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# 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=0}^infty(frac{x^2-1}{3})^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=0}^infty(frac{x^2-1}{3})^k

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Series asked 2020-12-09
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.)
$$\sum_{k=0}^\infty(\frac{x^2-1}{3})^k$$

## Answers (1) 2020-12-10
Given infinite series is:
$$\sum_{k=0}^\infty(\frac{x^2-1}{3})^k$$
The function representing the above series is:
$$f(x)=(\frac{x^2-1}{3})^k,k=0,1,2,3,...$$
To get the interval of convergence of the given series we will apply Ratio Test.
Applying Ratio Test we can say that the given series is convergent if:
$$\left|\frac{\left(\frac{x^2-1}{3}\right)^{k+1}}{\left(\frac{x^2-1}{3}\right)^k}\right|<1$$</span>
$$\left|\frac{x^2-1}{3}\right|<1$$</span>
$$\left|\frac{(x+1)(x-1)}{3}\right|<1$$</span>
$$\left|(x+1)(x-1)\right|<3$$</span>

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