Ratio Test — Question 8

*Question 8 Analyze $\displaystyle\sum_{n=1}^{\infty}\frac{2^{n^2}}{3^{n^2}}$. Rewrite its term using a single base, compute the consecutive-term ratio, and classify the series.

*Question 8 – Solution

$\displaystyle a_n=(2/3)^{n^2}$.

$$\frac{a_{n+1}}{a_n}=\left(\frac23\right)^{(n+1)^2-n^2}=\left(\frac23\right)^{2n+1}.$$ Because $0<2/3<1$, this expression tends to $0$.

The ratio limit is $L=0<1$, so the series converges. Its quadratic exponent makes the terms decay faster than those of $\sum(2/3)^n$.