How do you use the direct comparison test to determine if $\sum {{e^{ - {n^2}}}} $ from $\left[ {0,\infty } \right)$ is convergent or divergent?

Hint: By using the definition of series for direct comparison test we have if the infinite series $\sum {{b_n}} $ converges and $0 \leqslant {a_n} \leqslant {b_n}$ for all sufficiently large $n$ , the infinite series $\sum {{a_n}} $ also converges and in the same way $\sum {{b_n}} $ diverges and $0 \leqslant {b_n} \leqslant {a_n}$ for all sufficiently large $n$ , the infinite series $\sum {{a_n}} $ also diverges. By using these we can arrive at the solution by comparison.

Complete step by step answer:
Here in this question, they have asked to find whether the given series is convergent or divergent by using the direct comparison test.
For the direct comparison test, we have to know the conditions when the series will become convergent or divergent.
If the infinite series $\sum {{b_n}} $ converges and $0 \leqslant {a_n} \leqslant {b_n}$ for all sufficiently large $n$ , the infinite series $\sum {{a_n}} $ also converges and in the same way $\sum {{b_n}} $ diverges and $0 \leqslant {b_n} \leqslant {a_n}$ for all sufficiently large $n$ , the infinite series $\sum {{a_n}} $ also diverges.
Now we have $\sum\limits_{n = 0}^\infty {{e^{ - {n^2}}}} $ which can be written as $\sum\limits_{n = 0}^\infty {\dfrac{1}{{{e^{{n^2}}}}}} $ by looking at the above discussion, we can write ${a_n} = \dfrac{1}{{{e^{{n^2}}}}}$.
In order to compare, we make use of the comparison sequence as
${b_n} = \dfrac{1}{{{e^n}}} = {\left( {\dfrac{1}{e}} \right)^n} \geqslant {a_n}$ for all $n$ on $\left[ {0,\infty } \right)$ . Now by looking at the denominator we can see that the denominator is smaller which is because the removal of squared $n$ gives a larger sequence.
We know that \[\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{e}} \right)}^n}} \] converges as it is a geometric series with the absolute value of the common ratio $\left| r \right| = \dfrac{1}{e} < 1$.

From the above discussion as the larger series that is \[\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{e}} \right)}^n}} \] converges $\sum\limits_{n = 0}^\infty {{e^{ - {n^2}}}} $ also converges by direct comparison test.

Note:
Whenever they ask this type of problem, first we need to know the definition or the condition for the direct comparison for the convergence and the divergence. If you know when the series becomes convergent or divergent then you can easily arrive at the correct answer.