The coefficients of the power series $\displaystyle \sum_{n=0}^{\infty} a_n(x-2)^n$ satisfy $a_0=5$ and $a_n=\left(\dfrac{2n+1}{3n-1}\right)a_{n-1}$ for all $n\geq 1$. The radius of convergence of the series is

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We are given the recursive formula:

$$a_n=\left(\frac{2n+1}{3n-1} \right)a_{n-1}$$

This implies a geometric sequence with ratio $\dfrac{2n+1}{3n-1}$. This ratio approaches $\dfrac{2}{3}$ as $n$ approaches infinity.

Therefore,

$$\left| \frac{2}{3}(x-2) \right| \lt 1$$ $$-\frac{3}{2} \lt x-2 \lt \frac{3}{2}$$