# What is the radius of convergence of $\sum_{n=0}^{\infty} 2^{-n^2}z^{n}$?

So far I have used the ratio test and obtained \begin{align*} \frac{a_{n+1}}{a_n} = & \left\vert\frac{1}{2^{(n+1)^2}}.\frac{2^{n^2}}{1}\right\vert \\ = &\frac{2^{n^2}}{2^{(n+1)^2}}. \end{align*} This equals $\frac{1}{2}4^{-n}$.

So, we obtain $2/3$ as $n\rightarrow\infty$ if $\vert z\vert <1$ and $n\rightarrow \infty$ if $\vert z \vert >1$.

Thus, radius of convergence is $3/2$.

Is this correct as I am sure I have made some mistakes?

• For the ratio test, you only need to look at $a_n$, you don't care so much for $z$. Just look at what $\frac{1}{2}4^{-n}$ tends to as $n$ gets large and take the reciprocal to get your radius of convergence. – muffle Nov 15 '13 at 8:07
• It looks as if you found the sum of a geometric series. The reason for the summation is not clear to me, the Ratio Test does not involve sums. Root Test is more natural in this case. Roughly speaking, the coefficients go down like a bat out of h***. There is convergence for all $x$. – André Nicolas Nov 15 '13 at 8:08
• Is the radius of convergence for this series $1$? And thank you, I will get the root test a go. – Levi Nov 15 '13 at 8:12
• The radius is $\infty$. – André Nicolas Nov 15 '13 at 8:13

Use the Root Test. Note that $\sqrt[n]{2^{-n^2}}$ has limit $0$. Thus the series converges for all $z$. (The Ratio Test gives the same result, but involves slightly more work.)