We discuss convergence results for geometric series and telescoping series.
Suppose that we have a series \(\sum _{k=k_0}^{\infty } a_k\) and have to determine whether it converges or diverges. To answer this question, we define a new sequence \(\{s_n\}_{n=k_0}\) where \(s_n = \sum _{k=k_0}^{n}\) for all \(n \geq k_0\). We saw previously that
- the series \(\sum _{k=k_0}^\infty a_k\) converges if and only if \(\lim _{n\to \infty } s_n\) exists.
- the series \(\sum _{k=k_0}^\infty a_k\) diverges if and only if \(\lim _{n\to \infty } s_n\) does not exist.
The definitions above give us a way to determine whether a given series converges. In fact, to determine whether \(\sum _{k=k_0}^{\infty } a_k\) converges, we can do the following.
- Consider the associated sequence \(\{s_n\}\) of partial sums.
-
Try to find an explicit formula for the term \(s_n\). If you can find such a formula, analyze \(\lim _{n \to \infty } s_n\).
- If the limit exists, \(\sum _{k=k_0}^{\infty } a_k\) converges, and if we can determine that \(\lim _{n \to \infty } s_n =L\), then \(\sum _{k=k_0} a_k=L\).
- If \(\lim _{n \to \infty } s_n\) does not exist, then \(\sum _{k=k_0} a_k\) diverges.
- If an explicit formula for \(s_n\) cannot be found, further analysis is needed. We’ll expound on this in later sections.
1 A recursive formula for \(s_n\)
The most straightforward way to determine whether \(\lim _{n \to \infty } s_n\) exists is to have an explicit formula for the \(n\)-th term \(s_n\). Note that this is not an easy task; for example, can you find a formula for \(s_n\) for the series \(\sum _{k=1}^{\infty } \frac {1}{k}\)? It’s not too hard to write out the first several terms in the sequence \(\{s_n\}_{n=1}\), but try to find an explicit formula for \(s_n\)!
As it turns out, there is always a recursive formula for \(s_n\), and this will play an important role in later sections. Suppose that we want to consider \(\sum _{k=1}^{\infty } a_k\). Let’s write out the formula for \(s_n\).
\[ s_n = a_1+a_2+a_3+\ldots +a_{n-1}+a_n \]
We can make an observation by considering \(s_{n-1}\) in a similar way.
\[ s_{n-1} = a_1+a_2+a_3+\ldots +a_{n-1} \]
Now returning to our expression for \(s_n\), we can make an observation.
![[Picture]](digInSpecialSeries-8e8f754c164644a35cb3d798af5ef599.svg)
We thus have the formula
\[ s_n = s_{n-1}+a_n. \]
If we apply this to the series \(\sum _{k=1}^{\infty } \frac {1}{k}\), we have \(s_n = \sum _{k=1}^n \frac {1}{k}\) and \(a_n = \frac {1}{n}\). The recursive formula reads
\begin{align*} s_n &= s_{n-1} +a_n\\ s_n &= s_{n-1} + \frac {1}{n}. \end{align*}
This does not help us analyze whether \(\lim _{n \to \infty } s_n\) actually exists. Sometimes, however, we can find an explicit formula for \(s_n\), and we study two special types of series for which this is possible.
2 Two special types of series
3 Geometric series
Recall that a geometric sequence is a sequence for which the ratio of successive terms is constant. If \(\{a_n\}_{n=n_0}\) is such a sequence, then there are constants \(a \ne 0\) and \(r\) for which \(a_n = a\cdot r^n\).
We thus represent this sequence by the ordered list
\[ ar^{n_0} , ar^{n_0+1}, ar^{n_0+2}, \ldots \]
and we have a result that characterizes the behavior of this type of sequence, which we recall now.
Given a geometric sequence \(\{a_n\}_{n=n_0}\) where \(a_n = a \cdot r^{n}\),
\[ \lim _{n\to \infty } a_n = \begin{cases} 0 &\text {if $|r|<1$,}\\ 1 &\text {if $r=1$,}\\ \text {DNE} &\text {if $|r|>1$ or $r=-1$.} \end{cases} \]
We can now ask when we are able to sum the terms of a geometric sequence.
