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17calculus > infinite series > telescoping series

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How to Read and Do Proofs: An Introduction to Mathematical Thought Processes

Telescoping Series

on this page: ► notation     ► solving

Telescoping Series are series whose partial sums have terms that cancel and may eventually converge. They are not very common in mathematics but are interesting to study to get a feel for partial sums and how series work.

The idea with telescoping series is to arrange the terms in a form where you can see what is canceling, then to take the limit of what is left. The best way to learn how to solve telescoping series problems is by example. First, let us look at a bit of notation. It will help you to follow the discussion if you have the infinite series table in front of you. We will refer to it several times.

Notation

Notice in the infinite series table, the series column for a telescoping series uses a little different notation than the other rows. We use \( b_n \) instead of \( a_n \) for the terms in the series. This is deliberate. What this is saying is that \( a_n = b_n - b_{n+1} \). An example of this is

\(\displaystyle{ \sum_{n=1}^{\infty}{\left( \frac{1}{n} - \frac{1}{n+1}\right)} = \left( 1 - \frac{1}{2}\right) + \left( \frac{1}{2} - \frac{1}{3}\right) + \left( \frac{1}{3} - \frac{1}{4}\right) + . . . }\)
where \( \displaystyle{ a_n = \left( \frac{1}{n} - \frac{1}{n+1}\right); ~~~ b_n = \frac{1}{n}; ~~~ b_{n+1} = \frac{1}{n+1} }\)

This is just one example of a telescoping series. They come in many forms. For example, you might see \( a_n = b_n - b_{n+3} \) in which case, every third term might cancel. When you find what you think might be a telescoping series, write out some terms until you see a pattern. The number of terms is determined by how far apart a term repeats. This is the main technique for handling telescoping series.

A key idea derived from this discussion is to watch the notation carefully, and when a textbook or teacher changes notation, this can mean that something is going on that might not be explicitly stated. Ask questions until you understand.

Solving

The main technique when determining convergence of telescoping series is to get an expression for the partial sum and taking the limit of the partial sum. This is shown in the following practice problem. Determine whether the series converges or diverges. If the series converges, find the value to which it converges, if possible.

Practice 1

\(\displaystyle{\sum_{n=2}^{\infty}{\frac{1}{n^2-n}}}\)

answer

solution

You will probably not find a lot of discussion or videos on telescoping series since they do not show up very much. Your instructor may just touch on them in class but then may ask questions on the exam about them. Telescoping series are fairly straightforward but the main thing you need to watch for is the possibility of expansion by partial fractions. Most problems are not given as in the example above. You will probably have to do some algebra to get a fraction in telescoping form. After you do, the main technique is to write out several (usually many) terms until you see a pattern developing. It helps to build a table arranged to spot patterns. This will give you a feel for what is going on and enable you to answer questions about them. Check out the practice problems for examples and practice on working with these types of series.

A Note of Caution - In most series problems, you do not need to worry about where the series starts, i.e. it can start at \(0\), \(1\) or any other number. This is the case when you are trying to determine only if a series converges or diverges. However, when you are trying to determine what value a series converges to, you DO need to take into account what the starting value of the series is. So, keep track of this when you are working with telescoping series.

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Practice Problems

Instructions - - Unless otherwise instructed, determine whether the series converges or diverges. If the series converges, find the value to which it converges, if possible.

Basic Problems

Practice 2

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2+n}}}\)

answer

solution

Practice 3

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

answer

solution

Practice 4

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

answer

solution

Practice 5

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{3}{n^2+3n+2}}}\)

answer

solution

Practice 6

\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{4}{n(n+2)}} }\)

solution

Practice 7

\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n(n+1)}} }\)

solution

Practice 8

\(\displaystyle{\sum_{k=1}^{\infty}{[\ln(2k+4)-\ln(2k+2)]}}\)

solution

Intermediate Problems

Practice 9

\(\displaystyle{\sum_{n=1}^{\infty}{\ln(1+1/n)}}\)

answer

solution

Practice 10

\(\displaystyle{\frac{5}{1\times 3}+\frac{5}{2\times 4}+\frac{5}{3\times 5}+ . . . }\)

answer

solution

Practice 11

\(\displaystyle{\sum_{n=1}^{\infty}{\ln((n+2)/n)}}\)

answer

solution

Practice 12

\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{4n^2-1} } }\)

answer

solution

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