`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.

Comments

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