<s>[INST] $\frac37x + \frac{3\times6}{7\times10}x^2 + \frac{3\times6\times9}{7\times10\times13}x^3+ \cdots$ Convergence of the series for $x>0$:
$$\frac37x + \frac{3\times6}{7\times10}x^2 + \frac{3\times6\times9}{7\times10\times13}x^3+ \cdots$$
The general tem  is $a_n = \frac{3\times6\times \cdots \times(3+(n-1)3)}{7\times10 \times\cdots \times(7+(n-1)3)} x^n$.
I am thinking of using Raabe's Test. but cannot proceed.
We have to find $$\lim _{n \to \infty} n\left(\frac{a_n}{a_{n+1} } -1\right) = \lim _{n \to \infty} n \left(\frac{7+3n}{3+3n} \frac1x -1\right)$$
so the limit depends upon $x$.
 [/INST] $$a_n = \frac{3\times6\times \cdots \times(3+(n-1)3)}{7\times10 \times\cdots \times(7+(n-1)3)} x^n$$
Then $$\frac{a_{n+1}}{a_n} =\frac{3n+3}{3n+7} x = \frac{3+\frac3n}{3 + \frac7n}x$$
From Ratio test we can conclude that when $|x| <1$ the series is convergent and when $|x|>1$ the series is divergent. Ratio Test can't conclude anything for $x = 1$. Here we use Raabe's Test.
$$\lim _{n \to \infty} n\left(\frac{a_n}{a_{n+1} } -1\right) = \lim _{n \to \infty} n \left(\frac{7+3n}{3+3n}  -1\right) = \lim _{n \to \infty} \frac{4n}{3+3n} = \frac43 >1$$ and hence the series is convergent.
</s>