A geometric series is a series of the form \(\sum _{k=k_0}^\infty ar^k\) for some real numbers \(a \ne 0\) and \(r\).
Before exploring when such a series converges, note that sometimes, some preliminary algebra is necessary to recognize a series as geometric.
The series \(\sum _{k=4}^\infty \frac {2^{2k+1}}{3^k}\) isis not geometric since \(a_k =\frac {2^{2k+1}}{3^k}\) cancannot be brought into the form
\(a \cdot r^k\).
Using the laws of exponents shows us:
\[ \frac {2^{2k+1}}{3^k} = \frac {2^{2k} \cdot 2^1}{3^k}= 2 \cdot \frac {\left (2^{k}\right )^2}{3^k} = 2 \cdot \frac {4^k}{3^k} = 2 \cdot \left (\frac {4}{3}\right )^k. \]
Indeed, \(a= \answer [given]{2}\) and \(r = \answer [given]{\frac {4}{3}}\). The series \(\sum _{k=0}^\infty k^2 \left (\frac {1}{2}\right )^k\) isis not geometric since \(a_k = \answer [given]{k^2 \cdot \frac {1}{2}^k}\) cancannot be brought into the
form \(a \cdot r^k\). Indeed, the coefficient, \(k^2\), is not the same for each term in the series.
We can now try to determine when adding together the terms in such a series is possible; that is, we can explore for which values of \(a\) and \(r\) the series \(\sum _{k=n_0}^{\infty } a_k\) converges.
Let \(r \neq 1\) and consider the geometric series \(\sum _{k=0}^\infty a r^k\), and let \(s_n = \sum _{k=0}^{\infty } a r^k \). We find an explicit formula for the
term \(s_n\).
First, note that the sum above represents the attempt to add all of the terms in the
sequence \(\{a_n\}_{n=0}\), where \(a_n = r^n\). Let’s start by writing out \(s_n\).
\begin{align*} s_n &= 1 + r + r^2 + \dots + r^n\\ \end{align*}
The issue with finding a formula for \(s_n\) arises from the fact that we cannot perform the above for an unspecified value \(n\). However, to go from one term in the sum to the next, we multiply by \(r\), so let’s multiply both sides of the above equation by \(r\). \begin{align*} s_n &= 1 + r + r^2 + \dots + r^n\\ r s_n &= ~ \phantom { 1 + } r + r^2 + \dots + r^n + r^{n+1} \end{align*}
Now, we can subtract away the middle terms, which we show (with some slight abuse of notation) below.
\[ \begin{array}{rl} s_n &= 1 + \cancel { r + r^2 + \dots + r^n}\\ -\left ( \phantom { r^{n+1}} r s_n \right .&=~ \left . \phantom { 1 + } \cancel {r + r^2 + \dots + r^n} + r^{n+1}\right ) \\ \hline s_n - r s_n &= 1 \phantom { + r + r^2 + \dots + r^n } ~ - r^{n+1}\\ \end{array} \]
We can now solve for \(s_n\). \begin{align*} s_n - r s_n &= 1 - r^{n+1}\\ s_n(1-r) &= 1 - r^{n+1}\\ s_n &= \frac {1 - r^{n+1}}{1-r}. \end{align*}
Since \(r \ne 1\), there is no issue dividing by \(1-r\); we will treat the case \(r=1\) a bit later.
From our work above, we see that the \(n\)-th partial sum of the geometric series \(a_n = r^n\) is
\[ s_n = \sum _{k=0}^{n} r^k= \frac {1 - r^{n+1}}{1-r}. \]
We now have an explicit formula so we can determine for which values of \(r\) the limit \(\lim _{n \to \infty } s_n\) exists. First, note that by using the limit laws,
\[ \lim _{n \to \infty } s_n = \lim _{n \to \infty } \frac {1 - r^{n+1}}{1-r} = \frac {1}{1-r} - \frac {1}{1-r} \lim _{n \to \infty } r^{n+1}. \]
The existence of \(\lim _{n \to \infty } s_n \)is thus entirely determined by whether \(\lim _{n \to \infty } r^{n+1}\) exists, and this limit is the limit of a geometric sequence! In fact,
- if \(-1<r<1\), then \(\lim _{n \to \infty } r^{n+1}\) existsdoes not exist.
