power series

**Showcase_22** · May 16th 2009, 11:23 AM

Find the range of values of values of x for which the series $\sum_{n=0}^{\infty} \sin^{2n} (x)$ and $\sum_{n=0}^{\infty} 2n \sin ^{2n-1}(x)$ converge, and find an expression for their sums, carefully justifying your answers.

I've calculated the radius of convergence for $\sum_{n =0}^{\infty} \sin^{2n} x$ in two different ways and I get two different answers.

The first way:

Let $a_n= \sin^{2n} x$ .
Therefore $limsup_{n \rightarrow \infty} | \sin^{2n} x|^{\frac{1}{n} }=limsup_{n \rightarrow \infty} | \sin^2 x|=1$

Hence $R=\frac{1}{1}=1$ .

Alternatively, using the ratio test:

$\left| \frac{a_{n+1}}{a_n} \right|=\left|\frac{\sin^{2n+2} x}{ \sin^{2n} x} \right|=\left|\sin^2 (x) \right|< 1 \Leftrightarrow \sin x<1 \Rightarrow x<\frac{\pi}{2}$

This tells me that $R=\frac{\pi}{2}$ .
Which one is correct.

For $\sum_{n=0}^{\infty} 2n \sin^{2n-1} x$ :

$\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{2(n+1) \sin^{2n+1} x}{2n \sin^{2n-1} x } \right|= \left| \frac{n+1}{n} \sin^2 x \right| \rightarrow |sin^2 x|<1 \Rightarrow \ x< \frac{\pi}{2}$

The main problem I have now is trying to find the expression for the sums. Can someone show me how to do that?

**putnam120** · May 16th 2009, 12:27 PM

Well the first one is just a geometric series with common ration $\sin(x)$ . So you need $|\sin(x)|<1\Longrightarrow -\frac{\pi}{2}<x<\frac{\pi}{2}$ .

Additionally, the root test states that the series converges absolutely if $\limsup \sqrt[n]{|a_n|}<1$ . So using what you have above, you will arrive at the same result.

**HallsofIvy** · May 16th 2009, 02:19 PM

Originally Posted by Showcase_22

I've calculated the radius of convergence for $\sum_{n =0}^{\infty} \sin^{2n} x$ in two different ways and I get two different answers.

The first way:

Let $a_n= \sin^{2n} x$ .
Therefore $limsup_{n \rightarrow \infty} | \sin^{2n} x|^{\frac{1}{n} }=limsup_{n \rightarrow \infty} | \sin^2 x|=1$

Hence $R=\frac{1}{1}=1$ .

Alternatively, using the ratio test:

$\left| \frac{a_{n+1}}{a_n} \right|=\left|\frac{\sin^{2n+2} x}{ \sin^{2n} x} \right|=\left|\sin^2 (x) \right|< 1 \Leftrightarrow \sin x<1 \Rightarrow x<\frac{\pi}{2}$

This tells me that $R=\frac{\pi}{2}$ .
Which one is correct.

For $\sum_{n=0}^{\infty} 2n \sin^{2n-1} x$ :

$\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{2(n+1) \sin^{2n+1} x}{2n \sin^{2n-1} x } \right|= \left| \frac{n+1}{n} \sin^2 x \right| \rightarrow |sin^2 x|<1 \Rightarrow \ x< \frac{\pi}{2}$

The main problem I have now is trying to find the expression for the sums. Can someone show me how to do that?

**shawsend** · May 16th 2009, 04:54 PM

$\sum_{n=0}^{\infty}2n \sin^{2n-1}(x)\equiv \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n$

Now, what is:

$\sum_{n=0}^{\infty} n k^n$

**putnam120** · May 16th 2009, 07:28 PM

Well for $|x|<1$ you have that $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ . So doing the taking the derivative you have that $\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$ . From here you should see how to get to $\sum nx^n$

**Showcase_22** · May 17th 2009, 04:56 AM

Additionally, the root test states that the series converges absolutely if . So using what you have above, you will arrive at the same result.

