Question

Suppose |x| < 1. Prove that the series(from n= 0 to infinity) of x^(2n) converges and find it's sum.

Answer 1

Question for you: are you familiar with the fact that for \(|x|<1\), you have
\[\sum_{k=1}^\infty x^k=\frac{1}{1-x}~~?\]
Second question: What happens if you replace \(x\) with \(x^2\)?

Answer 2

I was not aware actually. I'm not entirely sure. I know that it will look similar to a particular integral (like how the integral of 1/(1+x^2) dx is arctan(x)

Answer 3

Correction to my first post: The series should start at \(k=0\), not \(k=1\).

Anyway, for the first part, you can consider any of the convergence tests you're familiar with. The ratio test would be particularly easy, so let's give that a shot. You know that a series \(\sum\limits_{n=0}^\infty a_n\) converges if \(\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1\). With \(a_n=x^{2n}\), you have
\[\lim_{n\to\infty}\left|\frac{x^{2(n+1)}}{x^{2n}}\right|=\lim_{n\to\infty}|x^2|=|x|^2\]When is this limit less than \(1\)? In other words, for which values of \(x\) is \(|x|^2<1\)?

Answer 4

To find its sum, knowing the particular result from my first post would simplify matters greatly, but if you're not familiar with it, there's no harm in deriving it.

Consider the series
\[\sum\limits_{n=0}^\infty x^{2n}=1+x^2+x^4+x^6+\cdots\]Since every term after the first contains a factor of \(x^2\), we can write
\[\sum\limits_{n=0}^\infty x^{2n}=1+x^2\left(1+x^2+\cdots\right)\]and the group of terms in the parentheses is exactly the original infinite sum, which means we have
\[\sum\limits_{n=0}^\infty x^{2n}=1+x^2\sum\limits_{n=0}^\infty x^{2n}\]and from here you can solve for \(\sum\limits_{n=0}^\infty x^{2n}\).