WMC 2018 Senior Problem 8

Calculus Level 2

Evaluate the limit below.

lim ⁡ n → ∞ ∑ k = 1 n 6 4 k 2 − 1 \large \lim _{n \to \infty}{ \sum _{ k=1 }^{ n } \frac { 6 }{ 4k^{ 2 }-1 } }

4 3.75 3 3.5 3.25

Tiger Ang by Tiger Ang , 20, Malaysia

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1 solution

Chew-Seong Cheong
Oct 11, 2018

S = ∑ k = 1 ∞ 6 4 k 2 − 1 = ∑ k = 1 ∞ 6 ( 2 k − 1 ) ( 2 k + 1 ) = 3 ∑ k = 1 ∞ ( 1 2 k − 1 − 1 2 k + 1 ) = 3 ( 1 1 − 1 3 + 1 3 − 1 5 + 1 5 − 1 7 + ⋯ ) = 3 ( 1 − 1 3 + 1 3 − 1 5 + 1 5 − 1 7 + ⋯ ) = 3 \begin{aligned} S & = \sum_{k=1}^\infty \frac 6{4k^2-1} \\ & = \sum_{k=1}^\infty \frac 6{(2k-1)(2k+1)} \\ & = 3 \sum_{k=1}^\infty \left(\frac 1{2k-1} - \frac 1{2k+1}\right) \\ & = 3 \left(\frac 11 - \frac 13 + \frac 13 - \frac 15 + \frac 15 - \frac 17 + \cdots \right) \\ & = 3 \left(1\color{#3D99F6} - \cancel{ \frac 13} + \cancel{\frac 13} \color{#D61F06} - \cancel{\frac 15} + \cancel{\frac 15} \color{#3D99F6} - \cancel{\frac 17} + \cdots \right) \\ & = 3 \end{aligned}

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