maths.free › Calculus › 5. Sequences and Series › Infinite Series
Infinite Series
Explain the meaning of the sum of an infinite series.
Sums and Series
An infinite series is a sum of infinitely many terms and is written in the form
\[\sum _{n=1}^{\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\text{\cdots }.\]But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the limit of partial sums. A partial sum of an infinite series is a finite sum of the form
\[\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\text{\cdots }+{a}_{k}.\]To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that \(1000\) gallons enters the lake the first week. During the second week, an additional \(500\) gallons of oil enters the lake. The third week, \(250\) more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after \(k\) weeks. Letting \({S}_{k}\) denote the amount of oil in the lake (measured in thousands of gallons) after \(k\) weeks, we see that
\[\begin{array}{l}{S}_{1}=1 \\ {S}_{2}=1+0.5=1+\frac{1}{2} \\ {S}_{3}=1+0.5+0.25=1+\frac{1}{2}+\frac{1}{4} \\ {S}_{4}=1+0.5+0.25+0.125=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \\ {S}_{5}=1+0.5+0.25+0.125+0.0625=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}.\end{array}\]Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after \(k\) weeks is
\[{S}_{k}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\text{\cdots }+\frac{1}{{2}^{k-1}}=\sum _{n=1}^{k}{(\frac{1}{2})}^{n-1}.\]We are interested in what happens as \(k\to \infty .\) Symbolically, the amount of oil in the lake as \(k\to \infty\) is given by the infinite series
\[\sum _{n=1}^{\infty }{(\frac{1}{2})}^{n-1}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\text{\cdots }.\]First, simplifying some of these partial sums, we see that
\[\begin{array}{l}{S}_{1}=1 \\ {S}_{2}=1+\frac{1}{2}=\frac{3}{2} \\ {S}_{3}=1+\frac{1}{2}+\frac{1}{4}=\frac{7}{4} \\ {S}_{4}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{15}{8} \\ {S}_{5}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{31}{16}.\end{array}\]| \(k\) | \(5\) | \(10\) | \(15\) | \(20\) |
| \({S}_{k}\) | \(1.9375\) | \(1.998\) | \(1.999939\) | \(1.999998\) |
Condensed — the full section is in OpenStax Calculus Volume 2.
Telescoping Series
Consider the series \(\sum _{n=1}^{\infty }\frac{1}{n(n+1)}.\) We discussed this series in , showing that the series converges by writing out the first several partial sums \({S}_{1},{S}_{2}\text{,\ldots },{S}_{6}\) and noticing that they are all of the form \({S}_{k}=\frac{k}{k+1}.\) Here we use a different technique to show that this series converges. By using partial fractions, we can write
\[\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.\]Therefore, the series can be written as
\[\sum _{n=1}^{\infty }[\frac{1}{n}-\frac{1}{n+1}]=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\text{\cdots }.\]Writing out the first several terms in the sequence of partial sums \(\{{S}_{k}\},\) we see that
\[\begin{array}{l}{S}_{1}=1-\frac{1}{2} \\ {S}_{2}=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})=1-\frac{1}{3} \\ {S}_{3}=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}.\end{array}\]In general,
\[{S}_{k}=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\text{\cdots }+(\frac{1}{k}-\frac{1}{k+1})=1-\frac{1}{k+1}.\]We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since \({S}_{k}=1-1\text{/}(k+1)\) and \(1\text{/}(k+1)\to 0\) as \(k\to \infty ,\) the sequence of partial sums converges to \(1,\) and therefore the series converges to \(1.\)
For example, any series of the form
\[\sum _{n=1}^{\infty }[{b}_{n}-{b}_{n+1}]=({b}_{1}-{b}_{2})+({b}_{2}-{b}_{3})+({b}_{3}-{b}_{4})+\text{\cdots }\]is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that
\[\begin{array}{l}{S}_{1}={b}_{1}-{b}_{2} \\ {S}_{2}=({b}_{1}-{b}_{2})+({b}_{2}-{b}_{3})={b}_{1}-{b}_{3} \\ {S}_{3}=({b}_{1}-{b}_{2})+({b}_{2}-{b}_{3})+({b}_{3}-{b}_{4})={b}_{1}-{b}_{4}.\end{array}\]\[{S}_{k}={b}_{1}-{b}_{k+1}.\]\[\sum _{n=1}^{\infty }[{b}_{n}-{b}_{n+1}]={b}_{1}-B.\]Condensed — the full section is in OpenStax Calculus Volume 2.
