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Evaluating a telescoping sum in closed form

A telescoping sum is one in which consecutive terms cancel out, leaving only a few terms at the beginning and end.

Consider the sum ∑k=1n(1k−1k+1).\sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right). Write out the first few terms and the last term of this sum, observe which parts cancel, and find a closed-form expression (in terms of nn, with no summation symbol) for this sum.

Hint

Write out the terms for k=1,2,3,…k=1, 2, 3, \dots up to k=nk=n explicitly. Notice that the −1/k-1/k part of one term cancels with the 1/k1/k part of a nearby term. What is left over after all the cancellation?

Step-by-step solution

Writing out the terms: for k=1k=1 we get 11−12\frac{1}{1}-\frac{1}{2}, for k=2k=2 we get 12−13\frac{1}{2}-\frac{1}{3}, for k=3k=3 we get 13−14\frac{1}{3}-\frac{1}{4}, and so on, up to k=nk=n which gives 1n−1n+1\frac{1}{n}-\frac{1}{n+1}. Adding these in order, each negative term −1k-\frac{1}{k} cancels with the positive term 1k\frac{1}{k} appearing in the next fraction.

After all the cancellation, only the very first positive term 11\frac{1}{1} and the very last negative term −1n+1-\frac{1}{n+1} survive. Therefore the sum simplifies to ∑k=1n(1k−1k+1)=1−1n+1.\sum_{k=1}^n \left(\frac{1}{k}-\frac{1}{k+1}\right) = 1 - \frac{1}{n+1}. This can also be written as nn+1\frac{n}{n+1}.

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