\[\lim_{n\to \infty} \large\sum_{k=1}^{n} \frac{k^{3}+6k^{2}+11k+5}{ (k+3 )!}\]

Now let the above sum be expressed in the form \[\frac{a}{b}\] , where \(a\) and \(b\) are integers which are co-prime. Then find \(a+b\).

If you think the summation may not converge to a finite limit then type \(22\) as your answer

This was a problem given in one of our tests.

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