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# 11.2: Series

• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College

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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

• Integrated by Justin Marshall.