1. ## convergent sequences

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.

I am also confused from reading wolfram articles how to evaluate a geometric series and how to tell whether one converges or not. Could someone maybe explain to me how to evalaute:

a) sum between 2&9 of 10^-n
&
b) sum between 0&ininfity of 3^(-n/2)
&
c) determine whether convergent or not:
Between 1&infinity [(n-1)/n]

many thanks

2. 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.
Actually, it isn't smaller- it is "less than or equal to". That is, $\displaystyle sin(x)/n\le 1/n$, for any x whatsoever, because $\displaystyle -1\le sin(x)\le 1$ for all x.

I guess if I understood this it would make sense that as n tends to infinity, the fn. tends to 0.

I am also confused from reading wolfram articles how to evaluate a geometric series and how to tell whether one converges or not. Could someone maybe explain to me how to evalaute:

a) sum between 2&9 of 10^-n
&
b) sum between 0&ininfity of 3^(-n/2)
&
Those are both "geometric" sequences:
$\displaystyle \sum_{i=0}^n r^i= \frac{1- r^{n-1}}{1- r}$
And, of course, $\displaystyle \sum_{i= 2}^9= \sum_{i=0}^9- \sum_{i=0}^2$.

c) determine whether convergent or not:
Between 1&infinity [(n-1)/n]
Do you mean the sum? it's fairly easy to see that $\displaystyle \frac{n-1}{n}= \frac{n}{n}- \frac{1}{n}= 1- \frac{1}{n}$ and that sequence converges to 1 as n goes to infinity.

Do you know this theorem: If $\displaystyle \sum{n=0}^\infty a_n$ converges, then $\displaystyle \lim_{n\to\infty} a_n= 0$?
If you you don't recognise that, realize that after n= 100000000000, say, you are still adding numbers close to 1.

many thanks

3. 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
$\displaystyle 1=\lim_{x\to 0}\;\frac{\sin(x)}{x}$
$\displaystyle =\lim_{n\to0}\frac{2}{\pi}\cdot\frac{\sin(n\pi/2)}{n}$
$\displaystyle =\frac{2}{\pi}\cdot\lim_{n\to0}\frac{\sin(n\pi/2)}{n}$
$\displaystyle \Rightarrow\lim_{n\to 0}\;\frac{\sin(n\pi/2)}{n}=\frac{\pi}{2}$

Hello Bryn[SIZE=3]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 should be $\displaystyle \lim_{n\to \color{blue}\infty}\;\frac{\sin(n\pi/2)}{n}$
It should be $\displaystyle \lim_{n\to \color{blue}\infty}\;\frac{\sin(n\pi/2)}{n}$
Ah yes, of course, you're right! I suppose I thought it was $\displaystyle n\to 0$ because it's not so obvious what the limit is in that case.