Question

Use the alternating series estimation theorm to determine how many terms should be used to estimate the sum of the entire series with an error of less than 0.001.

Answer 1

\[\sum_{n=1}^{\infty} (-1)^{n+1}\frac{ 1 }{ n+\sqrt{5n}}^{3}\]

Answer 2

@amistre64 :P

Answer 3

if we need to use a thrm, define the thrm

Answer 4

I only know the alternating series thrm, I dont understand the estimation one.

Answer 5

ive never heard of the esimating one so the definition will be required to see if i can even understand it :)

Answer 6

Lol okay, I'll figure it out. Thanks!

Answer 7

that link doesnt show for me, guess ill find it anothe rplace tho

Answer 8

https://www.youtube.com/watch?v=ilRkVcAxdCo

here he is on the youtube ...

Answer 9

thanks

Answer 10

looks to me that we simply see when the 'rule' is equal to .001 and solve for n

Answer 11

The answer i got is 0.0295

Answer 12

which is the first term..

Answer 13

\[ \left|\frac{ 1 }{ n+\sqrt{5n}}^{3}\right|<=.001\]
\[ \frac{ 1 }{ n+\sqrt{5n}}<=\sqrt[3]{.001}\]
\[ \frac{ 1 }{\sqrt[3]{.001} }<=n+\sqrt{5n}\]
... maybe

Answer 14

.0295 is not < .001

Answer 15

run some trials,

if n=10 < .001, try n=5
if n=5 > .001, try n=7

and so on to narrow in on it

Answer 16

Well what patrick did, was he calculated the first couple terms, and then whatever number was = to decimal places, he added the numbers before it.

Answer 17

exlcuding the number that fit the decimal place.

Answer 18

right, i watched the first example :) 10 minutes is too long lol

Answer 19

but the estimation thrm says that the error is at most the size of the first neglected term

Answer 20

i get n=5

Answer 21

or 4 depending on how you look at it

the first term less then .001 i when n=5

Answer 22

yeah brute math instead of solving algebraically

Answer 23

Alright, well thanks :P

Answer 24

the 5th term is equal to .001 :) so the first 4 terms shold have an error that is no more than .001

Answer 25

Okay, thanks so much!

Answer 26

youre welcome