GATE CS 2015 Set 1 Q.36 (CS Q.26): Calculus

\sum_{x=1}^{99} \frac{1}{x(x+1)}

The value of the summation is _____.

Your answer

Official answer and solution

Solution

Decompose the summand using partial fractions to obtain a telescoping sum.

The summand can be expanded using partial fractions as:

\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}.

Summing from $x = 1$ to $99$ :

\begin{aligned} \sum_{x=1}^{99} \frac{1}{x(x+1)} & = \sum_{x=1}^{99} \left( \frac{1}{x} - \frac{1}{x+1} \right) \\ & = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{99} - \frac{1}{100}\right). \end{aligned}

All intermediate terms cancel out in the telescoping sum:

S = 1 - \frac{1}{100} = \frac{99}{100} = 0.99.

The value is $0.99$ .

Final answer

0.99

Common trap

Approximating $\int 1/x^2 dx$ without computing the exact telescoping cancellation.

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