Key Concepts
- Given the infinite series
\[\sum _{n=1}^{\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\text{\cdots }\]
and the corresponding sequence of partial sums \(\{{S}_{k}\}\) where
\[{S}_{k}=\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\text{\cdots }+{a}_{k},\]
the series converges if and only if the sequence \(\{{S}_{k}\}\) converges. - The geometric series \(\sum _{n=1}^{\infty }a{r}^{n-1}\) converges if \(|r|<1\) and diverges if \(|r|\ge 1.\) For \(|r|<1,\)
\[\sum _{n=1}^{\infty }a{r}^{n-1}=\frac{a}{1-r}.\] - The harmonic series
\[\sum _{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\text{\cdots }\]
diverges. - A series of the form \(\sum _{n=1}^{\infty }[{b}_{n}-{b}_{n+1}]=[{b}_{1}-{b}_{2}]+[{b}_{2}-{b}_{3}]+[{b}_{3}-{b}_{4}]+\text{\cdots }+[{b}_{n}-{b}_{n+1}]+\text{\cdots }\)
is a telescoping series. The \(k\text{th}\) partial sum of this series is given by \({S}_{k}={b}_{1}-{b}_{k+1}.\) The series will converge if and only if \(\underset{k\to \infty }{\text{lim}}{b}_{k+1}\) exists. In that case,
\[\sum _{n=1}^{\infty }[{b}_{n}-{b}_{n+1}]={b}_{1}-\underset{k\to \infty }{\text{lim}}({b}_{k+1}).\]
Key Equations
| Harmonic series | \(\sum _{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{\cdots }\) |
| Sum of a geometric series | \(\sum _{n=1}^{\infty }a{r}^{n-1}=\frac{a}{1-r}\ \text{for}\ |r|<1\) |
Infinite Series
Using sigma notation, write the following expressions as infinite series.
Compute the first four partial sums \({S}_{1}\text{,\ldots },{S}_{4}\) for the series having \(n\text{th}\) term \({a}_{n}\) starting with \(n=1\) as follows.
In the following exercises, compute the general term \({a}_{n}\) of the series with the given partial sum \({S}_{n}.\) If the sequence of partial sums converges, find its limit \(S.\)
For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.
Suppose that \(\sum _{n=1}^{\infty }{a}_{n}=1,\) that \(\sum _{n=1}^{\infty }{b}_{n}=-1,\) that \({a}_{1}=2,\) and \({b}_{1}=-3.\) Find the sum of the indicated series.
State whether the given series converges and explain why.
For \({a}_{n}\) as follows, write the sum as a geometric series of the form \(\sum _{n=1}^{\infty }a{r}^{n}.\) State whether the series converges and if it does, find the value of \(\sum {a}_{n}.\)
Condensed — the full section is in OpenStax Calculus Volume 2.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.
- \(\sum _{n=1}^{\infty }\frac{n}{n+1}\)
- \(\sum _{n=1}^{\infty }{(-1)}^{n}\)
- \(\sum _{n=1}^{\infty }\frac{1}{n(n+1)}\)
Jawaby görkez
- The sequence of partial sums \(\{{S}_{k}\}\) satisfies
\[\begin{array}{l}{S}_{1}=\frac{1}{2} \\ {S}_{2}=\frac{1}{2}+\frac{2}{3} \\ {S}_{3}=\frac{1}{2}+\frac{2}{3}+\frac{3}{4} \\ {S}_{4}=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}.\end{array}\]
Notice that each term added is greater than \(1\text{/}2.\) As a result, we see that
\[\begin{array}{l}{S}_{1}=\frac{1}{2} \\ {S}_{2}=\frac{1}{2}+\frac{2}{3}>\frac{1}{2}+\frac{1}{2}=2(\frac{1}{2}) \\ {S}_{3}=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=3(\frac{1}{2}) \\ {S}_{4}=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=4(\frac{1}{2}).\end{array}\]
From this pattern we can see that \({S}_{k}>k(\frac{1}{2})\) for every integer \(k.\) Therefore, \(\{{S}_{k}\}\) is unbounded and consequently, diverges. Therefore, the infinite series \(\sum _{n=1}^{\infty }n\text{/}(n+1)\) diverges. - The sequence of partial sums \(\{{S}_{k}\}\) satisfies
