 # 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} remolatg 2020-11-10 Answered
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.)
$\sum _{k=1}^{\mathrm{\infty }}\frac{{x}^{2k}}{{4}^{k}}$
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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 $\left(-2,2\right)$

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