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We illustrate [link] in [link] . In particular, by representing the remainder R N = a N + 1 + a N + 2 + a N + 3 + as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by N f ( x ) d x and bounded below by N + 1 f ( x ) d x . In other words,

R N = a N + 1 + a N + 2 + a N + 3 + > N + 1 f ( x ) d x

and

R N = a N + 1 + a N + 2 + a N + 3 + < N f ( x ) d x .

We conclude that

N + 1 f ( x ) d x < R N < N f ( x ) d x .

Since

n = 1 a n = S N + R N ,

where S N is the N th partial sum, we conclude that

S N + N + 1 f ( x ) d x < n = 1 a n < S N + N f ( x ) d x .
This shows two graphs side by side of the same decreasing concave up function y = f(x) that approaches the x axis in quadrant 1. Rectangles are drawn with a base of 1 over the intervals N through N + 4. The heights of the rectangles in the first graph are determined by the value of the function at the right endpoints of the bases, and those in the second graph are determined by the value at the left endpoints of the bases. The areas of the rectangles are marked: a_(N + 1), a_(N + 2), through a_(N + 4).
Given a continuous, positive, decreasing function f and a sequence of positive terms a n such that a n = f ( n ) for all positive integers n , (a) the areas a N + 1 + a N + 2 + a N + 3 + < N f ( x ) d x , or (b) the areas a N + 1 + a N + 2 + a N + 3 + > N + 1 f ( x ) d x . Therefore, the integral is either an overestimate or an underestimate of the error.

Estimating the value of a series

Consider the series n = 1 1 / n 3 .

  1. Calculate S 10 = n = 1 10 1 / n 3 and estimate the error.
  2. Determine the least value of N necessary such that S N will estimate n = 1 1 / n 3 to within 0.001 .
  1. Using a calculating utility, we have
    S 10 = 1 + 1 2 3 + 1 3 3 + 1 4 3 + + 1 10 3 1.19753 .

    By the remainder estimate, we know
    R N < N 1 x 3 d x .

    We have
    10 1 x 3 d x = lim b 10 b 1 x 3 d x = lim b [ 1 2 x 2 ] N b = lim b [ 1 2 b 2 + 1 2 N 2 ] = 1 2 N 2 .

    Therefore, the error is R 10 < 1 / 2 ( 10 ) 2 = 0.005 .
  2. Find N such that R N < 0.001 . In part a. we showed that R N < 1 / 2 N 2 . Therefore, the remainder R N < 0.001 as long as 1 / 2 N 2 < 0.001 . That is, we need 2 N 2 > 1000 . Solving this inequality for N , we see that we need N > 22.36 . To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is N = 23 .
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For n = 1 1 n 4 , calculate S 5 and estimate the error R 5 .

S 5 1.09035 , R 5 < 0.00267

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Key concepts

  • If lim n a n 0 , then the series n = 1 a n diverges.
  • If lim n a n = 0 , the series n = 1 a n may converge or diverge.
  • If n = 1 a n is a series with positive terms a n and f is a continuous, decreasing function such that f ( n ) = a n for all positive integers n , then
    n = 1 a n and 1 f ( x ) d x

    either both converge or both diverge. Furthermore, if n = 1 a n converges, then the N th partial sum approximation S N is accurate up to an error R N where N + 1 f ( x ) d x < R N < N f ( x ) d x .
  • The p -series n = 1 1 / n p converges if p > 1 and diverges if p 1 .

Key equations

  • Divergence test
    If a n 0 as n , n = 1 a n diverges .
  • p -series
    n = 1 1 n p { converges if p > 1 diverges if p 1
  • Remainder estimate from the integral test
    N + 1 f ( x ) d x < R N < N f ( x ) d x

For each of the following sequences, if the divergence test applies, either state that lim n a n does not exist or find lim n a n . If the divergence test does not apply, state why.

a n = n 5 n 2 3

lim n a n = 0 . Divergence test does not apply.

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a n = ( 2 n + 1 ) ( n 1 ) ( n + 1 ) 2

lim n a n = 2 . Series diverges.

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a n = ( 2 n + 1 ) 2 n ( 3 n 2 + 1 ) n

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a n = 2 n 3 n / 2

lim n a n = (does not exist). Series diverges.

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a n = 2 n + 3 n 10 n / 2

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a n = e −2 / n

lim n a n = 1 . Series diverges.

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a n = tan n

lim n a n does not exist. Series diverges.

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a n = 1 cos 2 ( 1 / n ) sin 2 ( 2 / n )

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a n = ( 1 1 n ) 2 n

lim n a n = 1 / e 2 . Series diverges.

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a n = ( ln n ) 2 n

lim n a n = 0 . Divergence test does not apply.

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State whether the given p -series converges.

n = 1 1 n n

Series converges, p > 1 .

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n = 1 1 n 4 3

Series converges, p = 4 / 3 > 1 .

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Questions & Answers

do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
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Daniel
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Abigail
Do somebody tell me a best nano engineering book for beginners?
s. Reply
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Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
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what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
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Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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many many of nanotubes
Porter
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Yasmin
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I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
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You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
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Practice Key Terms 4

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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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