# Converging sequence.

• Nov 25th 2008, 10:52 AM
Showcase_22
Converging sequence.
Quote:

$\sum \frac{1}{n}sin \frac{n \pi}{2}$

Prove that this sequence converges.
Tackling the $sin \frac{n \pi}{2}$ first we get:

when n=1 $sin \frac{ \pi}{2}=1$

n=2 $sin \pi=0$

n=3 $sin \frac{3 \pi}{2}=-1$

n=4 $sin 2 \pi=0$

When n is even the term being added onto the sequence is 0.

$sin \frac{n \pi}{2}=1$ when n=1,5,.....,4n-3

$sin \frac{n \pi}{2}=-1$ when n=3,7,..........,4n-1

I have all this information but I don't know what to do with it!

I was thinking of substituting 4n-3 and 4n-1 instead of $sin \frac{n \pi}{2}$ but this only results in a divergent sequence. =(
• Nov 25th 2008, 11:09 AM
Moo
Hello,

Well, that's a good point oO
So note that it's 0 if n is even.
Hence you can keep only the odd ones.

So write $n=2k+1$

Now, if k is even, it's in the form 4k'+1 and $\sin \frac{n \pi}{2}=1=(-1)^{\color{red}even}$
and if k is odd, it's in the form 4k'-1 and $\sin \frac{n \pi}{2}=-1=(-1)^{\color{red}odd}$

can you see where I want to come ?

Try to see what $\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$ gives (Nod)
• Nov 25th 2008, 11:19 AM
Showcase_22
I can and I have succesfully completed question 6 out of 17!!

I would never have seen the substitution n=2k+1. It is a little ingenious.(Nerd)

Quote:

Originally Posted by Moo

Try to see what $\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$ gives (Nod)

Right, we haven't been shown yet how to determine what it converges to. I can show you it converges though (which is a little easy, I know):

$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$= $-\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}$

and by the alternating series test this converges since $\frac{1}{2k+1} \rightarrow 0 \ as \ k \rightarrow 0$

So how do you figure out what it converges to?
• Nov 25th 2008, 11:26 AM
Moo
Quote:

Originally Posted by Showcase_22
I can and I have succesfully completed question 6 out of 17!!

I would never have seen the substitution n=2k+1. It is a little ingenious.(Nerd)

Right, we haven't been shown yet how to determine what it converges to. I can show you it converges though (which is a little easy, I know):

$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$= $-\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}$

and by the alternating series test this converges since $\frac{1}{2k+1} \rightarrow 0 \ as \ k \rightarrow 0$

and that it is decreasing ^^

Quote:

So how do you figure out what it converges to?
I don't know, you're only asked to prove it converges (Surprised)

I said "try to see what..." to give you which series you were looking for :p
• Nov 25th 2008, 11:44 AM
Plato
Look at the series for the $\arctan(x)$ when $x=1$.
• Nov 25th 2008, 11:49 AM
Mathstud28
Quote:

Originally Posted by Showcase_22
I can and I have succesfully completed question 6 out of 17!!

I would never have seen the substitution n=2k+1. It is a little ingenious.(Nerd)

Right, we haven't been shown yet how to determine what it converges to. I can show you it converges though (which is a little easy, I know):

$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$= $-\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}$

and by the alternating series test this converges since $\frac{1}{2k+1} \rightarrow 0 \ as \ k \rightarrow 0$
so what does the series converge to?

Note that $\forall{x}\backepsilon|x|<1~\frac{1}{1+x^2}=\sum_{ n=0}^{\infty}(-1)^nx^{2n}$

So notice that your sum is equivalent to

\begin{aligned}\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}&=\sum_{n=0}^{\infty}\int_0^1(-1)^nx^{2n}dx\\
&=\int_0^1\sum_{n=0}^{\infty}(-1)^nx^{2n}dx\\
&=\int_0^1\frac{dx}{1+x^2}
\end{aligned}

EDIT: Plato beat me...sorry