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Concepts and Vocabulary
Multiple Choice If \(a_{1}\), \(a_{2}\), \(\ldots ,\) \(a_{n},\) \(\ldots\) is an infinite collection of numbers, the expression \(\sum\limits_{k\,=\,1}^{\infty }a_{k}=a_{1}+ a_{2}+ \cdots + a_{n}+ \cdots\) is called [(a) an infinite sequence, (b) an infinite series, (c) a partial sum].
Multiple Choice If \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) is an infinite series, then the sequence \(\{S_{n}\}\) where \(S_{n}=\sum\limits_{k\,=\,1}^{n}a_{k},\) is called the sequence of [(a) fractional parts, (b) early terms, (c) completeness, (d) partial sums] of the infinite series.
True or False A series converges if and only if its sequence of partial sums converges.
True or False A geometric series \(\sum\limits_{k\,=\,1}^{\infty }ar^{k-1},\) \(a\neq 0,\) converges if \(\vert r\vert \leq 1.\)
The sum of a convergent geometric series \(\sum\limits_{k\,=\,1}^{\infty }ar^{k-1},\) \(a\neq 0,\) is \(S= \)______________.
True or False The harmonic series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\) converges because \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{1}{n}=0.\)
Skill Building
In Problems 7–10, find the fourth partial sum of each series.
\(\sum\limits_{k=1}^{\infty }\left( \dfrac{3}{4}\right) ^{k-1}\)
\(\sum\limits_{k=1}^{\infty }\dfrac{(-1)^{k+1}}{3^{k-1}}\)
\(\sum\limits_{k=1}^{\infty }k\)
\(\sum\limits_{k=1}^{\infty }\ln k\)
In Problems 11–16, find the sum of each telescoping series.
\(\sum\limits_{k=1}^{\infty }\left( \dfrac{1}{k+2}-\dfrac{1}{k+3}\right)\)
\(\sum\limits_{k=1}^{\infty }\left[ \dfrac{1}{k^{2}}-\dfrac{1}{(k+1)^{2}}\right]\)
\(\sum\limits_{k=1}^{\infty }\left( \dfrac{1}{3^{k+1}}-\dfrac{1}{3^{k}}\right)\)
\(\sum\limits_{k=1}^{\infty }\left( \dfrac{1}{4^{k+1}}-\dfrac{1}{4^{k}}\right)\)
\(\sum\limits_{k=1}^{\infty }\dfrac{1}{4k^{2}-1}\) \(\left[Hint: \dfrac{1}{4k^{2}-1}=\dfrac{1}{2}\left( \dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right) \right]\)
\(\sum\limits_{k=1}^{\infty }\dfrac{1}{k(k+1)(k+2)}\) \(\left[Hint: \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2}\left( \dfrac{1}{k(k+1)}-\dfrac{1}{(k+1)(k+2)}\right) \right]\)
In Problems 17–38, determine whether each geometric series converges or diverges. If it converges, find its sum.
