Ratio Test — Question 3

*Question 3 Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ converges or diverges.

1. Compute and simplify $a_{n+1}/a_n$.

2. Apply the Ratio Test.

3. Explain how the limit compares polynomial and exponential growth.

*Question 3 – Solution

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{2^{n+1}}\frac{2^n}{n^2}=\frac12\left(1+\frac1n\right)^2.$$

$\displaystyle L=\frac12(1)^2=\frac12<1$.

The series converges absolutely. Its consecutive terms eventually behave as though multiplied by $1/2$. The polynomial factor changes the ratio by a factor tending to $1$, while $2^n$ supplies the decisive factor $1/2$.