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Telescoping Series

Calculus 2 · Chapter 6: Sequences & Introduction to SeriesSection 6.4 · lesson 30 of 549 min · free

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In this lesson

When the partial sum collapses to two terms.

  • Recognise a term written as a difference b_{n}-b_{n+1}
  • Use partial fractions to expose that hidden structure
  • Collapse the partial sum and take its limit
  • Count how many terms survive when the gap widens

Transcript

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MAT-265 - Calculus 2 · Section 6.4

Telescoping Series

Almost no infinite series lets you compute its exact sum. Geometric series were one exception, in the last section. Telescoping series are the other. And they work by a completely different mechanism. The partial sum collapses — because consecutive pieces cancel each other out. The name comes from an old collapsible telescope. Many segments sliding into just two. And when that happens, you don't need a convergence test at all. You get the exact value directly.

The Mechanism

Interior Terms Cancel in Pairs

Here's the mechanism. Suppose the general term can be written as a difference of consecutive values of some sequence b. a n equals b n, minus b n plus one. Write out the partial sum. You get b one minus b two, plus b two minus b three, plus b three minus b four, and so on. Now look at what happens. Every interior piece appears twice, with opposite signs. Minus b two from the first group cancels plus b two from the second. Minus b three cancels plus b three. And so on, all the way down the line. After all that cancellation, only two things remain. The very first piece, b one. And the very last piece, minus b n plus one. So s n equals b one, minus b n plus one. A partial sum of n terms collapses to just two. That's the signature of a telescoping series. Let's see it concretely. Take a k equals one over k, minus one over k plus one. The fourth partial sum is one minus one half, plus one half minus one third, plus one third minus one quarter, plus one quarter minus one fifth. Every interior fraction cancels with its neighbor. Only the first term, one, and the last term, minus one fifth, survive. So s four is one minus one fifth. Four fifths.

The Tool

Partial Fractions Reveal the Structure

Now here's the practical difficulty. Telescoping is rarely visible in the original form of a series. The term one over n times n plus one does not obviously look like a difference. The tool that exposes that hidden structure is partial fraction decomposition. From section one point five. Let's find the exact sum of the sum of one over n times n plus one. Step one. Decompose into partial fractions. Write one over n times n plus one as A over n, plus B over n plus one. Multiply both sides by n times n plus one. You get one equals A times n plus one, plus B n. Set n equals zero: one equals A. Set n equals minus one: one equals minus B, so B is minus one. Therefore one over n times n plus one equals one over n, minus one over n plus one. And that is exactly the form b n minus b n plus one, with b n equal to one over n. Step two. Write the partial sum and cancel. All interior terms cancel, leaving s n equals one, minus one over n plus one. Step three. Take the limit of the partial sums. By the definition of a series sum from section six point two, the sum is the limit of s n. One minus one over n plus one goes to one minus zero. One. The exact sum is one. No approximation. No convergence test estimate. An exact value, delivered by the collapse.

The General Result

Sum of a Telescoping Series

The general procedure is now clear, and applies whenever a series telescopes. Once the term is written as b n minus b n plus one — often after partial fractions — the partial sum reduces to its surviving boundary terms. And the limit gives the sum. So the sum equals b one, minus the limit of b n plus one. Provided that limit exists. And in fact, the series converges exactly when that limit exists. Now here's something worth thinking about. The Divergence Test from section six point two confirms that one over n times n plus one goes to zero. So the series might converge. But that test alone can never produce the sum. It can only rule out divergence. Telescoping is different in kind. By exposing the partial sum's closed form — s n equals one minus one over n plus one — it delivers the exact total. So why can so few series be summed this way? Because most series have no simple closed form for s n. Telescoping works precisely when the algebra of the terms makes the partial sum collapse to something a limit can be taken of directly.

