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.