# Using the ratio test in Real Analysis

I am trying to prove that the following series diverges when $1\le |x|$

$$\sum_{n=0}^{\infty}(-1)^nx^{2n}$$

So applying the ratio test I got that,

$$\lim_{n\to\infty}|\frac{x^{2n+2}}{x^{2n}}|$$ $$\lim_{n\to\infty}|x^2|=x^2$$

So then how can I prove that it converges only on $(-1,1)?$

Ratio test says that if $$\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=L$$ Then

If $L<1$ the series converges absolutely

If $L=1$ the series can converge or diverge

If $L>1$ the series diverges.

Now using what you got, when is $L<1$, and if $L=1$ does the series converge?

• if x is equal to 1 then isn't it just $(-1)^n$ Mar 24, 2017 at 15:41
• @user104 You're right,is $\sum_{n=0}^\infty (-1)^n$ convergent? Mar 24, 2017 at 15:43
• no since it goes back and forth between 1 and -1, so then how about if x=-1? Mar 24, 2017 at 15:45
• @user104 You're right the general term oscillate between 1 and -1,if x=-1 then you get $(-1)^n\cdot(-1)^{2n}=(-1)^n$ Mar 24, 2017 at 15:48
• so then again in ossillates between those 2 values. I understand what you mean now, thanks a lo t for your help Mar 24, 2017 at 15:52

You need to apply the convergence condition. To guarantee convergence, that limit must be less than 1. So you have $x^2 < 1$. Go from there.

• I think the correct, general argument is: "to converge, the limit must be less than or equalt to 1. Mar 24, 2017 at 17:36
• That's a bit misleading, as this series certainly does not converge for L=1, and in general there is no guarantee that any arbitrary series will converge for L=1. Mar 25, 2017 at 18:00
• @el On the contrary: I think your argument is misleading as you wrote "must be less than one", which in general is false. It could be one and still converge. This is what I meant. Mar 25, 2017 at 18:22
• I have updated my wording to be more precise. Mar 25, 2017 at 19:37
• Nicely put this time. +1 Mar 25, 2017 at 20:12

Using the ratio test you will get $$\lim_\limits{n\to\infty}\bigg{|}\frac{(-1)^{n+1}x^{2(n+1)}}{(-1)^nx^{2n}}\bigg{|}$$ which simplifies to $$\lim_\limits{n\to\infty}|(-1)x^{2}|=\lim_\limits{n\to\infty}|x^{2}|=L$$

We know that $x$ is in the domain $(-1,1)$, so we can say the following: $$x\in(-1,1)\Rightarrow|x|<1\Rightarrow|x^2|<<1$$

Using what @kingW3 mentioned for the value of L: $$\lim_\limits{n\to\infty}|x^{2}|=L<<1$$

So, the original sum converges absolutely on the domain $(-1,1)$.

I think it is way simpler:

$$|x|\ge1\implies x^{2n}\ge1\implies (-1)^nx^{2n}\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0\implies \sum_{n=0}^\infty(-1)^nx^{2n}$$

cannot converge.