Question

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Answer 1

see https://en.wikipedia.org/wiki/Alternating_series#Approximating_sums
for the bounds on the error you get when adding up an alternating series.

the error you get from summing m terms is ≤ | \(a_{m+1} \) |
where "a" for your problem is
\[ \frac{n}{n^3+1} \]

in other words we want to find where
\[ \frac{n}{n^3+1} ≤ 0.01\]
the number of terms we need will be one less than the n that we find.

Answer 2

one way to solve for n is trial and error. use a calculator and test various n's

Answer 3

Why is it (n)/((n^3)+1)<=0.01 and not [((-1)^n)*n]/((n^3)+1)<=0.01?

Answer 4

we do not care if the term is + or - . we just want its absolute value to be less than 0.01

Answer 5

I've just tried trial and error and I keep getting the wrong answer; I was wondering if there was another way to do it. It's just confusing to me

Answer 6

what do you get when n=5 ?

Answer 7

Wouldn't it be 0.04 approximately? When put into the equation you first wrote. Which is over 0.01

Answer 8

if we type
5/(5^3+1)=
into google it tells us 0.0396825397 or about 0.04 as you said
too big

try n=10

Answer 9

0.00999001

Answer 10

so the 10th term is smaller than 0.01

just to be sure, check x=9

Answer 11

0.0123287671

Answer 12

which is too big. so according to wikipedia, we need 9 terms to get the error less than 0.01 (the formula says if the 10th term is < error, we need 9 terms)

Answer 13

But didn't you just say that 9 is too big?

Answer 14

the formula says the sum of the first n terms will have an error less than the size of the n+1 term

Answer 15

we found the 10 term is smaller than 0.01
which means the sum of the first 9 terms will have an error less than 0.01

Answer 16

I notice the series starts at n=2, so we should be clear about how many terms we are using. the terms are for n=2,3,...8, 9 which means 8 terms total

Answer 17

Oh, I was just going to ask about that. Because the second part of this question tells me to add the number of terms together that I have computed in the first part of the question (which would be 8, right?) to give the estimate. So would I start adding all the terms together using the initial series equation, so, plugging in n=2, 3, ... , 8, 9 and all that added together would equal the estimate?

Answer 18

yes. that sounds right, although a bit painful.

Answer 19

Haha yeah, it kind of is. So if I would add them all together, it would be less than 0.01, is that correct?

Answer 20

you will get a number (not 0.01 I am sure)
but if you continue to add more terms (for ever if you are ambitious) the final answer will not be more than 0.01 different (bigger or smaller) than what you get using the first 8 terms

Answer 21

But the number of terms to add together to guarantee that the estimate has an error of no more than 0.01 (8 terms) is still correct?

Answer 22

yes, you need to add up 8 terms (n=2 to 9) to guarantee that the sum is within 0.01 of truth (which you get by adding up all the terms from n=2 to infinity)

Answer 23

Okay, thank you so much :)

Answer 24

if you add up the terms n=2 to 9 you should get
0.147501994