Let \(S_n\) be the sum of the first \(n\) terms of the sequence \(\{1/2^n\}\text{.}\) From the above, we see that \(S_1=1/2\text{,}\)\(S_2 = 3/4\text{,}\) etc. Our formula at the end shows that \(S_n = 1-1/2^n\text{.}\)
Now consider the following limit: \(\lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\big(1-1/2^n\big) = 1\text{.}\) This limit can be interpreted as saying something amazing: the sum of all the terms of the sequence \(\{1/2^n\}\) is 1.
Let \(S_n\) denote the sum of the first \(n\) terms in the sequence \(\{a_n\}\text{,}\) known as the \(n\)th partial sum of the sequence. We can then define the sequence \(\{S_n\}\) of partial sums of \(\{a_n\}\text{.}\)
If the sequence \(\{S_n\}\) converges to \(L\text{,}\) we say the series \(\ds \sum_{n=k}^\infty a_n\) converges to \(L\text{,}\) and we write \(\ds \sum_{n=k}^\infty a_n = L\text{.}\)
Note that in the definition above, we do not necessarily assume that our sum begins with \(n=1\text{.}\) In fact, it is quite common to have a series beginning at \(n=0\text{,}\) and in some cases we may need to consider other values as well. The \(n\)th partial sum \(S_n\) will always denote the sum of the first \(n\) terms: For example, \(\infser 1/n\) has
Since \(\lim\limits_{n\to\infty}S_n = \infty\text{,}\) we conclude that the series \(\ds \infser n^2\) diverges. It is instructive to write \(\ds \infser n^2=\infty\) for this tells us how the series diverges: it grows without bound. A scatter plot of the sequences \(\{a_n\}\) and \(\{S_n\}\) is given in Figure 9.2.4.(a). The terms of \(\{a_n\}\) are growing, so the terms of the partial sums \(\{S_n\}\) are growing even faster, illustrating that the series diverges.
The sequence \(\{b_n\}\) starts with 1, \(-1\text{,}\) 1, \(-1\text{,}\)\(\ldots\text{.}\) Consider some of the partial sums \(S_n\) of \(\{b_n\}\text{:}\)
This pattern repeats; we find that \(S_n = \begin{cases} 1 \amp n\, \text{ is odd } \\, 0 \amp n\, \text{ is even } \end{cases}\text{.}\) As \(\{S_n\}\) oscillates, repeating 1, 0, 1, 0, \(\ldots\text{,}\) we conclude that \(\lim\limits_{n\to\infty}S_n\) does not exist, hence \(\ds\infser (-1)^{n+1}\) diverges. A scatter plot of the sequence \(\{b_n\}\) and the partial sums \(\{S_n\}\) is given in Figure 9.2.4.(b). When \(n\) is odd, \(b_n = S_n\) so the marks for \(b_n\) are drawn oversized to show they coincide.
The scatter plot shows the first 10 terms in the sequence \(a_n=n^2\text{,}\) which follow the graph \(y=x^2\text{.}\) On the same plot, the first 10 terms in the sequence of partial sums \(S_n\) are shown. As one might expect, the value of \(S_n\) grows much more rapidly than that of \(a_n\text{.}\)
This scatter plot shows the terms in the sequence \(b_n\text{,}\) along with the corresponding partial sums. Since the value of \(b_n\) alternates between \(-1\) and \(1\text{,}\) we see one sequence of dots along the line \(y=1\text{,}\) and another along the line \(y=-1\text{.}\)
When \(n\) is even, \(S_n=0\text{,}\) so we also see a sequence of dots along the \(n\) axis. When \(n\) is odd, \(S_n=a_n\text{,}\) so the dots for \(b_n\) and \(S_n\) overlap.
While it is important to recognize when a series diverges, we are generally more interested in the series that converge. In this section we will demonstrate a few general techniques for determining convergence; later sections will delve deeper into this topic.
We started this section with a geometric series, although we dropped the first term of \(1\text{.}\) One reason geometric series are important is that they have nice convergence properties.
We begin by proving the formula for the simplied form for the partial sums. Consider the \(n\)th partial sum of the geometric series, \(S_n=\sum_{i=0}^n r^i\text{:}\)
Now, examining the partial sums, we consider five cases to determine when \(S_n\) converges:
If \(\abs{r}\lt 1\text{,}\) then \(r^n \to 0\) as \(n \to \infty\text{,}\) so we have \(\inflim S_n=\frac{1-0}{1-r}=\frac{1}{1-r}\text{,}\) a convergent sequence of partial sums.
If \(r \lt -1\text{,}\) then \(r^n\) will oscillate between large positive and large negative values as \(n\) increases. The same will be true of \(S_n\text{,}\) so \(\inflim S_n\) does not exist.
