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}^inftyfrac{x^{2k}}{4^k}

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.)
$\sum _{k=1}^{\mathrm{\infty }}\frac{{x}^{2k}}{{4}^{k}}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Margot Mill

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

Not exactly what you’re looking for?
Jeffrey Jordon

Answer is given below (on video)

Jeffrey Jordon