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 $\{a_n\}_{n=0}^\infty$ is a sequence then the associated series is

$$\sum_{i=0}^\infty a_n=a_0+a_1+a_2+\cdots$$

Associated with a series is a second sequence, called the sequence of partial sums:

$$\{s_n\}_{n=0}^\infty$$

with

$$s_n=\sum_{i=0}^n a_i.$$

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

Example 11.2.1: Geometric Series

If $a_n=kx^n$, then

$$\sum_{n=0}^\infty a_n$$

is called a geometric series. A typical partial sum is

$$s_n=k+kx+kx^2+kx^3+\cdots+kx^n=k(1+x+x^2+x^3+\cdots+x^n).$$

We note that

$$\eqalign{ s_n(1-x)&=k(1+x+x^2+x^3+\cdots+x^n)(1-x)\cr &=k(1+x+x^2+x^3+\cdots+x^n)1-k(1+x+x^2+x^3+\cdots+x^{n-1}+x^n)x\cr &=k(1+x+x^2+x^3+\cdots+x^n-x-x^2-x^3-\cdots-x^n-x^{n+1})\cr &=k(1-x^{n+1})\cr }$$

so

$$\eqalign{ s_n(1-x)&=k(1-x^{n+1})\cr s_n&=k{1-x^{n+1}\over 1-x}.\cr }$$

If $|x| < 1$, $\lim_{n\to\infty}x^n=0$ so

$$\lim_{n\to\infty}s_n=\lim_{n\to\infty}k{1-x^{n+1}\over 1-x}= k{1\over 1-x}.$$

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

$$s_n={1-(1/2)^{n+1}\over 1-1/2}={2^{n+1}-1\over 2^n}=2-{1\over 2^n} \quad\hbox{and}\quad \sum_{n=0}^\infty {1\over 2^n} = {1\over 1-1/2} = 2.$$

We began the chapter with the series $\sum_{n=1}^\infty {1\over 2^n},$ 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, $$\sum_{n=1}^\infty {1\over 2^n}=1.$$

It is not hard to see that the following theorem follows from theorem 11.1.2.

Theorem 11.2.2

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

1. $\sum ca_n$ is convergent and $\sum ca_n=c\sum a_n$
2. $\sum (a_n+b_n)$ is convergent and $\sum (a_n+b_n)=\sum a_n+\sum b_n$.

The two parts of this theorem are subtly different. Suppose that $\sum a_n$ diverges; does $\sum ca_n$ also diverge if $c$ is non-zero? Yes: suppose instead that $\sum ca_n$ converges; then by the theorem, $\sum (1/c)ca_n$ converges, but this is the same as $\sum a_n$, which by assumption diverges. Hence $\sum ca_n$ also diverges. Note that we are applying the theorem with $a_n$ replaced by $ca_n$ and $c$ replaced by $(1/c)$.

Now suppose that $\sum a_n$ and $\sum b_n$ diverge; does $\sum (a_n+b_n)$ also diverge? Now the answer is no: Let $a_n=1$ and $b_n=-1$, so certainly $\sum a_n$ and $\sum b_n$ diverge. But

$$\sum (a_n+b_n)=\sum(1+-1)=\sum 0 = 0.$$

Of course, sometimes $\sum (a_n+b_n)$ will also diverge, for example, if $a_n=b_n=1$, then $$\sum (a_n+b_n)=\sum(1+1)=\sum 2$$ diverges.

In general, the sequence of partial sums $s_n$ is harder to understand and analyze than the sequence of terms $a_n$, and it is difficult to determine whether series converge and if so to what. Sometimes things are relatively simple, starting with the following.

Theorem 11.2.3

If

$$\sum a_n$$

converges then

$$\lim_{n\to\infty}a_n=0.$$

Proof.

Since $\sum a_n$ converges, $\lim_{n\to\infty}s_n=L$ and $\lim_{n\to\infty}s_{n-1}=L$, because this really says the same thing but "renumbers'' the terms. By theorem 11.1.2,

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

But

$$s_{n}-s_{n-1}=(a_0+a_1+a_2+\cdots+a_n)-(a_0+a_1+a_2+\cdots+a_{n-1}) =a_n,$$

so as desired $\lim_{n\to\infty}a_n=0$.

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

Example 11.2.4

Show that

$$\sum_{n=1}^\infty {n\over n+1}$$

diverges.

Solution

We compute the limit: $$\lim _{n\to\infty}{n\over n+1}=1\not=0.$$ Looking at the first few terms perhaps makes it clear that the series has no chance of converging:

$${1\over2}+{2\over3}+{3\over4}+{4\over5}+\cdots$$

will just get larger and larger; indeed, after a bit longer 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 11.2.5: Harmonic Series

Show that

$$\sum_{n=1}^\infty {1\over n}$$

diverges.

Solution

Here the theorem does not apply: $\lim _{n\to\infty} 1/n=0$, so it looks like perhaps the series converges. Indeed, if you have the fortitude (or the software) to add up the first 1000 terms you will find that $$\sum_{n=1}^{1000} {1\over n}\approx 7.49,$$ 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:

$$1+{1\over 2}+{1\over 3}+{1\over 4} > 1+{1\over 2}+{1\over 4}+{1\over 4} = 1+{1\over 2}+{1\over 2}$$

$$1+{1\over 2}+{1\over 3}+{1\over 4}+ {1\over 5}+{1\over 6}+{1\over 7}+{1\over 8} > 1+{1\over 2}+{1\over 4}+{1\over 4}+{1\over 8}+{1\over 8}+{1\over 8}+{1\over 8} = 1+{1\over 2}+{1\over 2}+{1\over 2}$$

$$1+{1\over 2}+{1\over 3}+\cdots+{1\over16}> 1+{1\over 2}+{1\over 4}+{1\over 4}+{1\over 8}+\cdots+{1\over 8}+{1\over16}+\cdots +{1\over16} =1+{1\over 2}+{1\over 2}+{1\over 2}+{1\over 2}$$

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

$$1+{1\over 2}+{1\over 3}+\cdots+{1\over 2^n} > 1+{n\over 2},$$

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 $4\cdot 10^{59}$ terms. This series, $\sum (1/n)$, is called the harmonic series.

Contributors

• David Guichard (Whitman College)

• Integrated by Justin Marshall.