Wemazi

Topic 10.1 · BC

Simplify a telescoping sum

Builds on recovering terms or tails from partial sums. 8 minutes.

Drops that add up to one long drop

A series has the terms \(a_n=\frac{1}{2n-1}-\frac{1}{2n+1}\). The first terms are \(\frac23\), \(\frac2{15}\) and \(\frac2{35}\), shown as the bars on the left.

The column on the right shows the heights \(1,\frac13,\frac15,\frac17,\dots\). Each orange segment goes from one height down to the next, so its length is one term.

The segments join end to end, so the first three go from height 1 down to height \(\frac17\). The teal bracket spans them, and its length \(1-\frac17=\frac67\) is \(S_3\).

The heights \(\frac13\) and \(\frac15\) in the middle each end one segment and start the next, so they cancel out of the total.

Move the slider to add terms. The total depends only on the first height and the last height, and the last height approaches 0, so the partial sums approach 1.

Only the two ends count

The heights are \(b_1, b_2, b_3,\dots\), and each term is a drop from one height to the next: \(a_n=b_n-b_{n+1}\).

Add the first \(n\) drops. Each height from \(b_2\) to \(b_n\) is subtracted once and added once, so it cancels.

Only \(b_1\) and \(b_{n+1}\) remain, so the partial sum is the whole drop from the first height to the last height.

As \(n\) grows, the last height \(b_{n+1}\) approaches the limit of the sequence \(b_n\), so that limit decides the sum.

Notation

  1. A series telescopes when each term is the difference of two neighbors in a list \(b_n\).

    \[\cy{a_n}=b_n-b_{n+1}\]

  2. Write out the partial sum. The middle heights cancel in pairs.

    \[\begin{aligned}S_n&=(b_1-b_2)+(b_2-b_3)\\ &\quad+\cdots+(b_n-b_{n+1})\end{aligned}\]

  3. Only the first height and the last height remain. Keep the last height until you take the limit.

    \[\cb{S_n}=b_1-\cy{b_{n+1}}\]

  4. If \(b_n\) approaches a number \(B\), the series has the sum \(b_1-B\).

    \[\sum_{n=1}^{\infty}a_n=b_1-\lim_{n\to\infty}b_n\]

When it fails

The terms are now \(a_n=\sqrt{n+1}-\sqrt n\). They become smaller: about 0.41, 0.32 and 0.27.

The partial sum is \(S_n=\sqrt{n+1}-1\). The last height \(\sqrt{n+1}\) grows without bound, as the red arrow at the top shows, so the series has no sum. Terms that shrink are not enough.

Example 1: A difference of neighbors

Find the sum of \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\).

  1. Each term has the form \(b_n-b_{n+1}\), with \(b_n=\frac{1}{n+1}\).

  2. Write the partial sum. The middle fractions cancel in pairs.

    \[\begin{aligned}S_n&=\left(\tfrac12-\tfrac13\right)+\left(\tfrac13-\tfrac14\right)\\ &\quad+\cdots+\left(\tfrac1{n+1}-\tfrac1{n+2}\right)\\ &=\tfrac12-\tfrac1{n+2}\end{aligned}\]

  3. As a check, \(S_3=\frac12-\frac15=\frac3{10}\), and the first three terms are \(\frac16+\frac1{12}+\frac1{20}=\frac3{10}\).

  4. The last height \(\frac1{n+2}\) approaches 0, so the partial sums approach \(\frac12\).

    \[\sum_{n=1}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=\frac12\]

Example 2: Terms that need splitting

Find the sum of \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+3)}\).

  1. The terms are not differences yet. Split each term into two fractions that subtract. To check, combine the fractions on the right.

    \[\frac{1}{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right)\]

  2. Write the first terms of the partial sum. Each negative fraction cancels the positive fraction three terms later.

    \[\begin{aligned}S_n=\frac13\Big[&\left(1-\tfrac14\right)+\left(\tfrac12-\tfrac15\right)\\ &+\left(\tfrac13-\tfrac16\right)+\left(\tfrac14-\tfrac17\right)\\ &+\cdots\Big]\end{aligned}\]

  3. A tempting wrong move: Keep only the first fraction and the last fraction, as in the earlier examples.

    \[S_n=\frac13\left(1-\frac{1}{n+3}\right)\]

    Test \(n=2\). The terms are \(\frac14\) and \(\frac1{10}\), so \(S_2=0.35\), the height of the teal column. This formula gives about 0.27, a red ring below it. The fractions \(\frac12\) and \(\frac13\) at the start do not cancel with any other fraction.

  4. Each term reaches three places ahead, so three fractions survive at the start and three at the end.

    \[\begin{aligned}S_n=\frac13\Big(&1+\tfrac12+\tfrac13\\ &-\tfrac1{n+1}-\tfrac1{n+2}-\tfrac1{n+3}\Big)\end{aligned}\]

  5. The last three fractions approach 0, so the sum is one third of \(1+\frac12+\frac13=\frac{11}{6}\). The dashed line marks this sum.

    \[\sum_{n=1}^{\infty}\frac{1}{n(n+3)}=\frac13\cdot\frac{11}{6}=\frac{11}{18}\]

Practice

Find the sum of \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+1}}\right)\).

On the figure, the segments join end to end. Name the first height and the height where the last segment ends.

Solution

  1. Each term is \(b_n-b_{n+1}\) with \(b_n=\frac{1}{\sqrt n}\). The middle terms cancel, so only the first height and the last height remain.

    \[S_n=b_1-b_{n+1}=1-\frac{1}{\sqrt{n+1}}\]

  2. The last height approaches 0, so the partial sums approach 1.

    \[\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+1}}\right)=1\]

Questions

Questions 1 and 2 use the figure. The heights on the column are \(b_n=\frac6n\). Each orange segment is a term \(a_n=b_n-b_{n+1}\), and the segments join end to end.

  1. What is the partial sum \(S_3\)?

    • 4.5
    • 4
    • 1.5
    Answer

    4.5

  2. As \(n\) grows, which number do the partial sums approach?

    • 6
    • 0
    • 4.5
    Answer

    6

  3. What is \(\displaystyle\sum_{n=3}^{\infty}\left(\frac1n-\frac1{n+1}\right)\)?

    • \(\dfrac13\)
    • 1
    • \(\dfrac14\)
    • 0
    Answer

    \(\dfrac13\)

Summary

You sum a telescoping series by writing each term as a difference and keeping only the heights that do not cancel.

Each height is added once and subtracted once, except the first and the last, so the partial sum is the first height minus the last.

You use it when the terms are differences of neighbors, or when partial fractions turn them into differences.

This is one of five skills in topic 10.1 of Which & Why: AP Calculus.