\(\sum\limits_{k=1}^{\infty }(\sqrt{2})^{k-1}\)
\(\sum\limits_{k=1}^{\infty }(0.33)^{k-1}\)
\(\sum\limits_{k=1}^{\infty }5 \left(\dfrac{1}{6}\right)^{k-1}\)
\(\sum\limits_{k=1}^{\infty }4\left( 1.1\right) ^{k-1} \)
\(\sum\limits_{k\,=\,0}^{\infty }7 \left( \dfrac{1}{3}\right) ^{k}\)
\(\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{7}{4}\right)^{k}\)
\(\sum\limits_{k=1}^{\infty }( -0.38)^{k-1}\)
\(\sum\limits_{k=1}^{\infty }(-0.38)^{k}\)
\(\sum\limits_{k\,=\,0}^{\infty }\dfrac{2^{k+1}}{3^{k}}\)
\(\sum\limits_{k\,=\,0}^{\infty }\dfrac{5^{k}}{6^{k+1}}\)
\(\sum\limits_{k\,=\,0}^{\infty }\dfrac{1}{4^{k+1}}\)
\(\sum\limits_{k\,=\,0}^{\infty }\dfrac{4^{k+1}}{3^{k}}\)
\(\sum\limits_{k=1}^{\infty }\sin ^{k-1} \left(\dfrac{\pi }{2}\right)\)
\(\sum\limits_{k=1}^{\infty}\tan ^{k-1} \left( \dfrac{\pi }{4}\right)\)
\(\sum\limits_{k=1}^{\infty }\left( -\dfrac{3}{2}\right)^{k-1}\)
\(\sum\limits_{k=1}^{\infty }\left( -\dfrac{2}{3}\right)^{k-1}\)
\(1+\dfrac{1}{3}+\dfrac{1}{9}+\cdots +\left( \dfrac{1}{3}\right)^{n}+\cdots\)
\(1+\dfrac{1}{4}+\dfrac{1}{16}+\cdots +\left( {\dfrac{1}{4}}\right)^{n}+\cdots\)
\(1+2+4+\cdots +2^{n}+\cdots\)
\(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\cdots + \dfrac{(-1)^{n-1}}{2^{n-1}}+\cdots\)
\(\left( \dfrac{1}{7}\right) ^{2}+\left( \dfrac{1}{7}\right)^{3}+\cdots +\left( \dfrac{1}{7}\right) ^{n}+\cdots\)
\(\left( \dfrac{3}{4}\right) ^{5}+\left( \dfrac{3}{4}\right)^{6}+\cdots +\left( \dfrac{3}{4}\right) ^{n}+\cdots\)
In Problems 39–58, determine whether each series converges or diverges. If it converges, find its sum.
\(\sum\limits_{k\,=\,0}^{\infty}\dfrac{1}{k+1}\)
\(\sum\limits_{k=4}^{\infty}k^{-1}\)
\(\sum\limits_{k=1}^{\infty}\dfrac{1}{100^{k}}\)
\(\sum\limits_{k=1}^{\infty}e^{-k}\)
\(\sum\limits_{k=1}^{\infty} (-10k)\)
\(\sum\limits_{k=1}^{\infty}\dfrac{3k}{5}\)
\(\sum\limits_{k=1}^{\infty}\cos^{k-1} \left( \dfrac{2\pi }{3}\right)\)
\(\sum\limits_{k=1}^{\infty}\sin^{k-1} \left( \dfrac{\pi }{6}\right)\)
\(\sum\limits_{k=1}^{\infty}\dfrac{\tan ^{k}\left(\dfrac{\pi }{4}\right) }{k}\)
\(\sum\limits_{k=1}^{\infty}\dfrac{\sin ^{k} \left(\dfrac{\pi }{2}\right) }{k}\)
\(\sum\limits_{k=1}^{\infty }\cos (\pi k)\)
\(\sum\limits_{k=1}^{\infty}\sin \left( \dfrac{\pi k}{2}\right)\)
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\(\sum\limits_{k=1}^{\infty}2^{-k}3^{k+1}\)
\(\sum\limits_{k=1}^{\infty}3^{1-k}2^{1+k}\)
\(\sum\limits_{k=1}^{\infty}\left( -\dfrac{1}{3}\right)^{k}\)
\(\sum\limits_{k=1}^{\infty}\dfrac{\pi }{3^{k}}\)
\(\sum\limits_{k=1}^{\infty}\ln \dfrac{k}{k+1}\)
\(\sum\limits_{k=1}^{\infty}\left[e^{2k^{-1}}-e^{2(k+1) ^{-1}}\right]\)
\(\sum\limits_{k=1}^{\infty }\left(\sin \dfrac{1}{k}-\sin \dfrac{1}{k+1}\right)\)
\(\sum\limits_{k=1}^{\infty }\left( \tan \dfrac{1}{k}-\tan \dfrac{1}{k+1}\right)\)
In Problems 59–62, express each repeating decimal as a rational number by using a geometric series.
