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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.

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n^2-n} } }\) converges to 1.

Problem Statement

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

Solution

There are several ways to determine convergence. First, we can use one of the comparison tests and compare with \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n^2}} }\)
This will certainly work and it tells us about convergence or divergence. However, there is no way to determine what it converges to and the comparison tests don't help with this.
So, if we look at this carefully, we can do a little bit of algebra and get a telescoping series. Then, if we look at the first few terms, we can determine what the sum converges to. Here's how this works.
\(\displaystyle{ \frac{1}{n^2-n} = \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} }\)
The last step was accomplished by the method of partial fractions.
Notice the sum starts at n=2. So looking at the first few terms, we have
\(\displaystyle{\left( \frac{1}{1} - \frac{1}{2}\right) + \left( \frac{1}{2} - \frac{1}{3}\right) + \left( \frac{1}{3} - \frac{1}{4}\right) + \left( \frac{1}{4} - \frac{1}{5}\right) + ... \left( \frac{1}{N-1} - \frac{1}{N}\right) ... }\)
So, we can see that we end up with \(\displaystyle{ 1 - \frac{1}{N} }\) and since \(\displaystyle{ \lim_{N \to \infty}{\frac{1}{N}} = 0 }\), the sum goes to \(1-0 = 1\).

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n^2-n} } }\) converges to 1.

close 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.

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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

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

Problem Statement

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

Final Answer

The finite telescoping series \(\displaystyle{ \sum_{n=1}^{k}{ \left[ \frac{1}{n+2} - \frac{1}{n} \right] } }\) reduces to \(\displaystyle{ \frac{-3}{2} + \frac{1}{k+1} + \frac{1}{k+2} }\)

Problem Statement

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

Solution

258 solution video

video by PatrickJMT

Final Answer

The finite telescoping series \(\displaystyle{ \sum_{n=1}^{k}{ \left[ \frac{1}{n+2} - \frac{1}{n} \right] } }\) reduces to \(\displaystyle{ \frac{-3}{2} + \frac{1}{k+1} + \frac{1}{k+2} }\)

close solution

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{2^n} - \frac{1}{2^{n+1}} } }\) converges to \(1/2\).

Problem Statement

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

Solution

262 solution video

video by MIP4U

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{2^n} - \frac{1}{2^{n+1}} } }\) converges to \(1/2\).

close solution

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

Problem Statement

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

Final Answer

This telescoping series converges to \(1/2\).

Problem Statement

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

Solution

This series looks like a telescoping series. To see if that is what we have, we will build a table containing the first few values and see if we can get some cancellation.

k

ak

1

\(\displaystyle{ \frac{1}{2} - \frac{1}{3} }\)

2

\(\displaystyle{ \frac{1}{3} - \frac{1}{4} }\)

3

\(\displaystyle{ \frac{1}{4} - \frac{1}{5} }\)

.
.
.

n-1

\(\displaystyle{ \frac{1}{n} - \frac{1}{n+1} }\)

n

\(\displaystyle{ \frac{1}{n+1} - \frac{1}{n+2} }\)

We can see from this table that the first term in each row is canceled by the second term in the previous row. This means that if we add all the terms together from \(k=1\) to \(k=n\) we will have \(1/2\) from the first row and \(-1/(n+2)\) in the last row, giving us the partial sum \(\displaystyle{ S_n = \frac{1}{2} - \frac{1}{n+2} }\).

Taking the limit of \(S_n\) will give us the value to which the series converges.
\(\displaystyle{ \lim_{n\to\infty}{S_n} = }\) \(\displaystyle{ \lim_{n\to\infty}{\left[ \frac{1}{2} - \frac{1}{n+2}\right]} = }\) \(\displaystyle{ \frac{1}{2} - 0 = \frac{1}{2} }\)

Therefore the telescoping series converges to \(1/2\).

Final Answer

This telescoping series converges to \(1/2\).

close solution

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+n} } }\) converges to 1.

Problem Statement

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

Solution

256 solution video

video by PatrickJMT

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+n} } }\) converges to 1.

close solution

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{3}{n^2+3n+2} } }\) converges to \(3/2\).

Problem Statement

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

Solution

263 solution video

video by MIP4U

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{3}{n^2+3n+2} } }\) converges to \(3/2\).

close solution

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

Problem Statement

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

Solution

1212 solution video

video by MIP4U

close solution

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

Problem Statement

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

Solution

1871 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

1214 solution video

video by MIP4U

close solution

Intermediate Problems

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\ln( 1+1/n ) } }\) diverges.

Problem Statement

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

Solution

This is a great video showing how to use partial sums to determine that the series diverges.
Notice at the first of the video he tries the nth-Term Test (or Divergence Test) to see if it can tell us that the series diverges. Since the limit is zero, the Divergence Test is inconclusive and so he has try something else. Don't let this important fact go past you. Think about it for a minute. If \(\displaystyle{ \lim_{n\to\infty}{a_n} = 0 }\), then the Divergence Test tells us nothing. The series could converge or diverge. We just don't know.
Then he starts writing out terms of the series to see if he can determine what is going on. This is a great way to get a feel for a series when nothing jumps out at you and can sometimes lead to a quick solution.

257 solution video

video by PatrickJMT

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\ln( 1+1/n ) } }\) diverges.

close solution

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{ \ln( (n+2)/n ) } }\) diverges.

Problem Statement

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

Solution

260 solution video

video by PatrickJMT

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{ \ln( (n+2)/n ) } }\) diverges.

close solution

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{4n^2-1} } }\) converges to \(1/2\).

Problem Statement

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

Solution

The hardest part of this problem is performing the partial fraction expansion. In this video she did not take the limit of the terms to show that they go to zero, which is a necessary part of evaluting telescoping series, but her final answer is correct.

261 solution video

video by Krista King Math

Final Answer

The telescoping series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{4n^2-1} } }\) converges to \(1/2\).

close solution

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

Problem Statement

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

Final Answer

The telescoping series \(\displaystyle{\frac{5}{1 \times 3} + \frac{5}{2 \times 4} + \frac{5}{3 \times 5} + \ldots }\) converges to \(\displaystyle{ \frac{15}{4} }\)

Problem Statement

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

Solution

259 solution video

video by PatrickJMT

Final Answer

The telescoping series \(\displaystyle{\frac{5}{1 \times 3} + \frac{5}{2 \times 4} + \frac{5}{3 \times 5} + \ldots }\) converges to \(\displaystyle{ \frac{15}{4} }\)

close solution
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