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**Prove It** 1. Your series is $\displaystyle \displaystyle \sum_{n = 0}^{\infty}\left[\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n\right]$.

Using the ratio test

$\displaystyle \displaystyle \lim_{n \to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)[4(n+1)+1]}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)[2(n+1)+3]}x^{n+1}}{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n}\right|$

$\displaystyle \displaystyle = \lim_{n\to \infty}\left|\left[\frac{4(n+1)+1}{2(n+1)+3}\right]x\right|$

$\displaystyle \displaystyle = \lim_{n \to \infty}\left|\left(\frac{4n + 5}{2n + 5}\right)x\right|$

$\displaystyle \displaystyle = \lim_{n \to \infty}\left|\left(\frac{4 + \frac{5}{n}}{2 + \frac{5}{n}}\right)x\right|$

$\displaystyle \displaystyle = \left|\left(\frac{4}{2}\right)x\right|$

$\displaystyle \displaystyle = \left|2x\right|$.

The series is convergent when this limit $\displaystyle \displaystyle <1$. So

$\displaystyle \displaystyle |2x| < 1$

$\displaystyle \displaystyle -1 < 2x < 1$

$\displaystyle \displaystyle -\frac{1}{2} < x < \frac{1}{2}$.

So the radius of convergence is $\displaystyle \displaystyle \frac{1}{2}$, not $\displaystyle \displaystyle \frac{1}{6}$...

Now substitute each of the endpoints of this interval of convergence into your original series, and test the convergence of those series...