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

1 Answer
May 13, 2018

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.