\[\begin{array}{l}{S}_{1}=-1 \\ {S}_{2}=-1+1=0 \\ {S}_{3}=-1+1-1=-1 \\ {S}_{4}=-1+1-1+1=0.\end{array}\]
From this pattern we can see the sequence of partial sums is
\[\{{S}_{k}\}=\{-1,0,-1,0\text{,\ldots }\}.\]
Since this sequence diverges, the infinite series \(\sum _{n=1}^{\infty }{(-1)}^{n}\) diverges. - The sequence of partial sums \(\{{S}_{k}\}\) satisfies
\[\begin{array}{l}{S}_{1}=\frac{1}{1\cdot 2}=\frac{1}{2} \\ {S}_{2}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}=\frac{1}{2}+\frac{1}{6}=\frac{2}{3} \\ {S}_{3}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{3}{4} \\ {S}_{4}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}=\frac{4}{5} \\ {S}_{5}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}=\frac{5}{6}.\end{array}\]
From this pattern, we can see that the \(k\text{th}\) partial sum is given by the explicit formula
\[{S}_{k}=\frac{k}{k+1}.\]
Since \(k\text{/}(k+1)\to 1,\) we conclude that the sequence of partial sums converges, and therefore the infinite series converges to \(1.\) We have
\[\sum _{n=1}^{\infty }\frac{1}{n(n+1)}=1.\]
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Determine whether the series \(\sum _{n=1}^{\infty }(n+1)\text{/}n\) converges or diverges.
Jawaby görkez
The series diverges because the \(k\text{th}\) partial sum \({S}_{k}>k.\)
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Evaluate
\[\sum _{n=1}^{\infty }[\frac{3}{n(n+1)}+{(\frac{1}{2})}^{n-2}].\]Jawaby görkez
We showed earlier that
\[\sum _{n=1}^{\infty }\frac{1}{n(n+1)}=1\]and
\[\sum _{n=1}^{\infty }{(\frac{1}{2})}^{n-1}=2.\]Since both of those series converge, we can apply the properties of 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
\[\begin{array}{ll}\sum _{n=1}^{\infty }\frac{3}{n(n+1)}+\sum _{n=1}^{\infty }{(\frac{1}{2})}^{n-2} & =3\sum _{n=1}^{\infty }\frac{1}{n(n+1)}+{(\frac{1}{2})}^{-1}\sum _{n=1}^{\infty }{(\frac{1}{2})}^{n-1} \\ & =3(1)+{(\frac{1}{2})}^{-1}(2)=3+2(2)=7.\end{array}\] -
Evaluate \(\sum _{n=1}^{\infty }\frac{5}{{2}^{n-1}}.\)
Jawaby görkez
\(10.\)
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Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.
- \(\sum _{n=1}^{\infty }\frac{{(-3)}^{n+1}}{{4}^{n-1}}\)
- \(\sum _{n=1}^{\infty }{e}^{2n}\)
Jawaby görkez
- Writing out the first several terms in the series, we have
\[\begin{array}{ll}\sum _{n=1}^{\infty }\frac{{(-3)}^{n+1}}{{4}^{n-1}} & =\frac{{(-3)}^{2}}{{4}^{0}}+\frac{{(-3)}^{3}}{4}+\frac{{(-3)}^{4}}{{4}^{2}}+\text{\cdots } \\ & ={(-3)}^{2}+{(-3)}^{2}\cdot (\frac{-3}{4})+{(-3)}^{2}\cdot {(\frac{-3}{4})}^{2}+\text{\cdots } \\ & =9+9\cdot (\frac{-3}{4})+9\cdot {(\frac{-3}{4})}^{2}+\text{\cdots }.\end{array}\]
The initial term \(a=9\) and the ratio \(r=-3\text{/}4.\) Since \(|r|=3\text{/}4<1,\) the series converges to
\[\frac{9}{1-(-3\text{/}4)}=\frac{9}{7\text{/}4}=\frac{36}{7}.\] - Writing this series as
\[{e}^{2}\sum _{n=1}^{\infty }{({e}^{2})}^{n-1}\]
we can see that this is a geometric series where \(r={e}^{2}>1.\) Therefore, the series diverges.
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Determine whether the series \(\sum _{n=1}^{\infty }{(\frac{-2}{5})}^{n-1}\) converges or diverges. If it converges, find its sum.
