Textbook solution for Calculus 2012 Student Edition (by 4th Edition Ross L. Finney Chapter 11.3 Problem 54E. We have step-by-step solutions for your textbooks written by Bartleby experts! This lesson contains the following Essential Knowledge (EK) concepts for the.AP Calculus course.Click here for an overview of all the EK's in this course. EK 3.4D1. AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site. To Calculus 11.1 Introduction to Limits 11.2 Techniques for Evaluating Limits 11.3 The Tangent Line Problem 11.4 Limits at Infinity and Limits of Sequences 11.5 The Area Problem 11 0 100,000 0 6 Section 11.4, Example 3 Average Cost Andresr/iStockphoto.com.
While much more can be said about sequences, we now turn to ourprincipal interest, series. Recall that a series, roughly speaking, isthe sum of a sequence: if $ds{a_n}_{n=0}^infty$ is a sequence then theassociated 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 $ds{s_n}_{n=0}^infty$:$$s_n=sum_{i=0}^n a_i.$$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 $ds a_n=kx^n$, $dssum_{n=0}^infty a_n$ 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$$eqalign{ s_n(1-x)&=k(1+x+x^2+x^3+cdots+x^n)(1-x)cr &=k(1+x+x^2+x^3+cdots+x^n)1-k(1+x+x^2+x^3+cdots+x^{n-1}+x^n)xcr &=k(1+x+x^2+x^3+cdots+x^n-x-x^2-x^3-cdots-x^n-x^{n+1})cr &=k(1-x^{n+1})cr}$$so$$eqalign{ s_n(1-x)&=k(1-x^{n+1})cr s_n&=k{1-x^{n+1}over 1-x}.cr}$$If $|x|< 1$, $dslim_{ntoinfty}x^n=0$ so$$ lim_{ntoinfty}s_n=lim_{ntoinfty}k{1-x^{n+1}over 1-x}= k{1over 1-x}.$$ Thus, when $|x|< 1$ the geometric series converges to $k/(1-x)$. When, for example, $k=1$ and $x=1/2$:$$ s_n={1-(1/2)^{n+1}over 1-1/2}={2^{n+1}-1over 2^n}=2-{1over 2^n} quadhbox{and}quad sum_{n=0}^infty {1over 2^n} = {1over 1-1/2} = 2.$$We began the chapter with the series$$sum_{n=1}^infty {1over 2^n},$$namely, the geometric series without the first term $1$. Each partialsum 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 thevalue of the geometric series, that is,$$sum_{n=1}^infty {1over 2^n}=1.$$
It is not hard to see that the following theorem follows fromtheorem 11.1.2.
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Theorem 11.2.2 Suppose that $sum a_n$ and $sum b_n$ are convergent series,and $c$ is a constant. Then
1. $dssum ca_n$ is convergent and $dssum ca_n=csum a_n$
2. $dssum (a_n+b_n)$ is convergent and $dssum (a_n+b_n)=sum a_n+sum b_n$.
The two parts of this theorem are subtly different. Suppose that $suma_n$ 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 byassumption diverges. Hence $sum ca_n$ also diverges. Note that we areapplying 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 $ds s_n$ is harder to understandand analyze than the sequence of terms $ds a_n$, and it is difficultto determine whether series converge and if so to what. Sometimesthings are relatively simple, starting with the following.
Theorem 11.2.3 If $sum a_n$ converges then $dslim_{ntoinfty}a_n=0$.
Proof.
Since $sum a_n$ converges, $dslim_{ntoinfty}s_n=L$ and $dslim_{ntoinfty}s_{n-1}=L$, because this really says the samething but 'renumbers' the terms. Bytheorem 11.1.2, $$ lim_{ntoinfty} (s_{n}-s_{n-1})= lim_{ntoinfty} s_{n}-lim_{ntoinfty}s_{n-1}=L-L=0.$$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 $dslim_{ntoinfty}a_n=0$.
This theorem presents an easy divergence test: if given a series $suma_n$ the limit $dslim_{ntoinfty}a_n$ does not exist or has a valueother than zero, the series diverges. Note well that the converse isnot true: If $dslim_{ntoinfty}a_n=0$ then the series doesnot necessarily converge.
Example 11.2.4 Show that $dssum_{n=1}^infty {nover n+1}$ diverges.
We compute the limit:$$lim _{ntoinfty}{nover n+1}=1not=0.$$Looking at the first few terms perhaps makes it clear that the serieshas no chance of converging:$${1over2}+{2over3}+{3over4}+{4over5}+cdots$$will just get larger and larger; indeed, after a bit longer the seriesstarts to look very much like $cdots+1+1+1+1+cdots$, and of courseif we add up enough 1's we can make the sum as large as we desire.
Example 11.2.5 Show that $dssum_{n=1}^infty {1over n}$ diverges.
Here the theorem does not apply: $dslim _{ntoinfty} 1/n=0$, so itlooks like perhaps the series converges. Indeed, if you have thefortitude (or the software) to add up the first 1000 terms you will find that$$sum_{n=1}^{1000} {1over n}approx 7.49,$$so it might be reasonable to speculate that the series converges tosomething in the neighborhood of 10. But in fact the partial sums do goto infinity; they just get big very, very slowly. Consider thefollowing:
$ds 1+{1over 2}+{1over 3}+{1over 4} > 1+{1over 2}+{1over 4}+{1over 4} = 1+{1over 2}+{1over 2}$
$ds 1+{1over 2}+{1over 3}+{1over 4}+{1over 5}+{1over 6}+{1over 7}+{1over 8} > 1+{1over 2}+{1over 4}+{1over 4}+{1over 8}+{1over 8}+{1over 8}+{1over 8} = 1+{1over 2}+{1over 2}+{1over 2}$
$ds 1+{1over 2}+{1over 3}+cdots+{1over16}>1+{1over 2}+{1over 4}+{1over 4}+{1over 8}+cdots+{1over 8}+{1over16}+cdots +{1over16} =1+{1over 2}+{1over 2}+{1over 2}+{1over 2}$
and so on. By swallowing up more and more terms we can always manageto add at least another $1/2$ to the sum, and by adding enough ofthese we can make the partial sums as big as we like. In fact, it'snot hard to see from this pattern that$$1+{1over 2}+{1over 3}+cdots+{1over 2^n} > 1+{nover 2},$$so to make sure the sum is over 100, for example, we'd addup terms until we get to around $ds 1/2^{198}$, that is,about $ds 4cdot 10^{59}$ terms. This series, $sum (1/n)$, is called theharmonic series.
Exercises 11.2
Ex 11.2.1Explain why $dssum_{n=1}^infty {n^2over 2n^2+1}$diverges.(answer)
Ex 11.2.2Explain why $dssum_{n=1}^infty {5over 2^{1/n}+14}$diverges.(answer)
Ex 11.2.3Explain why $dssum_{n=1}^infty {3over n}$diverges.(answer)
Ex 11.2.4Compute $dssum_{n=0}^infty {4over (-3)^n}- {3over 3^n}$. (answer)
Ex 11.2.5Compute $dssum_{n=0}^infty {3over 2^n}+ {4over 5^n}$. (answer)
Ex 11.2.6Compute $dssum_{n=0}^infty {4^{n+1}over 5^n}$.(answer)
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Ex 11.2.7Compute $dssum_{n=0}^infty {3^{n+1}over 7^{n+1}}$.(answer)
Ex 11.2.8Compute $dssum_{n=1}^infty left({3over 5}right)^n$.(answer)
Ex 11.2.9Compute $dssum_{n=1}^infty {3^nover 5^{n+1}}$.(answer)