# How do you use the direct comparison test to determine if #sume^(-n^2)# from #[0,oo)# is convergent or divergent?

Converges by Direct Comparison Test.

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To use the direct comparison test to determine the convergence or divergence of the series ( \sum e^{-n^2} ) from ( n = 0 ) to infinity, we can compare it to another series whose convergence or divergence is known.

Since ( e^{-n^2} ) is always positive, we can compare it to ( \frac{1}{n^2} ), which is also always positive.

We know that ( e^{-n^2} < \frac{1}{n^2} ) for all ( n ) greater than or equal to 1, because ( e^{-n^2} ) decreases more rapidly than ( \frac{1}{n^2} ).

Since the series ( \sum \frac{1}{n^2} ) is convergent (it's a p-series with ( p = 2 > 1 )), and ( e^{-n^2} < \frac{1}{n^2} ) for all ( n \geq 1 ), by the direct comparison test, ( \sum e^{-n^2} ) is also convergent.

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