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Mathematics LibreTexts

11.2: 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 \[\{s_n\}_{n=0}^\infty\]: $$s_n=\sum_{i=0}^n a_i.$$ 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
If \(a_n=kx^n\), \(\sum_{n=0}^\infty a_n\) 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 $$\lim_{n\to\infty}a_n=0$$.


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.


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

Show that \(\sum_{n=1}^\infty {1\over n}) diverges.


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.