Jawaby görkez
\(5\text{/}7\)
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Use a geometric series to write \(3.\overset{\text{—}}{26}\) as a fraction of integers.
Jawaby görkez
Since \(3.\overset{\text{—}}{26}=3.262626\text{\ldots },\) first we write
\[\begin{array}{ll}3.262626\text{\ldots } & =3+\frac{26}{100}+\frac{26}{10,000}+\frac{26}{1,000,000}+\text{\cdots } \\ & =3+\frac{26}{{10}^{2}}+\frac{26}{{10}^{4}}+\frac{26}{{10}^{6}}+\text{\cdots }.\end{array}\]Ignoring the term 3, the rest of this expression is a geometric series with initial term \(a=26\text{/}{10}^{2}\) and ratio \(r=1\text{/}{10}^{2}.\) Therefore, the sum of this series is
\[\frac{26\text{/}{10}^{2}}{1-(1\text{/}{10}^{2})}=\frac{26\text{/}{10}^{2}}{99\text{/}{10}^{2}}=\frac{26}{99}.\]Thus,
\[3.262626\text{\ldots }=3+\frac{26}{99}=\frac{323}{99}.\] -
Write \(5.2\overset{\text{-}}{7}\) as a fraction of integers.
Jawaby görkez
\(475\text{/}90\)
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Define a sequence of figures \(\{{F}_{n}\}\) recursively as follows (). Let \({F}_{0}\) be an equilateral triangle with sides of length \(1.\) For \(n\ge 1,\) let \({F}_{n}\) be the curve created by removing the middle third of each side of \({F}_{n-1}\) and replacing it with an equilateral triangle pointing outward. The limiting figure as \(n\to \infty\) is known as Koch’s snowflake.
- Find the length \({L}_{n}\) of the perimeter of \({F}_{n}.\) Evaluate \(\underset{n\to \infty }{\text{lim}}{L}_{n}\) to find the length of the perimeter of Koch’s snowflake.
- Find the area \({A}_{n}\) of figure \({F}_{n}.\) Evaluate \(\underset{n\to \infty }{\text{lim}}{A}_{n}\) to find the area of Koch’s snowflake.
Jawaby görkez
- Let \({N}_{n}\) denote the number of sides of figure \({F}_{n}.\) Since \({F}_{0}\) is a triangle, \({N}_{0}=3.\) Let \({l}_{n}\) denote the length of each side of \({F}_{n}.\) Since \({F}_{0}\) is an equilateral triangle with sides of length \({l}_{0}=1,\) we now need to determine \({N}_{1}\) and \({l}_{1}.\) Since \({F}_{1}\) is created by removing the middle third of each side and replacing that line segment with two line segments, for each side of \({F}_{0},\) we get four sides in \({F}_{1}.\) Therefore, the number of sides for \({F}_{1}\) is
\[{N}_{1}=4\cdot 3.\]
Since the length of each of these new line segments is \(1\text{/}3\) the length of the line segments in \({F}_{0},\) the length of the line segments for \({F}_{1}\) is given by
\[{l}_{1}=\frac{1}{3}\cdot 1=\frac{1}{3}.\]
Similarly, for \({F}_{2},\) since the middle third of each side of \({F}_{1}\) is removed and replaced with two line segments, the number of sides in \({F}_{2}\) is given by
\[{N}_{2}=4{N}_{1}=4(4\cdot 3)={4}^{2}\cdot 3.\]
Since the length of each of these sides is \(1\text{/}3\) the length of the sides of \({F}_{1},\) the length of each side of figure \({F}_{2}\) is given by
\[{l}_{2}=\frac{1}{3}\cdot {l}_{1}=\frac{1}{3}\cdot \frac{1}{3}={(\frac{1}{3})}^{2}.\]
More generally, since \({F}_{n}\) is created by removing the middle third of each side of \({F}_{n-1}\) and replacing that line segment with two line segments of length \(\frac{1}{3}{l}_{n-1}\) in the shape of an equilateral triangle, we know that \({N}_{n}=4{N}_{n-1}\) and \({l}_{n}=\frac{{l}_{n-1}}{3}.\) Therefore, the number of sides of figure \({F}_{n}\) is
