5.2 Infinite series (Page 3/16)

Calculus volume 2 Page 3 / 16

Determine whether the series $\sum_{n = 1}^{\infty} (n + 1) / n$ converges or diverges.

The series diverges because the $k th$ partial sum $S_{k} > k .$

The harmonic series

A useful series to know about is the harmonic series . The harmonic series is defined as

\sum_{n = 1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ⋯ .

This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums ${S_{k}}$ approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence ${S_{k}}$ in the following table.

$k$	$10$	$100$	$1000$	$10,000$	$100,000$	$1,000,000$
$S_{k}$	$2.92897$	$5.18738$	$7.48547$	$9.78761$	$12.09015$	$14.39273$

Even after $1,000,000$ terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.

To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:

\begin{array}{l} S_{1} = 1 \\ S_{2} = 1 + \frac{1}{2} \\ S_{3} = 1 + \frac{1}{2} + \frac{1}{3} \\ S_{4} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} . \end{array}

Notice that for the last two terms in $S_{4},$

\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} .

Therefore, we conclude that

S_{4} > 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) = 1 + \frac{1}{2} + \frac{1}{2} = 1 + 2 (\frac{1}{2}) .

Using the same idea for $S_{8},$ we see that

\begin{matrix} S_{8} & = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) \\ = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1 + 3 (\frac{1}{2}) . \end{matrix}

From this pattern, we see that $S_{1} = 1,$ $S_{2} = 1 + 1 / 2,$ $S_{4} > 1 + 2 (1 / 2),$ and $S_{8} > 1 + 3 (1 / 2) .$ More generally, it can be shown that $S_{2^{j}} > 1 + j (1 / 2)$ for all $j > 1 .$ Since $1 + j (1 / 2) \to \infty,$ we conclude that the sequence ${S_{k}}$ is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since ${S_{k}}$ is unbounded, it diverges. Thus, the harmonic series diverges.

Algebraic properties of convergent series

Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.

Algebraic properties of convergent series

Let $\sum_{n = 1}^{\infty} a_{n}$ and $\sum_{n = 1}^{\infty} b_{n}$ be convergent series. Then the following algebraic properties hold.

The series $\sum_{n = 1}^{\infty} (a_{n} + b_{n})$ converges and $\sum_{n = 1}^{\infty} (a_{n} + b_{n}) = \sum_{n = 1}^{\infty} a_{n} + \sum_{n = 1}^{\infty} b_{n} .$ (Sum Rule)
The series $\sum_{n = 1}^{\infty} (a_{n} - b_{n})$ converges and $\sum_{n = 1}^{\infty} (a_{n} - b_{n}) = \sum_{n = 1}^{\infty} a_{n} - \sum_{n = 1}^{\infty} b_{n} .$ (Difference Rule)
For any real number $c,$ the series $\sum_{n = 1}^{\infty} c a_{n}$ converges and $\sum_{n = 1}^{\infty} c a_{n} = c \sum_{n = 1}^{\infty} a_{n} .$ (Constant Multiple Rule)

Using algebraic properties of convergent series

Evaluate

\sum_{n = 1}^{\infty} [\frac{3}{n (n + 1)} + {(\frac{1}{2})}^{n - 2}] .

We showed earlier that

\sum_{n = 1}^{\infty} \frac{1}{n (n + 1)}

and

\sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n - 1} = 2 .

Since both of those series converge, we can apply the properties of [link] to evaluate

\sum_{n = 1}^{\infty} [\frac{3}{n (n + 1)} + {(\frac{1}{2})}^{n - 2}] .

Using the sum rule, write

\sum_{n = 1}^{\infty} [\frac{3}{n (n + 1)} + {(\frac{1}{2})}^{n - 2}] = \sum_{n = 1}^{\infty} \frac{3}{n (n + 1)} {+ \sum}_{n = 1}^{\infty} {(\frac{1}{2})}^{n - 2} .

Then, using the constant multiple rule and the sums above, we can conclude that

\begin{array}{l} \sum_{n = 1}^{\infty} \frac{3}{n (n + 1)} + \sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n - 2} & = 3 \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} + {(\frac{1}{2})}^{−1} \sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n - 1} \\ = 3 (1) + {(\frac{1}{2})}^{−1} (2) = 3 + 2 (2) = 7. \end{array}

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Evaluate $\sum_{n = 1}^{\infty} \frac{5}{2^{n - 1}} .$

$10 .$

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Geometric series

A geometric series is any series that we can write in the form

a + a r + a r^{2} + a r^{3} + ⋯ = \sum_{n = 1}^{\infty} a r^{n - 1} .

Because the ratio of each term in this series to the previous term is r , the number r is called the ratio. We refer to a as the initial term because it is the first term in the series. For example, the series

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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