Question

Series to functions Find the function represented by the following series, and find the interval of convergence of the series. (Not all these series are power series.)
\(\sum_{k=1}^\infty\frac{x^{2k}}{4^k}\)

Answer 1

We know that \(\sum_{k=0}^\infty x^k=\frac{1}{1-x}\)
Given function is \(\sum_{k=0}^\infty\frac{x^{2k}}{4^k}\)
It can be written as \(\sum_{k=0}^\infty\frac{x^{2k}}{4^k}=\sum_{k=0}^\infty(\frac{x^2}{4})^k\)
Comparing this with equation, we may write
\(\sum_{k=0}^\infty(\frac{x^2}{4})^k=\frac{1}{1-\frac{x^2}{4}}\)
\(=\frac{4}{4-x^2}\) Now, for the function to be convergent,
\(|\frac{x^2}{4}|<1\)</span>
\(\Rightarrow|x^2|<4\)</span>
So the solution is \(-2
Therefore, the given series can be represented as \(\frac{4}{4-x^2}\) and the interval of convergence is \((-2,2)\)