\[{N}_{n}={4}^{n}\cdot 3\]
and the length of each side is
\[{l}_{n}={(\frac{1}{3})}^{n}.\]
Therefore, to calculate the perimeter of \({F}_{n},\) we multiply the number of sides \({N}_{n}\) and the length of each side \({l}_{n}.\) We conclude that the perimeter of \({F}_{n}\) is given by
\[{L}_{n}={N}_{n}\cdot {l}_{n}=3\cdot {(\frac{4}{3})}^{n}.\]
Therefore, the length of the perimeter of Koch’s snowflake is
\[L=\underset{n\to \infty }{\text{lim}}{L}_{n}=\infty .\] - Let \({T}_{n}\) denote the area of each new triangle created when forming \({F}_{n}.\) For \(n=0,\) \({T}_{0}\) is the area of the original equilateral triangle. Therefore, \({T}_{0}={A}_{0}=\sqrt{3}\text{/}4.\) For \(n\ge 1,\) since the lengths of the sides of the new triangle are \(1\text{/}3\) the length of the sides of \({F}_{n-1},\) we have
\[{T}_{n}={(\frac{1}{3})}^{2}{T}_{n-1}=\frac{1}{9}\cdot {T}_{n-1}.\]
Therefore, \({T}_{n}={(\frac{1}{9})}^{n}\cdot \frac{\sqrt{3}}{4}.\) Since a new triangle is formed on each side of \({F}_{n-1},\)
\[\begin{array}{ll}{A}_{n} & ={A}_{n-1}+{N}_{n-1}\cdot {T}_{n} \\ & ={A}_{n-1}+(3\cdot {4}^{n-1})\cdot {(\frac{1}{9})}^{n}\cdot \frac{\sqrt{3}}{4} \\ & ={A}_{n-1}+\frac{3}{4}\cdot {(\frac{4}{9})}^{n}\cdot \frac{\sqrt{3}}{4}.\end{array}\]
Writing out the first few terms \({A}_{0},{A}_{1},{A}_{2},\) we see that
\[\begin{array}{l} \\ {A}_{0}=\frac{\sqrt{3}}{4} \\ {A}_{1}={A}_{0}+\frac{3}{4}\cdot (\frac{4}{9})\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}+\frac{3}{4}\cdot (\frac{4}{9})\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}[1+\frac{3}{4}\cdot (\frac{4}{9})] \\ {A}_{2}={A}_{1}+\frac{3}{4}\cdot {(\frac{4}{9})}^{2}\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}[1+\frac{3}{4}\cdot (\frac{4}{9})]+\frac{3}{4}\cdot {(\frac{4}{9})}^{2}\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}[1+\frac{3}{4}\cdot (\frac{4}{9})+\frac{3}{4}\cdot {(\frac{4}{9})}^{2}].\end{array}\]
More generally,
\[{A}_{n}=\frac{\sqrt{3}}{4}[1+\frac{3}{4}(\frac{4}{9}+{(\frac{4}{9})}^{2}+\text{\cdots }+{(\frac{4}{9})}^{n})].\]
Factoring \(4\text{/}9\) out of each term inside the inner parentheses, we rewrite our expression as
\[{A}_{n}=\frac{\sqrt{3}}{4}[1+\frac{1}{3}(1+\frac{4}{9}+{(\frac{4}{9})}^{2}+\text{\cdots }+{(\frac{4}{9})}^{n-1})].\]
The expression \(1+(\frac{4}{9})+{(\frac{4}{9})}^{2}+\text{\cdots }+{(\frac{4}{9})}^{n-1}\) is a geometric sum. As shown earlier, this sum satisfies
\[1+\frac{4}{9}+{(\frac{4}{9})}^{2}+\text{\cdots }+{(\frac{4}{9})}^{n-1}=\frac{1-{(4\text{/}9)}^{n}}{1-(4\text{/}9)}.\]
Substituting this expression into the expression above and simplifying, we conclude that
\[\begin{array}{ll}{A}_{n} & =\frac{\sqrt{3}}{4}[1+\frac{1}{3}(\frac{1-{(4\text{/}9)}^{n}}{1-(4\text{/}9)})] \\ & =\frac{\sqrt{3}}{4}[\frac{8}{5}-\frac{3}{5}{(\frac{4}{9})}^{n}].\end{array}\]
Therefore, the area of Koch’s snowflake is
\[A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\frac{2\sqrt{3}}{5}.\]
-
Determine whether the telescoping series
\[\sum _{n=1}^{\infty }[\text{cos}(\frac{1}{n})-\text{cos}(\frac{1}{n+1})]\]converges or diverges. If it converges, find its sum.
