Just limiting sums

Calculus Level 5

lim ⁡ n → ∞ 1 ∑ 1 n r + 2 ∑ 1 n − 1 r + 3 ∑ 1 n − 2 r + ⋯ + n ⋅ 1 n 4 \lim_{n \to \infty} \frac{1 \displaystyle \sum_{1}^{n} r + 2 \displaystyle \sum_{1}^{n-1} r + 3 \displaystyle \sum_{1}^{n-2} r +\cdots + n \cdot 1}{n^{4}}

Find the reciprocal of the limit above.


The answer is 24.

Mohit Gupta by Mohit Gupta , 20, India

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2 solutions

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Ruofeng Liu
Nov 27, 2019

L = lim ⁡ n → ∞ ∑ k = 0 n k ( n + 1 − k ) ( n + 2 − k ) 2 n 4 L=\displaystyle{\lim_{n\to \infty}{\sum_{k=0}^{n}{\frac{k(n+1-k)(n+2-k)}{2n^4}}}}

= lim ⁡ n → ∞ 1 2 n ∑ k = 0 n k n ( 1 + 1 n − k n ) ( 1 + 2 n − k n ) =\displaystyle{\lim_{n\to \infty}{\frac{1}{2n}\sum_{k=0}^{n}{\frac{k}{n}(1+\frac{1}{n}-\frac{k}{n})(1+\frac{2}{n}-\frac{k}{n})}}}

= ∫ 0 1 x ( 1 − x ) 2 d x = 24 =\displaystyle{\int_0^1{x(1-x)^2 \,dx}}=24

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