# Find the interval of convergence of $\sum \frac{x^{2k+1}}{3^{k-1}}$

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} But I am unsure of where the square root has come from.

Can anyone help explain this to me?

Thank You

• 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. Jul 17, 2019 at 3:08

Hint: The convergence interval of $$\sum t^k$$ is $$(-1,1)$$. Letting $$t=\frac{x^2}3$$ gives the desired answer.
$$\sum \frac{x^{2k+1}}{3^{k-1}}=\sum \frac{x^{2k+1}}{{\sqrt3}^{2k-2}}=3\sqrt3\sum \left(\frac x {\sqrt 3}\right)^{2k+1}$$
Let $$a_k=\left(\frac x {\sqrt 3}\right)^{2k+1}\implies \frac{a_{k+1}}{a_k}=\frac {x^2}3 \implies R= ???$$
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 $$\vert z\vert\lt\sqrt3$$.