Advanced Math - Series Convergence Calculator, Telescoping Series Test

Last blog post, we went over what an alternating series is and how to determine if it converges using the alternating series test. In this blog post, we will discuss another infinite series, the telescoping series, and how to determine if it converges using the telescoping series test.

If it isn’t clear right away, telescoping is synonymous with the word collapsing. A telescoping series is a series where almost all the terms cancel with the preceding or following term leaving just the initial and final terms, i.e. a series that can be collapsed into a few terms.

Let’s see what this looks like . . .

                                     ∑_{n=1}^∞\frac{1}{n(n+1)}= ∑_{n=1}^∞\frac{1}{n}-\frac{1}{n+1}

                                   = (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+ ...+(\frac{1}{n}-\frac{1}{n+1})

                                                               =1-\frac{1}{n+1}

As you can see, we are able to cancel out all terms except the first and last.

Now that we’ve discussed what a telescoping series is, let’s go over the telescoping series test.

Telescoping Series Test:

For a finite upper boundary, ∑_{n=k}^N(a_{n+1}-a_n )=a_{N+1 }-a_k
For an infinite upper boundary, if a_n→0*, then ∑_{n=k}^∞(a_{n+1}-a_n )= -a_k
*If a_n doesn’t converge to 0, then the series diverges.

In regards to infinite series, we will focus on the infinite upper boundary scenario. In order to use this test, you will need to manipulate the series formula to equal a_{n+1}-a_n where you can easily identify what a_{n+1} and a_n are. Also, please note that if you are able to manipulate the series in this form, you can confirm that you have a telescoping series. With practice, this will come more naturally.

Let’s see some examples to better understand.

First example (click here):

                                                         ∑_{n=1}^∞\frac{5}{n}-\frac{5}{n+1}


1. Convert the series into the form a_{n+1}-a_n

                                                        \frac{5}{n}-\frac{5}{n+1}= -\frac{5}{n+1}-(-\frac{5}{n})

                                                               a_{n+1}=-\frac{5}{n+1}

                                                                      a_n=-\frac{5}{n}

2. Determine if a_n→0

                                                               a_n=-\frac{5}{n}= -5(\frac{1}{n})

                                              Since \frac{1}{n} converges to 0, -\frac{5}{n} converges to 0.

3. Calculate -a_k

                                                                         k=1

                                                                     a_n=-\frac{5}{n}

                                                             -a_k=-(-\frac{5}{1})=5
             The series converges to 5.

Next example (click here)

                                                          ∑_{n=1}^∞\frac{6}{(n+1)(n+2)}

1. Convert the series into the form a_{n+1}-a_n

                                   ∑_{n=1}^∞\frac{6}{(n+1)(n+2)}= 6∙∑_{n=1}^∞\frac{1}{(n+1)(n+2)}

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

                                                                 a_{n+1}=-\frac{1}{n+2}

                                                                     a_n=-\frac{1}{n+1}


2. Determine if a_n→0

                                                                     a_n=-\frac{1}{n+1}

                                                                    -\frac{1}{n+1}  →0

3. Calculate -a_k

                                                                            k=1

                                                                     a_n=-\frac{1}{n+1}

                                                           -a_k=-(-\frac{1}{1+1})=\frac{1}{2}


                                                   6∙∑_{n=1}^∞\frac{1}{(n+1)(n+2)} =6∙\frac{1}{2}=3

             The series converges to 3.

Last example (click here):

                                                                ∑_{n=1}^∞\frac{1}{4n^2-1}

1. Convert the series into the form a_{n+1}-a_n

                                                \frac{1}{4n^2-1}=-(\frac{1}{2(2n+1)} )-(-\frac{1}{2(2n-1)})

                                                                a_{n+1}= -(\frac{1}{2(2n+1)} )

                                                                    a_n=-\frac{1}{2(2n-1)}

2. Determine if a_n→0

                                                              a_n=-\frac{1}{2(2n-1)} =-\frac{1}{4n-1}

                                                                      -\frac{1}{4n-1}  →0


3. Calculate -a_k

                                                                               k=1

                                                                     a_n=-\frac{1}{2(2n-1)}

                                                           -a_k=-(-\frac{1}{2(2∙1-1)} )=\frac{1}{2}

              The series converges to \frac{1}{2}.

The trickiest part of this is manipulating the series formula into a_{n+1}-a_n. Once you’re able to do this, the rest should be pretty simple. The key thing to remember about a telescoping series is that all the terms will cancel out, except the first and last term.

For more help on telescoping series, check out Symbolab’s Practice. Next blog post, I’ll go over the convergence test for a radio series.

Until next time,

Leah