Jawaby görkez
By writing out terms in the sequence of partial sums, we can see that
\[\begin{array}{lll}{S}_{1} & = & \text{cos}(1)-\text{cos}(\frac{1}{2}) \\ {S}_{2} & = & (\text{cos}(1)-\text{cos}(\frac{1}{2}))+(\text{cos}(\frac{1}{2})-\text{cos}(\frac{1}{3}))=\text{cos}(1)-\text{cos}(\frac{1}{3}) \\ {S}_{3} & = & (\text{cos}(1)-\text{cos}(\frac{1}{2}))+(\text{cos}(\frac{1}{2})-\text{cos}(\frac{1}{3}))+(\text{cos}(\frac{1}{3})-\text{cos}(\frac{1}{4})) \\ & = & \text{cos}(1)-\text{cos}(\frac{1}{4}).\end{array}\]In general,
\[{S}_{k}=\text{cos}(1)-\text{cos}(\frac{1}{k+1}).\]Since \(1\text{/}(k+1)\to 0\) as \(k\to \infty\) and \(\text{cos}\ x\) is a continuous function, \(\text{cos}(1\text{/}(k+1))\to \text{cos}(0)=1.\) Therefore, we conclude that \({S}_{k}\to \text{cos}(1)-1.\) The telescoping series converges and the sum is given by
\[\sum _{n=1}^{\infty }[\text{cos}(\frac{1}{n})-\text{cos}(\frac{1}{n+1})]=\text{cos}(1)-1.\] -
Determine whether \(\sum _{n=1}^{\infty }[{e}^{1\text{/}n}-{e}^{1\text{/}(n+1)}]\) converges or diverges. If it converges, find its sum.
Jawaby görkez
\(e-1\)
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\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{\cdots }\)
Jawaby görkez
\(\sum _{n=1}^{\infty }\frac{1}{n}\)
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\(1-1+1-1+\text{\cdots }\)
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\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...\)
Jawaby görkez
\(\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n}\)
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\(\text{sin}\ 1+\text{sin}\ \frac{1}{2}+\text{sin}\ \frac{1}{3}+\text{sin}\ \frac{1}{4}+\text{\cdots }\)
-
\({a}_{n}=n\)
Jawaby görkez
\(1,3,6,10\)
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\({a}_{n}=1\text{/}n\)
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\({a}_{n}=\text{sin}(n\pi \text{/}2)\)
Jawaby görkez
\(1,1,0,0\)
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\({a}_{n}={(-1)}^{n}\)
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\({S}_{n}=1-\frac{1}{n},\) \(n\ge 2\)
Jawaby görkez
\({a}_{n}={S}_{n}-{S}_{n-1}=\frac{1}{n-1}-\frac{1}{n}.\) Series converges to \(S=1.\)
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\({S}_{n}=\frac{n(n+1)}{2},\) \(n\ge 1\)
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\({S}_{n}=\sqrt{n},n\ge 2\)
Jawaby görkez
\({a}_{n}={S}_{n}-{S}_{n-1}=\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n-1}+\sqrt{n}}.\) Series diverges because partial sums are unbounded.
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\({S}_{n}=2-(n+2)\text{/}{2}^{n},n\ge 1\)
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\(\sum _{n=1}^{\infty }\frac{n}{n+2}\)
Jawaby görkez
\({S}_{1}=1\text{/}3,\) \({S}_{2}=1\text{/}3+2\text{/}4>1\text{/}3+1\text{/}3=2\text{/}3,\) \({S}_{3}=1\text{/}3+2\text{/}4+3\text{/}5>3\cdot (1\text{/}3)=1.\) In general \({S}_{k}>k\text{/}3.\) Series diverges.
