Question

alternating series- How to find the error ....

Answer 1

According to the Remainder Estimate for the Integral Test, the error in the approximation s≈sn (where s is the value of the infinite sum and sn is the n-th partial sum) is \[\left| s-s _{n} \right|\le ?\]
Find the smallest value of n such that this upper bound is less than 0.00003 .

Answer 2

\[\sum_{n=1}^{\infty} \frac{ 1 }{ (4n+4)^3}\]

Answer 3

@darkigloo are you taking the ap calc bc exam?

Answer 4

no

Answer 5

is there a better way to do it than just to add and see which two come within .0003 of each other?

Answer 6

add what

Answer 7

the first few terms
it if is alternating, the error gets smaller and smaller
if you have two partial sums that are with in 0.003 of each other, then the actual sum is as well

Answer 8

oh yes, there is an easier way sorry

Answer 9

find \(n\)so that \[\frac{1}{(4n+3)^3}<0.00003\]

Answer 10

take the reciprocal, take the cubed root to find \(n\)

Answer 11

i got 7 but it says its wrong

Answer 12

hmm let me see what i get

Answer 13

looks like you rounded down
round up

Answer 14

i.e. try 8

Answer 15

thats incorrect as well

Answer 16

DAMN

Answer 17

wait a sec
is that an alternating series or no? you didn't write it as one, just asking

Answer 18

it is

Answer 19

thats how its written in my hw

Answer 20

does it have a \((-1)^n\) in it?

Answer 21

no

Answer 22

then it is not alternating, it is?

Answer 23

ahh no

Answer 24

think you are supposed to use this \[\int _m^{\infty}\frac{dx}{(4x+3)^3}\]

Answer 25

if it doesn't alternate, all that stuff i said above is nonsense

Answer 26

so you need the integral first, which is easy right?\[\int \frac{dx}{(4x+3)^3}=-\frac{1}{8(4x+3)^2}\]

Answer 27

at infinity it is 0, so what you really need to solve is \[0.00003<\frac{1}{8(4x+3)^2}\]

Answer 28

is it + 3 or +4 ?

Answer 29

oops typo

Answer 30

so maybe my integral was wrong too huh?

Answer 31

i got 15 and its wrong

Answer 32

let me check

Answer 33

first off my integral was wrong because i made a typo

Answer 34

second time i got \[\int \frac{dx}{(4x+4)^3}=-\frac{1}{128(x+1)^2}\]

Answer 35

then for the inequality i got \(x>15.13...\) so \(N>16\) should do it (i hope)

Answer 36

ah yes! how would i do sn - s = ?

Answer 37

there is nothing left to do
you have \[|S-Sn|<0.00003\] if \(n>16\)

Answer 38

it says: According to the Remainder Estimate for the Integral Test, the error in the approximation s≈sn (where s is the value of the infinite sum and sn is the n-th partial sum) is : |s-sn| <= ?
and then it asks for n

Answer 39

i am confused now
we have \(n>16\) right?

Answer 40

if they want you to estimate the actual sum to within that error, since \(n>16\) you have to actually add the first 16 terms to see what you get
looks annoying

Answer 41

i got it, they were asking for the integral