lim n → ∞ ∑ k = 1 n ( k + 3 )! k 3 + 6 k 2 + 11 k + 5 =

k 3 + 6 k 2 + 11 k + 5 = ( k + 1 ) ( k + 2 ) ( k + 3 ) − 1 , so ( k + 3 )! k 3 + 6 k 2 + 11 k + 5 = k ! 1 − ( k + 3 )! 1 . The sum telescopes: ∑ k = 1 n ( k ! 1 − ( k + 3 )! 1 ) = 1 ! 1 + 2 ! 1 + 3 ! 1 − ( ( n + 1 )! 1 + ( n + 2 )! 1 + ( n + 3 )! 1 ) → 1 + 2 1 + 6 1 = 3 5 .

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

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Solution:

$k^{3} + 6 k^{2} + 11 k + 5 = (k + 1) (k + 2) (k + 3) - 1$ , so $\frac{k ^{3} + 6 k ^{2} + 11 k + 5}{( k + 3 )!} = \frac{1}{k !} - \frac{1}{( k + 3 )!}$ . The sum telescopes: $\sum_{k = 1}^{n} (\frac{1}{k !} - \frac{1}{( k + 3 )!}) = \frac{1}{1 !} + \frac{1}{2 !} + \frac{1}{3 !} - (\frac{1}{( n + 1 )!} + \frac{1}{( n + 2 )!} + \frac{1}{( n + 3 )!}) \to 1 + \frac{1}{2} + \frac{1}{6} = \frac{5}{3}$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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