\(0.5555\ldots\)
\(0.727272\ldots\)
\(4.28555\ldots \) \((Hint: 4.28555\ldots =4.28+0.00555\ldots .)\)
\(7.162162\ldots\)
Applications and Extensions
Distance a Ball Travels A ball is dropped from a height of 18 ft. Each time it strikes the ground, it bounces back to two-thirds of the previous height. Find the total distance traveled by the ball.
Diminishing Returns A rich man promises to give you \($ 1000\) on January 1, 2015. Each day thereafter he will give you \(\dfrac{9}{10} \) of what he gave you the previous day.
Stocking a Lake Mirror Lake is stocked periodically with rainbow trout. In year \(n\) of the stocking program, the population is given by \(p_{n}=3000r^{n}+h\sum\limits_{k=1}^{n}r^{k-1},\) where \(h\) is the number of fish added by the program per year and \(r, 0 < r < 1\), is the percent of fish removed each year.
Marginal Propensity to Consume Suppose that individuals in the United States spend 90% of every additional dollar that they earn. Then according to economists, an individual’s marginal propensity to consume is 0.90. For example, if Jane earns an additional dollar, she will spend \(0.9(1)\) =\($0.90\) of it. The individual who earns Jane’s \($0.90\) will spend 90% of it or \(\$0.81.\)
The process of spending continues and results in the series \[ \sum\limits_{k=1}^{\infty }0.90^{k-1}=1+0.90+0.90^{2}+0.90^{3}+\ldots \]
The sum of this series is called the multiplier. What is the multiplier if the marginal propensity to consume is 90%?
Stock Pricing One method of pricing a stock is to discount the stream of future dividends of the stock. Suppose a stock currently pays \($P\) annually in dividends, and historically, the dividend has increased by \(i\%\) annually. If an investor wants an annual rate of return of \(r\%,\) this method of pricing a stock states that the stock should be priced at the present value of an infinite stream of payments: \[ \text{Price}= {\it P}+ {\it P} \dfrac{1+i}{1+r}+ {\it P}\left( \dfrac{1+i}{1+r}\right) ^{2}+ {\it P}\left( \dfrac{1+i}{1+r}\right) ^{3}+\ldots \]
Koch’s Snowflake The area inside the fractal known as the Koch snowflake can be described as the sum of the areas of infinitely many equilateral triangles. See the figure. For all but the center (largest) triangle, a triangle in the Koch snowflake is \(\dfrac{1}{9}\) the area of the next largest triangle in the fractal. Suppose the largest(center) triangle has an area of 1 square unit. Then the area of the snowflake is given by the series \[ 1+3 \left( \dfrac{1}{9}\right) +12 \left( \dfrac{1}{9}\right) ^{2}+48 \left( \dfrac{1}{9}\right) ^{3}+192 \left( \dfrac{1}{9}\right) ^{4}+\ldots \]
Find the area of the Koch snowflake by finding the sum of the series.
Zeno’s paradox is about a race between Achilles and a tortoise. The tortoise is allowed a certain lead at the start of the race. Zeno claimed the tortoise must win such a race. He reasoned that for Achilles to overtake the tortoise, at some time he must cover \(\dfrac{1}{2}\) of the distance that originally separated them. Then, when he covers another \(\dfrac{1}{4}\) of the original distance separating them, he will still have \(\dfrac{1}{4}\) of that distance remaining, and so on. Therefore by Zeno’s reasoning, Achilles never catches the tortoise. Use a series argument to explain this paradox. Assume that the difference in speed between Achilles and the tortoise is a constant \(v\) meters per second.