If \(r=1\text{,}\) then \(S_n = \frac{1-1^{n+1}}{1-1}\) is undefined. However, examining \(S_n = 1+r+r^2+\dots+r^n\) for \(r=1\text{,}\) we can see that the partial sums simplify to \(S_n=n\text{,}\) and this sequence diverges to \(\infty\text{.}\)
If \(r=-1\text{,}\) then \(S_n = \frac{1-(-1)^{n}}{2}\text{.}\) For even values of \(n\text{,}\) the partial sums are always \(0\text{.}\) For odd values of \(n\text{,}\) the partial sums are always \(1\text{.}\) So the sequence of partial sums diverges.
converges as \(r=1/2 \lt 1\text{,}\) and \(\ds \infser[0] \frac{1}{2^n} = \frac{1}{1-1/2} = 2\text{.}\) This concurs with our introductory example; while there we got a sum of 1, we skipped the first term of 1.
However, note the subscript of the summation in the given series: we are to start with \(n=2\text{.}\) Therefore we subtract off the first two terms, giving:
Scatter plots for both \(a_n = (3/4)^n\) and \(S_n\) are shown together in the same image. The point for \(a_2\) is not visible, as it is covered up by \(S_2\text{,}\) which has the same value.
The points in the plot for \(a_n\) illustrate a sequence that decreases toward 0, while the points in the plot for \(S_n\) show that the partial sums are increasing toward a value slightly larger than 2.
The partial sums of this series are plotted in Figure 9.2.10. Note how the partial sums are not purely increasing as some of the terms of the sequence \(\{(-1/2)^n\}\) are negative.
Scatter plots for \(a_n = (-1/2)^n\) and the corresponding partial sums are shown together in the same image. The numbers \(a_n\) alternate in sign between positive and negative values, but get closer to 0 as \(n\) increases.
The \(y\) coordinates for the points in the scatter plot for \(S_n\) also oscillate, but the oscillations get steadily smaller, and appear to settle down toward a common \(y\) value.
Scatter plots for the sequence \(a_n=3^n\) and the corresponding partial sums are shown together in the same image. Both plots follow curves that appear to be exponential in nature, with the points for \(S_n\) slightly above those for \(a_n\text{.}\)
This is a \(p\)-series with \(p=1\text{.}\) By Theorem 9.2.13, this series diverges. This series is a famous series, called the Harmonic Series, so named because of its relationship to harmonics in the study of music and sound.
This is a \(p\)-series with \(p=2\text{.}\) By Theorem 9.2.13, it converges. Note that the theorem does not give a formula by which we can determine what the series converges to; we just know it converges. A famous, unexpected result is that this series converges to \(\ds{\pi^2}/{6}\text{.}\)
This is not a \(p\)-series; the definition does not allow for alternating signs. Therefore we cannot apply Theorem 9.2.13. (Another famous result states that this series, the Alternating Harmonic Series, converges to \(\ln(2)\text{.}\))
Later sections will provide tests by which we can determine whether or not a given series converges. This, in general, is much easier than determining what a given series converges to. There are many cases, though, where the sum can be determined.
Note how most of the terms in each partial sum are canceled out! In general, we see that \(\ds S_n = 1-\frac{1}{n+1}\text{.}\) The sequence \(\{S_n\}\) converges, as \(\lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\left(1-\frac1{n+1}\right) = 1\text{,}\) and so we conclude that \(\ds \infser \left(\frac1n-\frac1{n+1}\right) = 1\text{.}\) Partial sums of the series are plotted in Figure 9.2.16.
A scatter plot for the sequence \(a_n = \frac1n-\frac{1}{n+1}=\frac{1}{n(n+1)}\) is shown. These points begin at \((1,1/2)\) and then descend toward the \(n\) axis.
Also shown is the scatter plot for the sequence \(S_n\) of partial sums of this series. These terms are shown to be given by \(S_n = 1-\frac{1}{n+1}\text{,}\) and we see that the points on the scatter plot begin at \((1,1/2)\) and then rise toward the line \(y=1\text{.}\)
The series in Example 9.2.15 is an example of a telescoping series. Informally, a telescoping series is one in which most terms cancel with preceding or following terms, reducing the number of terms in each partial sum. The partial sum \(S_n\) did not contain \(n\) terms, but rather just two: 1 and \(1/(n+1)\text{.}\)
When possible, seek a way to write an explicit formula for the \(n\)th partial sum \(S_n\text{.}\) This makes evaluating the limit \(\lim\limits_{n\to\infty} S_n\) much more approachable. We do so in the next example.
(See Section 6.4, Partial Fraction Decomposition, to recall how this is done, if necessary.) Expressing the terms of \(\{S_n\}\) is now more instructive:
We again have a telescoping series. In each partial sum, most of the terms cancel and we obtain the formula \(\ds S_n = 1+\frac12-\frac1{n+1}-\frac1{n+2}\text{.}\) Taking limits allows us to determine the convergence of the series:
We can conclude that \(\{S_n\} = \big\{\ln(n+1)\big\}\text{.}\) This sequence does not converge, as \(\lim\limits_{n\to\infty}S_n=\infty\text{.}\) Therefore \(\ds\infser \ln\left(\frac{n+1}{n}\right)=\infty\text{;}\) the series diverges. Note in Figure 9.2.18.(b) how the sequence of partial sums grows slowly; after 100 terms, it is not yet over 5. Graphically we may be fooled into thinking the series converges, but our analysis above shows that it does not.