That's one of my problems.

Is it true that $limsup_{n \rightarrow \infty} \sin^2 x =1$ ?

Therefore the radius of convergence R should be $R=\frac{1}{1}=1$ .

However, when I use the ratio test (as seen in previous posts) I get $\frac{\pi}{2}$ . It seems odd that i'm getting two different answers

**putnam120** · May 17th 2009, 08:29 AM

Originally Posted by Showcase_22

That's one of my problems.

Is it true that $limsup_{n \rightarrow \infty} \sin^2 x =1$ ?

Therefore the radius of convergence R should be $R=\frac{1}{1}=1$ .

However, when I use the ratio test (as seen in previous posts) I get $\frac{\pi}{2}$ . It seems odd that i'm getting two different answers

No, you are using $\limsup \sqrt[n]{|a_n|}=1$ , this is different from $\limsup\sqrt[n]{|a_n|}<1$ , that is where you problem is coming from.

**Showcase_22** · May 17th 2009, 11:34 AM

No, you are using , this is different from , that is where you problem is coming from.

Sorry but I can't see the difference

On a lighter note, he's my working on the sums:

$\sum_{n=0}^{\infty}\sin^{2n} x=\frac{1}{1-\sin^2 x}=\frac{1}{\cos^2 x}=\sec^2 x$

$\sum_{n=0}^{\infty} 2n \sin^{2n-1} x=\frac{d}{dx} \left(\frac{1}{\cos^2 x}\right)=\frac{\cos^2 x (0)+2 \cos x \sin x}{\cos^4 x}=\frac{2 \sin x}{\cos^3 x}=2 \tan x \sec^2 x$

Finally, Shawsend i'm quite interested in solving what you've put. Unfortunately, what is the "a" in your $ \sum_{n=0}^{\infty}2n \sin^{2n-1}(x)\equiv \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n $ ?

**shawsend** · May 17th 2009, 01:59 PM

Originally Posted by Showcase_22

Finally, Shawsend i'm quite interested in solving what you've put. Unfortunately, what is the "a" in your $ \sum_{n=0}^{\infty}2n \sin^{2n-1}(x)\equiv \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n $ ?

$a=\sin(x)$

**putnam120** · May 17th 2009, 02:01 PM

In what you are using there is an equal sign, $=$ , however it should be replaces with less than, $<$ .

I am sure that we would both agree that the following are different:
$x=1$ and $x<1$ .

**shawsend** · May 17th 2009, 03:06 PM

I should have written:

$\sum_{n=0}^{\infty}2n \sin^{2n-1}(x)= \frac{2}{a}\sum_{n=0}^{\infty} n(a^2)^n;\quad a=\sin(x) $

**Showcase_22** · May 18th 2009, 09:06 AM

$ \sum_{n=0}^{\infty} n k^n $

We know that $ \sum_{n=0}^\infty x^n=\frac{1}{1-x} $ and $ \sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2} $ .

This implies that $ \sum_{n=0}^{\infty} n k^n =\frac{k}{(k-1)^2}$

**putnam120** · May 18th 2009, 10:25 AM

Well you might want to check the lower index on $\sum_{n=0}^\infty nx^{n-1}$ , but that's the general idea.

**Showcase_22** · May 18th 2009, 12:39 PM

why?

What's wrong with it?

**putnam120** · May 18th 2009, 02:39 PM

The case where x=0 would cause some problems since how would you define $0\cdot 0^{-1}$ ? But I suppose the same can be said about the original sum with its $0^0$ term.

Thread: power series

LinkBack

Thread Tools

Display

power series

Similar Math Help Forum Discussions

Exploiting geometric series with power series to find Taylors series

question about power series, taylor series, maclaurin series

Formal power series & Taylor series

Multiplying power series - Maclaurin series

Finding Power Series and Evaluating as a Power Series

Search Tags