# Convergence of a series

• Jul 29th 2010, 01:45 PM
rebecca
Convergence of a series
Hi, i need help in the following question:

I need to prove that there are a_0, a_1,... in R and r>0 such that the series ∑(a_n)x^n converge when |x|< r, and such that ∑(a_n)x^n = arctanx for every |x|< r

What should i do here? Use the Taylor series of arctanx or something?

Thanks for any help, and sorry for the mess :)
• Jul 29th 2010, 02:19 PM
Vlasev
I think you need to prove that Arctan(x) has a convergent power series on the given domain. For example when does a function have a Taylor series?
• Jul 29th 2010, 05:36 PM
Prove It
Well the way I would think about it is to note that

$\frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}$.

You should know that for a geometric series $\sum_{n = 0}^{\infty}ar^n$, its infinite sum is $\frac{a}{1 - r}$ provided $|r| < 1$.

I think you can see that if you write $\frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \frac{a}{1 - r}$ then you can see $a = 1, r = -x^2$.

So $\frac{1}{1 + x^2}$ is the closed form of $\sum_{n = 0}^{\infty}1(-x^2)^n = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$. This is provided of course that $|r| < 1$.

So $|-x^2| < 1$

$|x|^2 <1$

$|x| < 1$.

So we have

$\frac{1}{1 + x^2} = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$ for $|x| < 1$

$\frac{d}{dx}(\arctan{x}) = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$ for $|x| < 1$

$\arctan{x} = \int{\sum_{n = 0}^{\infty}(-1)^nx^{2n}\,dx}$ for $|x| < 1$

$\arctan{x} = \sum_{n = 0}^{\infty}\int{(-1)^nx^{2n}\,dx}$ for $|x|<1$

$\arctan{x} = \sum_{n = 0}^{\infty}\frac{(-1)^nx^{2n + 1}}{2n + 1}$ for $|x| < 1$.
• Jul 29th 2010, 05:55 PM
Also sprach Zarathustra
Great solution!
• Jul 30th 2010, 03:28 AM
Also sprach Zarathustra
Quote:

Originally Posted by Prove It
Well the way I would think about it is to note that

$\frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}$.

You should know that for a geometric series $\sum_{n = 0}^{\infty}ar^n$, its infinite sum is $\frac{a}{1 - r}$ provided $|r| < 1$.

I think you can see that if you write $\frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \frac{a}{1 - r}$ then you can see $a = 1, r = -x^2$.

So $\frac{1}{1 + x^2}$ is the closed form of $\sum_{n = 0}^{\infty}1(-x^2)^n = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$. This is provided of course that $|r| < 1$.

So $|-x^2| < 1$

$|x|^2 <1$

$|x| < 1$.

So we have

$\frac{1}{1 + x^2} = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$ for $|x| < 1$

$\frac{d}{dx}(\arctan{x}) = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$ for $|x| < 1$

$\arctan{x} +C= \int{\sum_{n = 0}^{\infty}(-1)^nx^{2n}\,dx}$ for $|x| < 1$ (*)

$\arctan{x} = \sum_{n = 0}^{\infty}\int{(-1)^nx^{2n}\,dx}$ for $|x|<1$

$\arctan{x} = \sum_{n = 0}^{\infty}\frac{(-1)^nx^{2n + 1}}{2n + 1}$ for $|x| < 1$.

Maybe... IN(*) you should prove that C=0 by substituting x=0 ?