## Related questions with answers

Question

Find a power series representation for the function and determine the interval of convergence. \

$f(x)=\dfrac{1}{1+9 x^2}$

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 2Let's rewrite the function as a more familiar expression:

$\begin{align*} \frac{1}{1+9x^2} = \frac{1}{1-(-9x^2)} = \sum_{n=0}^\infty \left(-9x^2\right)^n = \sum_{n=0}^\infty (-9)^n \, x^{2n} \end{align*}$

The series converges for:

$\begin{align*} \left|-9x^2\right| < 1 \iff 9|x|^2 < 1 \iff |x|^2 < \frac{1}{9} \iff |x| < \frac{1}{3} \end{align*}$

Therefore, the interval of convergence is $\left(-\frac{1}{3},\frac{1}{3}\right).$

$\textit{Note:}$ We do not need to check the endpoints because we know the series diverges for $|x| \geq \frac{1}{3}$.

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