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>

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

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>