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  • #22253
    Ashani Dasgupta
    Keymaster

    Limit of a sum

    #22254
    Ashani Dasgupta
    Keymaster

    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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