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How do I find the interval of convergence of this series; $$\sum \frac{x^{2k+1}}{3^{k-1}}$$

I have been told that the answer is $$\ -\sqrt{3}<x<\sqrt{3}$$ But I am unsure of where the square root has come from.

Can anyone help explain this to me?

Thank You

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    $\begingroup$ You can apply either the Root or the Ratio Test here. With the Ratio Test it may be a little easier to see what's going on, so I suggest you try it. $\endgroup$ – zipirovich Jul 17 at 3:08
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Hint: The convergence interval of $\sum t^k$ is $(-1,1)$. Letting $t=\frac{x^2}3$ gives the desired answer.

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$$\sum \frac{x^{2k+1}}{3^{k-1}}=\sum \frac{x^{2k+1}}{{\sqrt3}^{2k-2}}=3\sqrt3\sum \left(\frac x {\sqrt 3}\right)^{2k+1}$$

Let $$a_k=\left(\frac x {\sqrt 3}\right)^{2k+1}\implies \frac{a_{k+1}}{a_k}=\frac {x^2}3 \implies R= ???$$

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By the Cauchy-Hadamard theorem, $r=\dfrac1{\limsup_{k\to\infty}\sqrt[2k+1]{3^{-(k-1)}}}=\dfrac 1{\limsup_{k\to\infty}3^{\frac{-(k-1)}{2k+1}}}=\dfrac1{3^{-\frac12}}=\sqrt3$.

This holds in fact for any complex $z$ with $\vert z\vert\lt\sqrt3$.

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