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Tagged: calculus
Clearly \( a_n = \frac{n^2}{n!} = \frac{ n}{(n-1)!} = \frac{n-1 +1}{(n-1)!} = \frac{1}{(n-2)!} + \frac{1}{(n-1)!} \)
Hence \( \sum a_n = \sum \frac{1}{(n-2)!} + \sum \frac{1}{(n-1)!} = 2e \)
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