11.3: Series

While much more can be said about sequences, we now turn to our principal interest, series. Recall that a series, roughly speaking, is the sum of a sequence: if \(\bigg\{a_i\bigg\}_{i=0}^\infty\) is a sequence, then the associated series is \[\sum_{i=0}^\infty a_i= a_0+a_1+a_2+\cdots\nonumber\]

Associated with a series is a second sequence, called the sequence of partial sums: \(\bigg\{s_n\bigg\}_{n=0}^\infty\) \[s_n=\sum_{i=0}^n a_i.\nonumber\]

So \[s_0=a_0,\quad s_1=a_0+a_1,\quad s_2=a_0+a_1+a_2,\quad \ldots\nonumber\] A series converges if the sequence of partial sums converges, and otherwise the series diverges.

Example \(\PageIndex{1}\)

If \(a_n=kx^n\), then \(\displaystyle\sum_{n=0}^\infty a_n\) is called a geometric series. A typical partial sum is \[s_n= k+kx+kx^2+kx^3+\cdots+kx^n= k(1+x+x^2+x^3+\cdots+x^n),\nonumber\]

where \(k\) is some constant. We note that \[\eqalign{s_n\times(1-x)&= k(1+x+x^2+x^3+\cdots+x^n)(1-x)\cr &= k(1+x+x^2+x^3+\cdots+x^n)\;-\;kx(1+x+x^2+x^3+\cdots+x^{n-1}+x^n)\cr &= k(1+x+x^2+x^3+\cdots+x^n\quad\quad\quad-x-x^2-x^3-\cdots-x^n-x^{n+1})\cr &= k(1-x^{n+1})\cr}\nonumber\]so \[s_n=k\left(\dfrac{1-x^{n+1}}{1-x}\right).\nonumber\]

If \(|x| \lt 1\), \(\displaystyle\lim_{n\to\infty}x^n=0,\quad\) so \[\lim_{n\to\infty}s_n=\lim_{n\to\infty}k\left(\dfrac{1-x^{n+1}}{1-x}\right)= \dfrac{k}{1-x}.\nonumber\]

Thus, when \(|x| \lt 1\) the geometric series converges to \(k/(1-x)\). When, for example, \(k=1\) and \(x=1/2\):

\[ s_n=\dfrac{1-(1/2)^{n+1}}{1-1/2}= \dfrac{2^{n+1}-1}{2^n}= 2-\dfrac{1}{2^n}, \qquad\text{and}\qquad \sum_{n=0}^\infty\dfrac{1}{2^n} = \dfrac{1}{1-1/2} = 2.\nonumber\]

We began the chapter with the series \[\sum_{n=1}^\infty\dfrac{1}{2^n},\nonumber\] namely, the geometric series without the first term \(1\). Each partial sum of this series is \(1\) less than the corresponding partial sum for the geometric series, so of course the limit is also one less than the value of the geometric series, that is, \[\sum_{n=1}^\infty\dfrac{1}{2^n}=1.\nonumber\]

It is not hard to see that the following theorem follows from theorem 11.2.2.

Theorem \(\PageIndex{2}\)

Suppose that \(\sum a_n\) and \(\sum b_n\) are convergent series, and \(c\) is a constant. Then

\(\sum ca_n\) is convergent and \(\sum ca_n=c\sum a_n\);
\(\sum (a_n+b_n)\) is convergent and \(\sum (a_n+b_n)=\sum a_n+\sum b_n\).

The converses of the two parts of this theorem are subtly different. For the first part, suppose that \(\sum a_n\) diverges; does \(\sum ca_n\) also diverge if \(c\) is non-zero? Yes, because if (for a contradiction) one assumes instead that \(\sum ca_n\) converges; then by the theorem, \(\sum (1/c)ca_n\) also converges, but this is the same as \(\sum a_n\), which we have assumed is divergent. Hence \(\sum ca_n\) must also diverge. (Note that we are applying the theorem with \(a_n\) replaced by \(ca_n\) and \(c\) replaced by \((1/c)\)).

For the second part, suppose that \(\sum a_n\) and \(\sum b_n\) both diverge; must \(\sum (a_n+b_n)\) also be divergent? Now the answer is no, not necessarily. For example, let \(a_n=1\) and \(b_n=-1\), for all \(n\), so certainly \(\sum a_n\) and \(\sum b_n\) diverge. But \[\sum (a_n+b_n)=\sum(1+-1)=\sum 0 = 0.\nonumber\]

Of course, sometimes \(\sum (a_n+b_n)\) will also diverge, for example, if \(a_n=b_n=1\), then \(\sum (a_n+b_n)=\sum(1+1)=\sum 2\) which is clearly divergent.

In general, the sequence of partial sums \(s_n\) is harder to understand and analyze than the sequence of terms \(a_n\), and it can be difficult to determine whether a given series is convergent, and if so, to what. Sometimes things are relatively simple however, starting with the following.

Theorem \(\PageIndex{3}\)

If \(\displaystyle\;\;\sum a_n\;\) converges, then \(\displaystyle\lim_{n\to\infty}a_n=0.\)

Proof.

Since \(\displaystyle\sum a_n\) converges, \(\displaystyle\lim_{n\to\infty}s_n=L\) and also \(\displaystyle\lim_{n\to\infty}s_{n-1}=L\), because this really says the same thing but "renumbers" the terms. By theorem 11.2.2,

\[\lim_{n\to\infty} (s_{n}-s_{n-1})= \lim_{n\to\infty} s_{n}-\lim_{n\to\infty}s_{n-1}=L-L=0.\nonumber\]

But \[s_{n}-s_{n-1}= (a_0+a_1+a_2+\cdots+a_n)-(a_0+a_1+a_2+\cdots+a_{n-1}) =a_n,\nonumber\] so as desired, \(\displaystyle\; \lim_{n\to\infty}a_n=0\).

