 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/20/a/ii/solution

For

a_{n} = n^{2} / n!

, the ratio test gives

\frac{a _{n + 1}}{a _{n}} = \frac{( n + 1 ) ^{2}}{n ^{2} ( n + 1 )} = \frac{n + 1}{n ^{2}} ⟶ 0 < 1.

Thus the series converges absolutely. As a check,

n^{2} = n (n - 1) + n

and the exponential power series give the sum

e + e = 2 e

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