolimits_{k=1}^{n}{{{e}^{-i\frac{\pi k}{n}}}}=\sum\nolimits_{k=1}^{n}{{{\left( {{e}^{-i\frac{\pi }{n}}} \right)}^{k}}=\frac{{{e}^{-\frac{i\pi }{n}}}\left( 1-{{e}^{-i\pi }} \right)}{1-{{e}^{-\frac{i\pi }{n}}}}}$

But $1-{{e}^{i\pi }}=1-\cos \pi =2\,\,\And \,\,1-{{e}^{-i\pi }}=1-\cos \left( -\pi  \right)=1+1=2$

So  $\sum\nolimits_{k=1}^{n}{\sin \left( \frac{k\pi }{n} \right)}=\frac{1}{2i}\left[ \frac{2{{e}^{i\frac{\pi }{n}}}}{1-{{e}^{i\frac{\pi }{n}}}}-\frac{2{{e}^{-i\frac{\pi }{n}}}}{1-{{e}^{-i\frac{\pi }{n}}}} \right]=\frac{1}{i}\left[ \frac{{{e}^{i\frac{\pi }{n}}}}{1-{{e}^{i\frac{\pi }{n}}}}-\frac{{{e}^{-i\frac{\pi }{n}}}}{1-{{e}^{-i\frac{\pi }{n}}}} \right]$

                         $=\frac{1}{i}\left[ \frac{{{e}^{i\frac{\pi }{n}}}\left( 1-{{e}^{-i\frac{\pi }{n}}} \right)-{{e}^{-i\frac{\pi }{n}}}\left( 1-{{e}^{i\frac{\pi }{n}}} \right)}{\left( 1-{{e}^{i\frac{\pi }{n}}} \right)\left( 1-{{e}^{-i\frac{\pi }{n}}} \right)} \right]$

                         $=\frac{1}{i}\left[ \frac{{{e}^{i\frac{\pi }{n}}}-1-{{e}^{-i\frac{\pi }{n}}}+1}{\left( 1-{{e}^{i\frac{\pi }{n}}} \right)\left( 1-{{e}^{-i\frac{\pi }{n}}} \right)} \right]=\frac{1}{i}\left[ \frac{{{e}^{i\frac{\pi }{n}}}-{{e}^{-i\frac{\pi }{n}}}}{\left( 1-{{e}^{i\frac{\pi }{n}}} \right)\left( 1-{{e}^{-i\frac{\pi }{n}}} \right)} \right]$

But $\left( 1-{{e}^{i\frac{\pi }{n}}} \right)\left( 1-{{e}^{-i\frac{\pi }{n}}} \right)=1-{{e}^{-i\frac{\pi }{n}}}-{{e}^{i\frac{\pi }{n}}}+1=2-\left( {{e}^{-i\frac{\pi }{n}}}+{{e}^{i\frac{\pi }{n}}} \right)$

$=2-\left( \cos \frac{-\pi }{n}+i\sin \frac{-\pi }{n}+\cos \frac{\pi }{n}+i\sin \frac{\pi }{n} \right)=2\left( 1-\cos \frac{\pi }{n} \right)=4{{\sin }^{2}}\frac{\pi }{2n}$

But ${{e}^{i\frac{\pi }{n}}}-{{e}^{-i\frac{\pi }{n}}}=\cos \frac{\pi }{n}+i\sin \frac{\pi }{n}-\cos \frac{-\pi }{n}-i\sin \frac{-\pi }{n}=2i\sin \frac{\pi }{n}=4i\sin \frac{\pi }{2n}\cos \frac{\pi }{2n}$

Thus $\sum\nolimits_{k=1}^{n}{\sin \left( \frac{k\pi }{n} \right)}=\frac{4i\sin \frac{\pi }{2n}\cos \frac{\pi }{2n}}{i4{{\sin }^{2}}\frac{\pi }{2n}}=\cot \frac{\pi }{2n}$            Q.E.D