Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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
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 $\vert z\vert\lt\sqrt3$.
