how many terms of the series do we need to add in order to find the sum to the indicated accuracy?

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Amount of Terms Required to Notice a Sum at an Indicated Accuracy

• Thread starter NastyAccident
• Start engagement Oct xiv, 2009

Homework Statement

Testify that the serial is convergent. How many terms of the series practice we demand to add in order to observe the sum to the indicated accuracy?

[tex]\sum^{\infty}_{north=i}\frac{(-1)^{due north}}{n*9^{north}}[/tex]
(|error| < 0.0001)

Homework Equations

Alternating Series Examination
Full general cognition of adding series up...

The Attempt at a Solution

Alternating Serial Examination
one.) Limit 10->infinity
[tex]\frac{(-one)^{due north}}{due north*ix^{n}}[/tex] = 0
two.) Decreasing eventually for north (really immediately)
[tex]\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{(due north+1)*9^{north+1}} < \sum^{\infty}_{n=i}\frac{(-i)^{n}}{n*9^{n}}[/tex]

By the Alternating Series Exam, this series ([tex]\sum^{\infty}_{n=1}\frac{(-1)^{n}}{north*9^{n}}[/tex]) is convergent.

Well, I originally cull northward = 4 terms in social club to find the indicated accurateness. However, that was incorrect.

So, I'm sorta scratching my caput as to what I did wrong...

My arroyo:
[tex]\sum^{\infty}_{n=1}\frac{(-i)^{n}}{n*nine^{n}}[/tex]

I beginning list out a_northward's terms -
|a_n| < 0.0001
a_one = 0.11111111111
a₂ = 0.006172839506
a_iii = 0.000457247370
a₄ = 0.000038103947
a_v = 0.000003387017
a_six = 0.000000313612
a_vii = 0.000000029867

a₄ < 0.0001, so northward should equal iv terms.... However, it doesn't for some odd reason?

Whatever help/suggestions every bit to where I went incorrect are appreciated and will be thanked!

Sincerely,

NastyAccident

Last edited: October fourteen, 2009

Answers and Replies

Yous have an alternating series, and then every other a_northward (starting from a_a) should be negative. You show all of them as positive in your list.

You accept an alternating series, so every other a_n (starting from a_a) should exist negative. You prove all of them as positive in your listing.

So, in essence, I shouldn't of taken the absolute value of each of the a_{due north}'south and should of left it with the +/- signs?

If that is the case, and so the problem should be okay for north = 3?

Since:
[tex]\sum^{\infty}_{n=ane}\frac{(-one)^{due north}}{northward*9^{northward}}[/tex]
is within
[tex]\sum^{\infty}_{northward=i}\frac{(1)^{n+ane}}{(due north+1)*ix^{(n+1)}}[/tex]
which gives me:

a_n < 0.0001
a₁ = 0.061728395
a₂ = 0.0004572473708
a₃ = 0.000038103947
a_iv = 0.000003387017
a₅ = 0.000000313612
a₆ = 0.000000029867

So, n = 3 would be correct...

Last edited: Oct xiv, 2009

Correct. You don't want the abs. value of your terms. Just add them upwardly. Your estimate with a_one + a₂ + a_iii should exist within .0001 of the actual value of the infinite sum. You did add upward the 3 terms, right?

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