$z_n=n\left\{1-\cos\left(\frac{\theta}{n}\right)-i\sin\left(\frac{\theta}{n}\right)\right\}$ converges or diverges.

Question: Find out if the sequence $$z_n=n\left\{1-\cos\left(\frac{\theta}{n}\right)-i\sin\left(\frac{\theta}{n}\right)\right\},\hspace{0.2 cm}\theta\text{ fixed},$$ converges or diverges. If it converges find out its limit.

Solution: $$\forall n\in\mathbb{N}$$, $$z_n$$ can be written as $$z_n=n\left\{1-\cos\left(\frac{\theta}{n}\right)\right\}+i\left\{-n\sin \left(\frac{\theta}{n}\right)\right\},$$ which implies that $$x_n:=\Re(z_n)=n\left\{1-\cos\left(\frac{\theta}{n}\right)\right\}$$ and $$y_n:=\Im(z_n)=-n\sin \left(\frac{\theta}{n}\right), \forall n\in\mathbb{N}.$$

Now $$\lim_{n\to\infty}x_n=\lim_{n\to\infty}n\left\{1-\cos\left(\frac{\theta}{n}\right)\right\}=\lim_{n\to\infty}\frac{1-\cos\left(\frac{\theta}{n}\right)}{\frac{1}{n}}=\lim_{n\to\infty} \frac{\sin\frac{\theta}{n}\left(-\frac{\theta}{n^2}\right)}{-\frac{1}{n^2}}=0,$$ and

$$\lim_{n\to\infty}y_n=\lim_{n\to\infty}-n\sin \left(\frac{\theta}{n}\right)=-\lim_{n\to\infty}\frac{\sin\left(\frac{\theta}{n}\right)}{\frac{1}{n}}=-\lim_{n\to\infty}\frac{\cos\frac{\theta}{n}\left(-\frac{\theta}{n^2}\right)}{-\frac{1}{n^2}}=-\theta.$$

Therefore, since, both $$(x_n)_{n\ge 1}$$ and $$(y_n)_{n\ge 1}$$ are convergent sequences, implies that $$(z_n)_{n\ge 1}$$ is also convergent and $$\lim_{n\to\infty}z_n=\lim_{n\to\infty}x_n+i\lim_{n\to\infty}y_n=0+i(-\theta)=-i\theta.$$

Is this solution correct and rigorous enough? What are the other ways to solve this problem?

• You might want to justify $\lim x_n=0$ and $\lim y_n=-\theta$ in a bit more detail. Also you could write $z_n=n(1-e^{i\theta/n})$. – Angina Seng Jun 27 '20 at 5:21
• @AnginaSeng, I have justified the limits. Please check if they are enough or not. – Sanket Biswas Jun 27 '20 at 5:56

2 Answers

Note that\begin{align} z_n & = n\left(1-\cos\left(\frac{\theta}{n}\right)-i\sin\left(\frac{\theta}{n}\right)\right) \\ & = 2n\sin\left(\frac{\theta}{2n}\right)\left(\sin\left(\frac{\theta}{2n}\right)-i\cos\left(\frac{\theta}{2n}\right)\right) \\ & = -i\theta\dfrac{\sin\left(\theta/2n\right)}{\left(\theta/2n\right)}\exp\left(\frac{i\theta}{2n}\right) \end{align}

Hence $$\lim_{n\to\infty}z_n=-i\theta .$$

\begin{align} \lim_{n\to \infty}z_n & =\lim_{n\to \infty} n\left(1-\cos\left(\frac{\theta}{n}\right)-i\sin\left(\frac{\theta}{n}\right)\right) \\ & = \lim_{n\to \infty} \frac{\left(1-\cos\left(\frac{\theta}{n}\right)-i\sin\left(\frac{\theta}{n}\right)\right)}{\frac{1}{n}} \\ & = \lim_{t\to 0} \frac{\left(1-\cos\left({t\theta}\right)-i\sin\left({t\theta}\right)\right)}{t}\\ & =\lim_{t\to 0} {\left(\theta\sin\left({t\theta}\right)-i\theta\cos\left({t\theta}\right)\right)}\text{ (via L'Hospital's Rule)}\notag\\ & =-i\theta \end{align}