- if \(r>1\) or \(r\le -1\), then \(\lim _{n \to \infty } r^{n+1}\) existsdoes not exist.
The above formula covers every case except when \(r= 1\), but notice that
\[\sum _{k=0}^n 1 = n+1,\]
so if \(r=1\), \(s_n = \answer [given]{n+1}\) and \(\lim _{n \to \infty } s_n = \infty \), so \(\sum _{k=0}^{\infty } 1\) diverges.
When \(-1<r<1\), note \(\lim _{n \to \infty } r^{n+1}=0\), so in this case,
\[ \lim _{n\to \infty }\frac {1 - r^{n+1}}{1-r} = \frac {1-\answer [given]{0}}{1-r}. \]
By noting that \(\sum _{k=0}^n ar^k = a \sum _{k=0}^n r^k\), we can combine this observation with the above argument and write the result in a theorem.
The geometric series \(\sum _{k= 0}^\infty a \cdot r^k\)
- converges to \(\frac {a}{1-r}\) when \(|r| < 1\).
- diverges if \(|r| \geq 1\).
There is a useful trick that allows us to find the sum of a convergent geometric series when the lower index does not start at \(0\).
The series \(\sum _{k=3}^{\infty } \left (\frac {2}{3}\right )^k\) is a geometric series with \(r=\frac {2}{3}<1\), so it converges. To find the value to which it
converges, notice the following.
\begin{align*} \sum _{k=3}^{\infty } \left (\frac {2}{3}\right )^k &= \left (\frac {2}{3}\right )^3+ \left (\frac {2}{3}\right )^4+ \left (\frac {2}{3}\right )^5+\ldots \\ &= \left (\frac {2}{3}\right )^3 \cdot \left (1+ \frac {2}{3}+ \left (\frac {2}{3}\right )^2+\ldots \right ) \\ &= \frac {8}{27} \cdot \sum _{k=0}^{\infty }\left (\frac {2}{3}\right )^k \\ &= \sum _{k=0}^{\infty } \frac {8}{27} \cdot \left (\frac {2}{3}\right )^k \end{align*}
This is now a geometric series whose lower index is \(0\), so we can use the formula to find its value. Noting that \(a=\answer [given]{ \frac {8}{27} }\) and \(r= \frac {2}{3}\) gives:
\[ \sum _{k=3}^{\infty } \left (\frac {2}{3}\right )^k = \frac {8/27}{1-2/3} = \frac {8}{9}. \]
We can easily generalize this example and doing so allows us to write down a more comprehensive theorem about geometric series.
The geometric series \(\sum _{k= k_0}^\infty a \cdot r^k\)
- converges to \(\frac {ar^{k_0}}{1-r}\) when \(|r| < 1\).
- diverges if \(|r| \geq 1\).
Let’s take another look at the series that started off the section, \(\sum _{k=1}^{\infty } \left (\frac {1}{2}\right )^k\). Here, \(a=\answer [given]{1}\), \(r=\answer [given]{1/2}\) and \(k_0 = \answer [given]{1}\). Since
\(|r|<1\), the series convergesdiverges, and using the formula above, we have
that
\[ \sum _{k=1}^{\infty } \left (\frac {1}{2}\right )^k = \frac {ar^{k_0}}{1-r} = \frac {1\cdot \frac {1}{2}}{1-1/2} =1.\]
This matches the earlier result!
Although we have mentioned it before, we mention it again here:
The lower index in a series does not affect whether the series converges or diverges, but if the series converges, it can affect the value to which the series converges.
The formula listed above very explicitly shows exactly how the lower index \(k_0\) affects the value to which a convergent geometric series converges.
Now, try some questions to check your understanding of the above material.
Which of the following series converge?
Determine if the series \(\sum _{k=2}^{\infty } 2^{3-2k}\) converges or diverges. If it converges, give the value to which
it converges.
