374 /www/nrich/html/content/01/09/15plus2/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"><p>Prove that the sum of the reciprocals of the first $n$ triangular numbers gets closer and closer to $2$ as $n$ grows.</p> <p><br></br>  </p></mdoxml> <mdoxml version="1.0"><br></br> <p>This solution was sent by Etienne from Parramatta Highschool, NSW Australia.</p> <p>The $r$th triangular is $r(r+1)/2$ and it's reciprocal is $2/[r(r+1)]=2 \times [1/(r(r+1))]$</p> <p>Now $1/r(r+1) = 1/r - 1/(r+1)$. Take note of this, very useful technique!</p> <p>\[\frac{1}{r} - \frac{1}{r+1} = \frac{(r+1)-r}{r(r+1)} = \frac{1}{r(r+1)} \]</p> <p>The sum of the reciprocals of the first n triangular numbers</p> <p>\[\frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \cdots + \frac{2}{n(n+1)} \] \[ = 2 \{ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{n(n+1)} \} \] \[ = 2 \{ [\frac{1}{1} - \frac{1}{2}] + [\frac{1}{2} - \frac{1}{3}] + \cdots + [\frac{1}{n} - \frac{1}{(n+1)}] \} \]</p> <p>Surprise, you get terms that cancel out each other, ie $-1/2$ and $1/2$, $-1/3$ and $1/3$, $-1/n$ and $1/n$. This is called 'telescoping'.</p> <p>The sum thus equals \[ 2 \{ 1 - \frac{1}{(n+1)} \} \]</p> <p>When n is large, $1/(n+1)$ is very small, so the sum is approximately $2$.</p> <p>When $n$ tends to infinity, $1/(n+1)$ tends to $0$, and it turns out the infinite sum is exactly $2$.</p> <br></br></mdoxml> <mdoxml version="1.0"><p>The algebraic expression for $r$th triangular number is</p> <p>$$T_r = \frac{1}{2} r(r+1) $$</p> <p> </p> <p>The expression that you are trying to evaluate is $$\sum_{r=1}^{n} \frac{1}{T_r} = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3} + ... + \frac{1}{T_n} \cong 2 $$</p></mdoxml> 2 3 0 0 0 0 1 Prove that the sum of the reciprocals of the first n triangular numbers is approximately equal to 2 when n is large and tends to 2 as n tends to infinity. Advanced Algebra Summation of series Pre-Calculus and Calculus Limits Numbers and the Number System Triangle numbers