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 \((-2,2)\)

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 \((-2,2)\)