 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2/10/iii/solution

For

a_{n} = n^{2} e^{- α n}

, the ratio test gives

lim_{n \to \infty} \frac{a _{n + 1}}{a _{n}} = lim_{n \to \infty} (1 + \frac{1}{n})^{2} e^{- α} = e^{- α} .

This is less than one exactly when

α > 0

. For

α = 0

the terms are

n^{2}

, and for

α < 0

they grow exponentially; in both cases they fail to tend to zero. Consequently

convergent exactly for α > 0.

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