Question

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

Answer 1

Converges by Direct Comparison Test.

Explanation:

We have

`sum_(n=0)^ooe^(-n^2)=sum_(n=0)1/e^(n^2)`

We see `a_n=1/e^(n^2)`

For the comparison sequence, we'll use `b_n=1/e^n=(1/e)^n>=a_n` for all `n` on `[0, oo),` as we have a smaller denominator (due to removing the squared `n`) and therefore a larger sequence.

We know `sum_(n=0)^oo(1/e)^n` converges, as it's a geometric series with the absolute value of the common ratio `|r|=1/e<1`.

Thus, since the larger series converges, so does the smaller series `sum_(n=0)^ooe^(-n^2)` by the Direct Comparison Test.