Hello Bryn Originally Posted by

**Bryn** Hi,

Can someone explain to me why [sin(0.5*npi)]/n converges to 0.

The explanation I have been given is that since it is < 1/n and therefore can be made as small as we like for large enough n.

But I don't follow why it is smaller than 1/n. I guess if I understood this it would make sense that as n tends to infinity, the fn. tends to 0.

You are right to question this: it doesn't converge to $\displaystyle 0$; $\displaystyle \lim_{n\to 0}\;\frac{\sin(n\pi/2)}{n}=\frac{\pi}{2}$

It is well-known that $\displaystyle \lim_{x\to 0}\;\frac{\sin(x)}{x}=1$, so if we replace $\displaystyle x$ by $\displaystyle (n\pi/2), \frac{\sin(x)}{x}=\frac{\sin(n\pi/2)}{(n\pi/2)}= \frac{2}{\pi}\cdot\frac{\sin(n\pi/2)}{n}$, and we note that as $\displaystyle x \to 0, n\to0$.

Therefore $\displaystyle 1=\lim_{x\to 0}\;\frac{\sin(x)}{x}$

$\displaystyle =\lim_{n\to0}\frac{2}{\pi}\cdot\frac{\sin(n\pi/2)}{n}$

$\displaystyle =\frac{2}{\pi}\cdot\lim_{n\to0}\frac{\sin(n\pi/2)}{n}$

$\displaystyle \Rightarrow\lim_{n\to 0}\;\frac{\sin(n\pi/2)}{n}=\frac{\pi}{2}$

Grandad