# Homework Help: Amount of Terms Required to Find a Sum at an Indicated Accuracy

1. Oct 14, 2009

### NastyAccident

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
Alternating Series Test
General knowledge of adding series up...

3. 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)
$$\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{(n+1)*9^{n+1}} < \sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}$$

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 sorta 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 an's terms -
|an| < 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!

Sincerely,

NastyAccident

Last edited: Oct 14, 2009
2. Oct 14, 2009

### Staff: Mentor

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

3. Oct 14, 2009

### NastyAccident

So, in essence, I shouldn't of taken the absolute value of each of the an'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:

an < 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...

Last edited: Oct 14, 2009
4. Oct 14, 2009

### Staff: Mentor

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?