Question

If $z_n=\cos \frac{\pi }{(2n+1)(2n+3)}+i\sin \frac{\pi }{(2n+1)(2n+3)}$, then find the value of $\underset{n\to \infty }{lim}(z_1.z_2.z_3.........z_n)?$

Answer 1

We know that;

$cos\left(\theta \right)+isin\left(\theta \right)=e^{i\theta }$ -by euler's formula.

Now, for convenience let :

${\theta }_n=\frac{\pi }{(2n+1)(2n+3)}$

Hence,

$z_n=e^{i{\theta }_n}$

Now,

$\underset{n\to \infty }{lim}{(z}_1.z_2.z_3....z_n)=\underset{n\to \infty }{lim}{e}^{i{\theta }_1+i{\theta }_2+{i\theta }_3+...+i{\theta }_n}$

$=e^{i{\sum }_1^{\infty }{\theta }_n}$

Now all we need to do is solve for the summation of ${\theta }_n$.

${\sum }_1^{\infty }{\theta }_n={\sum }_1^{\infty }\frac{\pi }{(2n+1)(2n+3)}=\pi \sum \frac{1}{(2n+1)(2n+3)}=\frac{\pi }{2}{\sum }_1^{\infty }\frac{(2n+3)-(2n+1)}{(2n+1)(2n+3)}=\frac{\pi }{2}{\sum }_1^{\infty }\frac{1}{(2n+1)}-\frac{1}{(2n+3)}=\frac{\pi }{2}\left\{\right(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+.......\infty \}=\frac{\pi }{2}\{\frac{1}{3}+0+0+\mathrm{0.....}\}=\frac{\pi }{6}$

In the above calculations , we split the dfraction into partial dfractions and solve for the telescoping series as all terms after $\frac{1}{3}$ get cancelled out by the next term. Also, note that in the second step the numerator $(2n+3)-(2n+1)$ equals to $2$ which has been compensated by dividing the sum by $2$.

Now, let's substitute the value of the summation;

$\underset{n\to \infty }{lim}{(z}_1.z_2.z_3....z_n)=e^{i\frac{\pi }{6}}=cos\left(\frac{\pi }{6}\right)+isin\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}+i\frac{1}{2}$

Above, I have again re-applied the Euler's Formula.