Potential Telescoping Series

Calculus Level 3

S = 1 1 − 1 2 + 1 2 − 1 4 + 1 3 − 1 6 + 1 4 − … S=\frac { 1 }{ 1 } -\frac { 1 }{ 2 } +\frac { 1 }{ 2 } -\frac { 1 }{ 4 } +\frac { 1 }{ 3 } -\frac { 1 }{ 6 } +\frac { 1 }{ 4 } -\dots

Assuming the pattern continues, determine the value of S S .

The sum does not converge to a finite value. π \pi 2 \sqrt { 2 } The sum converges, but its exact value cannot be calculated. 1

Liam P by Liam P , 22, Canada

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1 solution

Chew-Seong Cheong
Apr 20, 2017

S = 1 1 − 1 2 + 1 2 − 1 4 + 1 3 − 1 6 + 1 4 − ⋯ = ∑ k = 1 ∞ ( 1 k − 1 2 k ) = ∑ k = 1 ∞ 1 2 k = lim ⁡ n → ∞ H n 2 where H n is the n th harmonic number. \begin{aligned} S & = \frac 11 - \frac 12 + \frac 12 - \frac 14 + \frac 13 - \frac 16 + \frac 14 - \cdots \\ & = \sum_{k=1}^\infty \left(\frac 1k - \frac 1{2k} \right) \\ & = \sum_{k=1}^\infty \frac 1{2k} \\ & = \lim_{n \to \infty} \frac {H_n}2 \quad \quad \small \color{#3D99F6} \text{where }H_n \text{ is the }n^\text{th} \text{ harmonic number.} \end{aligned}

Since lim ⁡ n → ∞ H n \displaystyle \lim_{n \to \infty} H_n does not converge, S S does not converge .

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