Telescoping Series
Let’s examine a special type of series where terms cancel out in a predictable way. Consider the series:
[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)}[/latex].
We can analyze this series using a technique called partial fractions. Notice that:
[latex]\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}[/latex] ✓
Using this decomposition, we can rewrite the series as:
Let’s calculate the first few partial sums [latex]{{S}_{k}}[/latex]:
Notice the pattern? The middle terms cancel each other out! In general:
The series collapses like a telescope, where most sections disappear into each other, leaving only the first and last terms visible. For this reason, we call a series that has this property a telescoping series.
For this series, since [latex]{S}_{k}=1 - \frac{1}{k+1}[/latex] and [latex]\frac{1}{k+1}\to 0[/latex] as [latex]k\to \infty[/latex], the sequence of partial sums converges to [latex]1[/latex]. Therefore, [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)}=1[/latex].
telescoping series
A telescoping series is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.
[latex]\\[/latex]
The key characteristic is that consecutive terms have parts that subtract to zero.
Any series of the form:
[latex]\displaystyle\sum _{n=1}^{\infty }\left[{b}_{n}-{b}_{n+1}\right]=\left({b}_{1}-{b}_{2}\right)+\left({b}_{2}-{b}_{3}\right)+\left({b}_{3}-{b}_{4}\right)+\cdots[/latex]
is a telescoping series. Let’s see why by examining the partial sums:
[latex]\begin{array}{l}{S}_{1}={b}_{1}-{b}_{2}\hfill \\ {S}_{2}=\left({b}_{1}-{b}_{2}\right)+\left({b}_{2}-{b}_{3}\right)={b}_{1}-{b}_{3}\hfill \\ {S}_{3}=\left({b}_{1}-{b}_{2}\right)+\left({b}_{2}-{b}_{3}\right)+\left({b}_{3}-{b}_{4}\right)={b}_{1}-{b}_{4}.\hfill \end{array}[/latex]
The pattern is clear: the [latex]k[/latex]th partial sum simplifies to:
[latex]{S}_{k}={b}_{1}-{b}_{k+1}[/latex].
Since we can express each partial sum as [latex]{S}{k}={b}{1}-{b}_{k+1}[/latex], the convergence behavior becomes straightforward to analyze.
convergence of general telescoping series
A telescoping series [latex]\displaystyle\sum {n=1}^{\infty }\left[{b}{n}-{b}{n+1}\right][/latex] converges if and only if the sequence [latex]{{b}{k+1}}[/latex] converges.
[latex]\\[/latex]
If [latex]{b}{k+1} \to B[/latex] as [latex]k \to \infty[/latex], then:
[latex]\displaystyle\sum {n=1}^{\infty }\left[{b}{n}-{b}{n+1}\right]={b}_{1}-B[/latex]
This gives us a powerful method for evaluating certain infinite series. Instead of struggling with complex partial sum formulas, we just need to:
- Identify the sequence [latex]{b_n}[/latex]
- Check if [latex]\underset{n\to \infty }{\text{lim}} b_n[/latex] exists
- Apply the formula [latex]{b}_{1}-\underset{n\to \infty }{\text{lim}} b_n[/latex]
In the next example, we show how to use these ideas to analyze a telescoping series of this form.
Determine whether the telescoping series
converges or diverges. If it converges, find its sum.