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How can I find the interval of convergence of:

$$\sum_{k=1}^{\infty} \dfrac {1\cdot 3\cdot 5\cdot \ldots \cdot (2k-1)}{2\cdot 5 \cdot 8 \cdot \ldots \cdot (3k-1)}x^k$$

Thought it can be solved with using integral but I couldn't solve.

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Person X
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    Welcome to stackexchange. Please [edit] the question to show us what you tried -perhaps the ratio test. No images - use mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker May 14 '19 at 22:17
  • Have you tried the ratio test? – user170231 May 14 '19 at 22:22

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Hint: By the ratio test,

$$\sum_{k\ge1}\frac{1\cdot3\cdot5\cdot\cdots\cdot(2k-1)}{2\cdot5\cdot8\cdot\cdots\cdot(3k-1)}x^k$$

converges if

$$\lim_{k\to\infty}\left|\frac{\frac{1\cdot3\cdot5\cdot\cdots\cdot(2k-1)\cdot(2k+1)}{2\cdot5\cdot8\cdot\cdots\cdot(3k-1)\cdot(3k+2)}x^{k+1}}{\frac{1\cdot3\cdot5\cdot\cdots\cdot(2k-1)}{2\cdot5\cdot8\cdot\cdots\cdot(3k-1)}x^k}\right|<1$$

user170231
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