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\(\sum _{n=1}^{\infty }(1-{(-1)}^{n}))\)
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\(\sum _{n=1}^{\infty }\frac{1}{(n+1)(n+2)}\) (Hint: Use a partial fraction decomposition like that for \(\sum _{n=1}^{\infty }\frac{1}{n(n+1)}.)\)
Jawaby görkez
\(\begin{array}{l}{S}_{1}=1\text{/}(2.3)=1\text{/}6=2\text{/}3-1\text{/}2, \\ {S}_{2}=1\text{/}(2.3)+1\text{/}(3.4)=2\text{/}12+1\text{/}12=1\text{/}4=3\text{/}4-1\text{/}2, \\ {S}_{3}=1\text{/}(2.3)+1\text{/}(3.4)+1\text{/}(4.5)=10\text{/}60+5\text{/}60+3\text{/}60=3\text{/}10=4\text{/}5-1\text{/}2, \\ {S}_{4}=1\text{/}(2.3)+1\text{/}(3.4)+1\text{/}(4.5)+1\text{/}(5.6)=10\text{/}60+5\text{/}60+3\text{/}60+2\text{/}60=1\text{/}3=5\text{/}6-1\text{/}2.\end{array}\)
The pattern is \({S}_{k}=(k+1)\text{/}(k+2)-1\text{/}2\) and the series converges to \(1\text{/}2.\)
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\(\sum _{n=1}^{\infty }\frac{1}{2n+1}\) (Hint: Follow the reasoning for \(\sum _{n=1}^{\infty }\frac{1}{n}.)\)
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\(\sum _{n=1}^{\infty }({a}_{n}+{b}_{n})\)
Jawaby görkez
\(0\)
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\(\sum _{n=1}^{\infty }({a}_{n}-2{b}_{n})\)
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\(\sum _{n=2}^{\infty }({a}_{n}-{b}_{n})\)
Jawaby görkez
\(-3\)
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\(\sum _{n=1}^{\infty }(3{a}_{n+1}-4{b}_{n+1})\)
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\(\sum _{n=1}^{\infty }\frac{1}{n+1000}\) (Hint: Rewrite using a change of index.)
Jawaby görkez
diverges, \(\sum _{n=1001}^{\infty }\frac{1}{n}\)
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\(\sum _{n=1}^{\infty }\frac{1}{n+{10}^{80}}\) (Hint: Rewrite using a change of index.)
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\(1+\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\text{\cdots }\)
Jawaby görkez
convergent geometric series, \(r=1\text{/}10<1\)
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\(1+\frac{e}{\pi }+\frac{{e}^{2}}{{\pi }^{2}}+\frac{{e}^{3}}{{\pi }^{3}}+\text{\cdots }\)
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\(1+\frac{\pi }{{e}^{2}}+\frac{{\pi }^{2}}{{e}^{4}}+\frac{{\pi }^{3}}{{e}^{6}}+\frac{{\pi }^{4}}{{e}^{8}}+\text{\cdots }\)
Jawaby görkez
convergent geometric series, \(r=\pi \text{/}{e}^{2}<1\)
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\(1-\sqrt{\frac{\pi }{3}}+\sqrt{\frac{{\pi }^{2}}{9}}-\sqrt{\frac{{\pi }^{3}}{27}}+\text{\cdots }\)
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\({a}_{1}=-1\) and \({a}_{n}\text{/}{a}_{n+1}=-5\) for \(n\ge 1.\)
Jawaby görkez
\(\sum _{n=1}^{\infty }5\cdot {(-1\text{/}5)}^{n},\) converges to \(-5\text{/}6\)
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\({a}_{1}=2\) and \({a}_{n}\text{/}{a}_{n+1}=1\text{/}2\) for \(n\ge 1.\)
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\({a}_{1}=10\) and \({a}_{n}\text{/}{a}_{n+1}=10\) for \(n\ge 1.\)
Jawaby görkez
\(\sum _{n=1}^{\infty }100\cdot {(1\text{/}10)}^{n},\) converges to \(100\text{/}9\)
Symbols used here
Add a_k for k = 1 up to n.
Not a number: "grows without bound" in limits and intervals.
Ratio of a circle's circumference to its diameter, 3.14159…
Inequalities that allow equality; < and > exclude it.
Least upper bound, greatest lower bound.
2.71828…, the base whose exponential is its own derivative.
Ratios of sides in a right triangle; coordinates on the unit circle.
The exponent b must be raised to for x; ln uses base e.
The value f(x) approaches as x approaches a.
Instantaneous rate of change; slope of the graph.
Antiderivative (indefinite) or signed area from a to b (definite).
Prime notation for derivatives with respect to x (or t).
Constants of integration fixed by initial conditions.
How to: Infinite Series
- Explain the meaning of the sum of an infinite series.
- Calculate the sum of a geometric series.
- Evaluate a telescoping series.
- The sequence of partial sums
- The sequence of partial sums
- The sequence of partial sums
- The series
- The series
Questions people ask
What is a derivative in one sentence?
The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.
Özüňi synla
Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
_Ýaşa Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests