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11.3: Series

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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:



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

    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


    We note that

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


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

    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


    \[\sum a_n\]

    converges then



    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,

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


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

    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

    \[\sum_{n=1}^\infty {n\over n+1}\]



    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:


    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

    \[\sum_{n=1}^\infty {1\over n}\]



    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.


    This page titled 11.3: Series is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Guichard via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.