First, note that the series isis not geometric since the laws of exponents allow us
to write the following.
\[ 2^{3-2k} = \frac {2^3}{2^{2k}} = \frac {8}{\left (2^2\right )^k} = 8 \cdot \frac {1}{4^k} = 8 \cdot \frac {1^k}{4^k} = 8 \cdot \left (\frac {1}{4}\right )^k \]
The series is geometric with \(r = \answer [given]{1/4}\), and using the result \(\sum _{k=k_0} ar^k = \frac {ar^{k_0}}{1-r}\) gives:
\[ \sum _{k=2}^{\infty } 2^{3-2k} = 8 \cdot \left (\frac {1}{4}\right )^k = \frac {8 \cdot (1/4)^2}{1-1/4} = \answer [given]{\frac {2}{3}}. \]
4 Telescoping series
A second type of series for which we can find an explicit formula for \(s_n\) are “telescoping series”. Rather than try to give a formal definition, we think of telescoping series are infinite sums for which the required addition required to find a formula for \(s_n\) can be done so many of the intermediate terms naturally cancel. An example will make this point more clear.
Evaluate the sum
\[ \sum _{k=1}^\infty \left (\frac {1}{k}-\frac {1}{k+1}\right ). \]
It will help to write down the first few partial sums for this series.
![[Picture]](digInSpecialSeries-0276eb7e9b3338f9e4262cf774c754ac.svg)
For \(s_2\) and
beyond, note how most of the intermediate terms in each partial sum cancel out! In
general, we can notice from pattern recognition (specifically by looking at the
denominator in each expression and comparing it to the index) that \(s_n =\) \(1-\frac {1}{n}\),\(1-\frac {1}{n+1}\). The
sequence \(\{s_n\}_{n=1}\) thus convergesdiverges since \(\lim _{n \to \infty } s_n\) existsdoes not exist. Furthermore,
since \(\lim _{n\to \infty }s_n = \lim _{n\to \infty }\left (1-\frac 1{n+1}\right ) = \answer [given]1\), we conclude that \(\sum _{n=1}^\infty \left (\frac 1n-\frac 1{n+1}\right ) = 1\).
Finding the above formula required us to use pattern recognition. Validating that the
pattern must hold for all terms in the sequence can be done formally by using an
idea called mathematical induction. We leave it to the curious reader to explore this
idea further if desired.
We’ve just seen an example of a telescoping series. Informally, a telescoping series is one in which the partial sums reduce to just a finite sum of terms. In the last example, the partial sum \(s_n\) only was the sum of two nonzero terms:
\[ s_n = 1 - \frac {1}{n-1}. \]
Determine if the series \(\sum _{k=1}^\infty \ln \left (\frac {k+1}{k}\right )\) converges or diverges.
We begin by writing the first few partial sums of the series: \begin{align*} s_1 &= \ln \left (2\right ) \\ s_2 &= \ln \left (2\right )+\ln \left (\frac 32\right ) \\ s_3 &= \ln \left (2\right )+\ln \left (\frac 32\right )+\ln \left (\frac 43\right ) \\ s_4 &= \ln \left (2\right )+\ln \left (\frac 32\right )+\ln \left (\frac 43\right )+\ln \left (\frac 54\right ) \end{align*}
![[Picture]](digInSpecialSeries-4170e72dbd4994413303852a58f324aa.svg)
At first, it doesn’t look like we will have much luck writing this as a telescoping series, but noting that \( \ln \left (\frac {n+1}{n}\right ) = \ln (n+1)-\ln (n)\) allows us to write out terms of \(s_n\) in a more convenient way.
We can conclude that \(s_n =\answer [given]{\ln (n+1)}\) and analyze \(\lim _{n \to \infty } s_n\).
Since \(\lim _{n\to \infty }s_n=\answer [given]{\infty }\), \(\sum _{k=1}^\infty \ln \left (\frac {k+1}{k}\right )\) diverges.
5 Summary
Now that we have seen two special types of series for which we can find an explicit formula for the \(n\)-th term in the sequence of partial sums, it helps to summarize the logic that we employed.
- Consider the associated sequence \(\{s_n\}\) of partial sums.
-
Try to find an explicit formula for the term \(s_n\). If you can find such a formula, analyze \(\lim _{n \to \infty s_n}\).
- If the limit exists, \(\sum _{k=k_0}^{\infty } a_k\) converges, and if we can determine that \(\lim _{n \to \infty } s_n =L\), then \(\sum _{k=k_0} a_k=L\).
- If \(\lim _{n \to \infty } s_n\) does not exist, then \(\sum _{k=k_0} a_k\) diverges.