Worked Example 2

A Wider Gap in the Denominator

Let's do one with a wider gap in the denominator. Find the sum of two over the quantity two n minus one, times two n plus one. Decompose. Set it equal to A over two n minus one, plus B over two n plus one. Clearing denominators gives two equals A times two n plus one, plus B times two n minus one. Choosing n equals one half gives two equals two A, so A is one. Choosing n equals minus one half gives two equals minus two B, so B is minus one. So the term becomes one over two n minus one, minus one over two n plus one. Collapse the partial sum, with b n equal to one over two n minus one. You get one minus one third, plus one third minus one fifth, plus one fifth minus one seventh, and so on. Every interior term cancels. Leaving one, minus one over two n plus one. Take the limit. One over two n plus one goes to zero. So the sum is exactly one.

Worked Example 3

When the Terms Sit Two Apart

Now a case that catches students out. What if the terms sit two apart? Find the sum of one over n times n plus two. Decompose. You get A over n plus B over n plus two, and clearing gives one equals A times n plus two, plus B n. Set n equals zero: one equals two A, so A is one half. Set n equals minus two: one equals minus two B, so B is minus one half. So the term is one half, times the quantity one over n, minus one over n plus two. Now watch the cancellation pattern carefully, because it's different. Because the gap is two apart, each negative piece cancels the positive piece two terms later. Not one term later. So two terms survive at each end. Not one. Write out the partial sum. The minus one third cancels the later plus one third. The minus one quarter cancels the later plus one quarter. And so on. The terms that never get cancelled are the first two positives — one and one half — and the last two negatives. So s n is one half, times the quantity one plus one half, minus one over n plus one, minus one over n plus two. Take the limit. Both of those last fractions vanish. You get one half, times three halves. Three quarters. And here's the lesson. When the difference spans a gap of g terms, exactly g terms survive at each end. So always write out enough of the partial sum to see precisely which ones remain. Assuming only two survive is the classic error in this section.

Key Takeaways

What to Carry Forward

Five takeaways. A series telescopes when its term can be written as b n minus b n plus one — or a wider-gap difference — so that consecutive pieces cancel. Partial fraction decomposition, from section one point five, is the standard tool for revealing that hidden difference structure in rational terms. After cancellation, the partial sum collapses to its surviving boundary terms. For a one-apart gap, that's b one minus b n plus one. The exact sum is the limit of s n, computed directly from those surviving terms. Which builds on the partial-sum definition from section six point two. And when the difference spans a gap of g terms, g terms survive at each end. Write out enough of the partial sum to identify exactly which ones remain, before taking the limit. Next up is section six point five — Properties of Convergent Series. The algebra that lets you combine and manipulate them.

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FAQ

Questions students ask us

About Telescoping Series

What is a telescoping series?
A telescoping series is one whose general term can be written as a difference b_n - b_{n+1}, so consecutive pieces cancel in the partial sum. Writing out s_n shows every interior term appearing twice with opposite signs, leaving only s_n = b_1 - b_{n+1}. The sum is then b_1 - \lim_{n\to\infty} b_{n+1}, and the series converges exactly when that limit exists.
How do you use partial fractions to find the sum of a telescoping series?
Decompose the rational term into partial fractions to expose the hidden difference b_n - b_{n+1}, then collapse the partial sum and take its limit. For example, \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, so the interior terms cancel and s_n = 1 - \frac{1}{n+1}. Since \frac{1}{n+1} \to 0, the sum is \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1 exactly.
What is the sum of the series 1/(n(n+2)) from n = 1 to infinity?
The sum of \sum_{n=1}^{\infty} \frac{1}{n(n+2)} is \frac{3}{4}. Partial fractions give \frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right). Because the terms sit two apart, two terms survive at each end, so s_n = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right). Both trailing fractions go to zero, leaving \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}.
What is the most common mistake when summing a telescoping series?
The classic mistake is assuming only one term survives at each end of the partial sum, so that s_n = b_1 - b_{n+1} always holds. When the difference spans a gap of g terms, such as \frac{1}{n} - \frac{1}{n+2}, exactly g terms survive at each end. Always write out enough of the partial sum to see precisely which terms remain uncancelled before taking the limit.

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