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

[NastyAccident]

Homework Statement

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}
(|error| < 0.0001)

Homework Equations

Alternating Series Test
General knowledge of adding series up...

The Attempt at a Solution

Alternating Series Test
1.) Limit x->infinity
\frac{(-1)^{n}}{n*9^{n}} = 0
2.) Decreasing eventually for n (really immediately)
\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{(n+1)*9^{n+1}} &lt; \sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}

By the Alternating Series Test, this series (\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}) is convergent.

Well, I originally choose n = 4 terms in order to find the indicated accuracy. However, that was wrong.

So, I'm sort of scratching my head as to what I did wrong...

My approach:
\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}

I start listing out a_n's terms -
|a_n| < 0.0001
a₁ = 0.11111111111
a₂ = 0.006172839506
a₃ = 0.000457247370
a₄ = 0.000038103947
a₅ = 0.000003387017
a₆ = 0.000000313612
a₇ = 0.000000029867

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

Any help/suggestions as to where I went wrong are appreciated and will be thanked!
NastyAccident

 

[Mark44]

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

 

[NastyAccident]

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

So, in essence, I shouldn't of taken the absolute value of each of the a_n's and should of left it with the +/- signs?

If that is the case, then the problem should be okay for n = 3?

Since:
\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}
is within
\sum^{\infty}_{n=1}\frac{(1)^{n+1}}{(n+1)*9^{(n+1)}}
which gives me:

a_n < 0.0001
a₁ = 0.061728395
a₂ = 0.0004572473708
a₃ = 0.000038103947
a₄ = 0.000003387017
a₅ = 0.000000313612
a₆ = 0.000000029867

So, n = 3 would be correct...

 

[Mark44]

Right. You don't want the abs. value of your terms. Just add them up. Your estimate with a₁ + a₂ + a₃ should be within .0001 of the actual value of the infinite sum. You did add up the three terms, right?