The scatter plots given appear very similar to the previous example. Again, points plotted for the terms in the sequence \(a_n\text{,}\) which begin at \((1,2/3)\text{,}\) descend toward the \(n\) axis, while the points for the sequence of partial sums ascend from the same point, in this case getting closer and closer to the line \(y=3/2\text{.}\)
Scatter plots are shown for both \(a_n = \ln(1+1/n)\) and \(S_n = \ln(n+1)\text{,}\) for \(n=10,20,\ldots, 100\text{.}\) The points for the sequence \(a_n\) are all very close to the \(n\) axis. The points for the sequence \(S_n\) of partial sums follow the graph \(y=\ln(x+1)\text{;}\) although this may appear to be bounded, we know that the value of the logarithm will eventually approach infinity, even if it does so very slowly.
This looks very similar to the series that involves \(e\) in Key Idea 9.2.20. Note, however, that the series given in this example starts with \(n=1\) and not \(n=0\text{.}\) The first term of the series in the Key Idea is \(1/0! = 1\text{,}\) so we will subtract this from our result below:
The points in the scatter plot for the sequence \(a_n\) appear to alternate betwee positive and negative \(y\) values, and they converge toward the \(n\) axis from both sides.
Scatter plots are shown for the sequence \(a_n = 1000/n!\text{,}\) and the corresponding partial sums. The value of \(a_n\) approaches zero quite rapidly, and as a result, we also see that the points in the plot for \(S_n\) quickly converge toward the limiting value.
The denominators in each term are perfect squares; we are adding \(\ds \sum_{n=4}^\infty \frac{1}{n^2}\) (note we start with \(n=4\text{,}\) not \(n=1\)). This series will converge. Using the formula from Key Idea 9.2.20, we have the following:
It may take a while before one is comfortable with this statement, whose truth lies at the heart of the study of infinite series: it is possible that the sum of an infinite list of nonzero numbers is finite. We have seen this repeatedly in this section, yet it still may “take some getting used to.”
Important! This theorem does not state that if \(\ds \lim_{n\to\infty} a_n = 0\) then \(\ds \sum_{n=1}^\infty a_n\) converges. The standard example of this is the Harmonic Series, as given in Key Idea 9.2.20. The Harmonic Sequence, \(\{1/n\}\text{,}\) converges to 0; the Harmonic Series, \(\ds \sum_{n=1}^\infty \frac1n\text{,}\) diverges.
Looking back, we can apply this theorem to the series in Example 9.2.3. In that example, the \(n\)th terms of both sequences do not converge to 0, therefore we can quickly conclude that each series diverges.
One can rewrite Theorem 9.2.23 to state “If a series converges, then the underlying sequence converges to 0.” While it is important to understand the truth of this statement, in practice it is rarely used. It is generally far easier to prove the convergence of a sequence than the convergence of a series.
Consider once more the Harmonic Series \(\ds\infser \frac1n\) which diverges; that is, the sequence of partial sums \(\{S_n\}\) grows (very, very slowly) without bound. One might think that by removing the “large” terms of the sequence that perhaps the series will converge. This is simply not the case. For instance, the sum of the first 10 million terms of the Harmonic Series is about 16.7. Removing the first 10 million terms from the Harmonic Series changes the \(n\)th partial sums, effectively subtracting 16.7 from the sum. However, a sequence that is growing without bound will still grow without bound when 16.7 is subtracted from it.
The equations below illustrate this. The first line shows the infinite sum of the Harmonic Series split into the sum of the first 10 million terms plus the sum of “everything else.” The next equation shows us subtracting these first 10 million terms from both sides. The final equation employs a bit of “psuedo-math”: subtracting 16.7 from “infinity” still leaves one with “infinity.”
it is not apparent that the partial sums diverge. Indeed they do diverge, but very, very slowly. (If you graph them on a logarithmic scale however, you can clearly see the divergence of the partial sums.) Instead, we will consider the partial sums, indexed by powers of \(2\text{.}\) That is, we will consider \(S_2,S_4, S_8, S_{16}, \dots\text{.}\)
Generally, we can see that \(S_{2^n} \gt 1+\frac{n}2\text{.}\) (In order to really show this, we should employ proof by induction.) Since the sequence of partial sums clearly diverges, so does the series \(\infser 1/n\text{.}\)
This section introduced us to series and defined a few special types of series whose convergence properties are well known: we know when a \(p\)-series or a geometric series converges or diverges. Most series that we encounter are not one of these types, but we are still interested in knowing whether or not they converge. The next three sections introduce tests that help us determine whether or not a given series converges.
Explain why (a) and (b) demonstrate that the series of odd terms is convergent, if, and only if, the series of even terms is also convergent. (That is, show both converge or both diverge.)