KindiakMath

An Absurd Puzzle

Problem 1. Evaluate

\displaystyle S := \frac 1{\sqrt 1 + \sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \cdots + \frac 1{\sqrt{2024} + \sqrt{2025}}.

(Click for Solution)

Solution. By rationalising the denominator,

\begin{aligned} \frac 1{\sqrt n + \sqrt{n+1}} &= \frac 1{\sqrt n + \sqrt{n+1}} \cdot \frac{ \sqrt n - \sqrt{n+1} }{ \sqrt n - \sqrt{n+1} } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ (\sqrt n)^2 - (\sqrt{n+1})^2 } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ n - (n+1) } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ -1 } \\ &= \sqrt{n+1} - \sqrt{n}. \end{aligned}

Hence, summing the terms,

\begin{aligned} S &=\frac 1{\sqrt 1 + \sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \cdots + \frac 1{\sqrt{2023} + \sqrt{2024}} + \frac 1{\sqrt{2024} + \sqrt{2025}} \\ &= \sqrt 2 + \sqrt 3 + \cdots + \sqrt{2024} + \sqrt{2025} \\ &\phantom{=} - \sqrt{1} - \sqrt 2 - \cdots - \sqrt{2023} - \sqrt{2024} \\ &= \sqrt{2025} - 1 \\ &= 45 - 1 = 44. \end{aligned}

Problem 2. Evaluate

\displaystyle T := \frac 1{1 \times 2} + \frac 1{2 \times 3} + \cdots + \frac 1{n(n+1)}

in terms of n.

(Click for Solution)

Solution. For any k,

\begin{aligned} \frac 1{k(k+1)} &= \frac {(k+1)-k}{k(k+1)} \\ &= \frac{k+1}{k(k+1)} - \frac k{k(k+1)} \\ &= \frac 1k - \frac 1{k+1}. \end{aligned}

Hence,

\begin{aligned} T &= \frac 1{1 \times 2} + \frac 1{2 \times 3} + \cdots + \frac 1{(n-1)n} + \frac 1{n(n+1)} \\ &= \frac 11 + \frac 12 + \cdots + \frac 1{n-1} + \frac 1n \\ &\phantom{=} - \frac 12 - \frac 13 - \cdots - \frac 1n - \frac 1{n+1} \\ &= 1 - \frac 1{n+1} = \frac{n}{n+1}. \end{aligned}

Remark 1. In both problems, we used the method of differences or a telescoping sum (or more fancifully, the fundamental theorem of discrete calculus), which states that if u_n = v_{n+1} - v_n, then

u_1 + u_2 + \cdots + u_n = v_{n+1} - v_1.

—Joel Kindiak, 19 Jan 26, 2204H

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joelkindiak

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