# How do you use the direct comparison test to determine if sume^(-n^2) from [0,oo) is convergent or divergent?

May 13, 2018

Converges by Direct Comparison Test.

#### Explanation:

We have

${\sum}_{n = 0}^{\infty} {e}^{- {n}^{2}} = {\sum}_{n = 0} \frac{1}{e} ^ \left({n}^{2}\right)$

We see ${a}_{n} = \frac{1}{e} ^ \left({n}^{2}\right)$

For the comparison sequence, we'll use ${b}_{n} = \frac{1}{e} ^ n = {\left(\frac{1}{e}\right)}^{n} \ge {a}_{n}$ for all $n$ on $\left[0 , \infty\right) ,$ as we have a smaller denominator (due to removing the squared $n$) and therefore a larger sequence.

We know ${\sum}_{n = 0}^{\infty} {\left(\frac{1}{e}\right)}^{n}$ converges, as it's a geometric series with the absolute value of the common ratio $| r | = \frac{1}{e} < 1$.

Thus, since the larger series converges, so does the smaller series ${\sum}_{n = 0}^{\infty} {e}^{- {n}^{2}}$ by the Direct Comparison Test.