Question

Calculus II Question

Answer 1

With some rewriting, you get

$\displaystyle \sum_{k=0}^\infty (-1)^k\frac{x^{k+2}}{4^k} = x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k$

Recall that for |x| < 1, you have

$\displaystyle \frac1{1-x} = \sum_{k=0}^\infty x^k$

So as long as |-x/4| = |x/4| < 1, or |x| < 4, your series converges to

$\displaystyle x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k = \frac{x^2}{1-\left(-\frac x4\right)} = \frac{x^2}{1+\frac x4} = \boxed{\frac{4x^2}{4+x}}$