sum_(n=0)^oo e^(-n^2) Use the Direct Comparison Test to determine the convergence or divergence of the series.

Direct comparison test is applicable when suma_n and sumb_n are both positive series for all n , such that a_n<=b_n . It follows that:
If sumb_n converges then suma_n converges
If suma_n diverges then sumb_n diverges.
sum_(n=0)^ooe^(-n^2)=sum_(n=0)^oo1/e^(n^2)
Let a_n=1/e^(n^2) and b_n=1/e^n=(1/e)^n
1/e^n>=1/e^(n^2)>0
sum_(n=0)^oo(1/e)^n is a geometric series with ratio r=1/e<1  
If |r|<1 then the geometric series converges.
Thus, by comparing the series sum_(n=0)^ooe^(-n^2) with the convergent geometric series sum_(n=0)^oo(1/e)^n , it converges.