How do you use the direct comparison test to determine if sume^(-n^2)en2 from [0,oo)[0,) 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)n=0en2=n=01en2

We see a_n=1/e^(n^2)an=1en2

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

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

Thus, since the larger series converges, so does the smaller series sum_(n=0)^ooe^(-n^2)n=0en2 by the Direct Comparison Test.