# Math Help - Convergence

1. ## Convergence

Find the interval of convergence for f(x) = x-(x^2/2)+(x^3/3)-(x^4/4)+.....

2. Hello,

Originally Posted by jules027
Find the interval of convergence for f(x) = x-(x^2/2)+(x^3/3)-(x^4/4)+.....
$f(x)=-\sum_{n=1}^\infty \frac{(-1)^n}{n} x^n$

According to the alternate series test :
- $\frac{|x|^n}{n}=a_n$ should be decreasing
- $\lim_{n \to \infty} \frac{|x|^n}{n}=0$

The first one... We want : $\frac{a_{n+1}}{a_n}<1$

$\frac{|x|^{n+1}}{n+1} \cdot \frac{n}{|x|^n} <1$

$|x| \cdot \frac{n}{n+1} <1$

$|x|<\frac{n+1}{n}$

But $\frac{n+1}{n}$ has a minimum in 1, when $n \to \infty$.

Hence $|x|$ has to be $<1$

And this satisfies the second point.

3. Originally Posted by Moo
Hello,

$f(x)=-\sum_{n=1}^\infty \frac{(-1)^n}{n} x^n$

According to the alternate series test :
- $\frac{|x|^n}{n}=a_n$ should be decreasing
- $\lim_{n \to \infty} \frac{|x|^n}{n}=0$

The second one is always true, whatever x is.

Now, the first one... We want : $\frac{a_{n+1}}{a_n}<1$

$\frac{|x|^{n+1}}{n+1} \cdot \frac{n}{|x|^n} <1$

$|x| \cdot \frac{n}{n+1} <1$

$|x|<\frac{n+1}{n}$

But $\frac{n+1}{n}$ has a minimum in 1, when $n \to \infty$.

Hence $|x|$ has to be $<1$
Going off of Moo's work we knwo so far that this is convergent on

$(-1,1)$

so we must test the endpoitns....putting int netgative one yields

$\sum_{n=1}^{\infty}\frac{(-1)^n\cdot(-1)^n}{n}=\sum_{n=1}^{\infty}\frac{1}{n}$

which diverges by either the integral or the p series test

$\int_1^{\infty}\frac{dx}{x}=\infty$

inputting 1 we get

$\sum_{n=1}^{\infty}\frac{(-1)^n1^n}{n}=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$

and since $\forall{n}>0,a_n>a_{n+1}$

and $\lim_{n\to\infty}a_n=0$

this converges

so the interval of convergence is

$(-1,1]$

4. Thank you so much guys!