This theorem presents an easy divergence test: if given a series \(\sum a_n\) the limit \(\displaystyle\;\lim_{n\to\infty}a_n\;\) does not exist, or has a value other than zero, the series diverges. Note well that the converse is not true: If \(\displaystyle\;\lim_{n\to\infty}a_n=0,\;\) then the series does not necessarily converge; it may, or may not.

Example \(\PageIndex{4}\)

Show that \(\;\displaystyle\sum_{n=1}^\infty \dfrac{n}{n+1}\;\) diverges.

Solution

We compute the limit: \[\lim _{n\to\infty}\dfrac{n}{n+1}=1\not=0.\nonumber\]
Looking at the first few terms perhaps makes it clear that the series has no chance of converging: \[{1\over 2}+{2\over 3}+{3\over 4}+{4\over 5}+\cdots\nonumber\] The partial sums just get larger and larger; indeed, after a bit further the series starts to look very much like \(\;\cdots+1+1+1+1+\cdots\;\), and of course if we add up enough \(1\)'s we can make the sum as large as we desire.

Example \(\PageIndex{5}\): Harmonic Series

Show that \(\displaystyle\;\sum_{n=1}^\infty \dfrac{1}{n}\;\) diverges.

Solution

Here, theorem \(\PageIndex{3}\) is not conclusive: \(\displaystyle\lim _{n\to\infty} 1/n=0\), so it looks like perhaps the series may be convergent. Indeed, if you have the fortitude (or the software) to add up the first 1000 terms you will find that \(\displaystyle\sum_{n=1}^{1000}\dfrac{1}{n}\approx 7.49,\) so it might be reasonable to speculate that the series converges to something in the neighborhood of \(10\). But in fact the partial sums do go to infinity; they just get big very, very slowly. Consider the following:

\[ 1+{1\over 2}+{1\over 3}+{1\over 4} \gt 1+{1\over 2}+{1\over 4}+{1\over 4} = 1+{1\over 2}+{1\over 2}\nonumber\]
\[ 1+{1\over 2}+{1\over 3}+{1\over 4}+ {1\over 5}+{1\over 6}+{1\over 7}+{1\over 8} \gt 1+{1\over 2}+{1\over 4}+{1\over 4}+{1\over 8}+{1\over 8}+{1\over 8}+{1\over 8} = 1+{1\over 2}+{1\over 2}+{1\over 2}\nonumber\]
\[ 1+{1\over 2}+{1\over 3}+\cdots+{1\over 16} \gt 1+{1\over 2}+{1\over 4}+{1\over 4}+{1\over 8}+\cdots+{1\over 8}+{1\over 16}+\cdots +{1\over 16} = 1+{1\over 2}+{1\over 2}+{1\over 2}+{1\over 2}\nonumber\]

and so on. By swallowing up more and more terms we can always manage to add at least another \(1/2\) to the sum, and by adding enough of these we can make the partial sums as big as we like. In fact, it's not hard to see from this pattern that \[1+{1\over 2}+{1\over 3}+\cdots+{1\over 2^n} \gt 1+{n\over 2},\nonumber\]

so to make sure the sum is over \(100\), for example, we'd add up terms until we get to around \(1/2^{198}\), that is, about \(4\times 10^{59}\) terms. This series, \(\sum (1/n)\), is called the harmonic series.

Exercises \(\PageIndex{}\)

Exercise \(\PageIndex{1}\)

Explain why \(\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2}{2n^2+1}\) diverges.

Answer

\(\displaystyle\lim_{n\to{\infty}}\dfrac{n^2}{2n^2+1}= \dfrac{1}{2},\quad \ne 0.\)

Exercise \(\PageIndex{2}\)

Explain why \(\displaystyle\sum_{n=1}^{\infty}\dfrac{5}{2^{1/n}+14}\) diverges.

Answer

\(\displaystyle\lim_{n\to{\infty}}\dfrac{5}{2^{1/n}+14}= \dfrac{1}{3},\quad \ne 0.\)

Exercise \(\PageIndex{3}\)

Explain why \(\displaystyle\sum_{n=1}^{\infty}\dfrac{3}{n}\) diverges.

Answer

This series is three times the harmonic series \(\sum 1/n\), which diverges.

Exercise \(\PageIndex{4}\)

Compute \(\displaystyle\sum_{n=0}^{\infty}\left(\dfrac{4}{(-3)^n}-\dfrac{3}{3^n}\right).\)

Answer

\(-\dfrac{3}{2}\)

Exercise \(\PageIndex{5}\)

Compute \(\displaystyle\sum_{n=0}^{\infty}\left(\dfrac{3}{2^n}+\dfrac{4}{5^n}\right).\)

Answer

\(11\)

Exercise \(\PageIndex{6}\)

Compute \(\displaystyle\sum_{n=0}^{\infty}\dfrac{4^{n+1}}{5^n}.\)

Answer

\(20\)

Exercise \(\PageIndex{7}\)

Compute \(\displaystyle\sum_{n=0}^{\infty}\dfrac{3^{n+1}}{7^{n+1}}.\)

Answer

\(\dfrac{3}{4}\)

Exercise \(\PageIndex{8}\)

Compute \(\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{3}{5}\right)^n.\)

Answer

\(\dfrac{3}{2}\)

Exercise \(\PageIndex{9}\)

Compute \(\displaystyle\sum_{n=1}^{\infty}\dfrac{3^n}{5^{n+1}}.\)

Answer

\(\dfrac{3}{10}\)

Contributors

• David Guichard (Whitman College)

• Integrated by Justin Marshall.