Probability A coin-flipping game involves two people who successively flip a coin. The first person to obtain a head is the winner. In probability, it turns out that the person who flips first has the probability of winning given by the series below. Find this probability. \[ \frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots +\frac{1}{2^{2n-1}}+\cdots \]
Controlling Salmonella Salmonella is a common enteric bacterium infecting both humans and farm animals with salmonellosis. Barn surfaces contaminated with salmonella can be the major source of salmonellosis spread in a farm. While cleaning barn surfaces is used as a control measure on pig and cattle farms, the efficiency of cleaning has been a concern. Suppose, on average, there are \(p\) kilograms (kg) of feces produced each day and cleaning is performed with the constant efficiency \(e,\) \(0<e<1.\) At the end of each day, \((1-e)\) kg of feces from the previous day is added to the amount of the present day. Let \(T(n)\) be the total accumulated fecal material on day \(n.\) Farmers are concerned when \(T(n)\) exceeds the threshold level \(L.\)
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Source: R. Gautam, G. Lahodny, M. Bani-Yaghoub, R. Ivanek Based on their paper “Understanding the role of cleaning in the control of Salmonella Typhimurium in a grower finisher pig herd: a modeling approach.”
Show that \(0.9999 \ldots\,=1.\)
In Problems 73 and 74 use a geometric series to prove the given statement.
\(\dfrac{x}{x-1}=\sum\limits_{k=1}^{\infty }\dfrac{1}{x^{k-1}}\) for \(\vert x\vert >1\)
\(\dfrac{1}{1+x}=\sum\limits_{k\,=\,0}^{\infty }(-1)^{k}x^{k}\) for \(\vert x\vert <1\)
Find the smallest number \(n\) for which \(\sum\limits_{k=1}^{n}\dfrac{1}{k}\geq 3\).
Find the smallest number \(n\) for which \(\sum\limits_{k=1}^{n}\dfrac{1}{k}\geq 4\).
Show that the series \(\sum\limits_{k=1}^{\infty }\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}\) converges and has the sum \(1\).
Show that \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k(k+2)}=\dfrac{3}{4}\).
Show that \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{4}\).
Show that \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k(k+1)(k+2)(k+3)}=\dfrac{1}{18}\).
Show that \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k(k+1)(k+2)\cdots (k+a)}=\dfrac{1}{a}\left( \dfrac{1}{a!}\right)\); \(a\geq 1\) is an integer.
Solve for \(x\): \(\dfrac{x}{2+2x}=x+x^{2}+x^{3}+\cdots, |x| < 1\).
Show that the sum of any convergent geometric series whose first term and common ratio are rational is rational.
The sum \(S_{n}\) of the first \(n\) terms of a geometric series is given by the formula \[ S_{n}=a + ar + ar^{2} + \cdots + ar^{n-1}=\frac{a(r^{n}-1)}{r-1}\qquad a>0,\quad r\neq 1 \]
Find \(\lim\limits_{r\rightarrow 1}\dfrac{a(r^{n}-1)}{r-1}\) and compare the result with a geometric series in which \(r=1\).
Challenge Problems
The following discussion relates to Problems 85 and 86.
An interesting relationship between the \(n\)th partial sum of the harmonic series and \(\ln n\) was discovered by Euler. In particular, he showed that \[ \gamma =\lim\limits_{n\rightarrow \infty }\left( 1+\frac{1}{2}+\frac{1}{3} +\cdots +\frac{1}{n}-\ln n\right) \]
exists and is approximately equal to \(0.5772\). Euler’s number, as \(\gamma\) is called, appears in many interesting areas of mathematics. For example, it is involved in the evaluation of the exponential integral, \(\int_{x}^{\infty }\dfrac{e^{-t}}{t}\,dt\), which is important in applied mathematics. It is also related to two special functions—the gamma function and Riemann’s zeta function (see Challenge Problem 75, Section 8.3). Surprisingly, it is still unknown whether Euler’s number is rational or irrational.
The harmonic series diverges quite slowly. For example, the partial sums \(S_{10}, S_{20}, S_{50}\), and \(S_{100}\) have approximate values \(2.92897\), \(3.59774\), \(4.49921\), and \(5.18738\), respectively. In fact, the sum of the first million terms of the harmonic series is about \(14.4\). With this in mind, what would you conjecture about the rate of convergence of the limit defining \(\gamma \)? Test your conjecture by calculating approximate values for \(\gamma\) by using the partial sums given above.
Use the approximate value of Euler’s number \(0.5772\) to approximate \[ 1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{1{,}000{,}000{,}000} \]
Show that a real number has a repeating decimal if and only if it is rational.