Calc 2 Questions

Question

Hi,

 

We are currently finishing our chapter on Series in class, and I am having a little trouble with the homework, especially these types of problems (error bound problems). I really do not know where to start.. Can anyone help? Thank you so much.

 

1. Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10^-4.

 

Σ k=1 to infinity of ((-1)^(k+1))/(k^6)

 

2.  Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10^-4.

Σ k=0 to infinity of (6(-1)^k)/(k!)

 

3. Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10^-4.

Σ k=1 to infinity of (9(-1)^(k+1))/k

 

Answer 1

All of these are alternating series (the sign of the terms go from negative to positive).  To find the kth term that is less than the remainder, you set up the inequality and solve for k.

 

 

1) |(-1)^k+1/k⁶| < 10^-4

 

10⁴ < k⁶

 

10^2/3 = 4.64 < k

 

Round up to get k = 5.

 

 

Use the same set-up for 2 & 3.

Answer 2

For alternating series like these, the error is always bounded by the absolute value of the first term that is not included.

 

1. ∑_k=1^∞ (-1)^k+1/k⁶ will have

E = 1/(n+1)⁶

where n is the last term in the sum. To solve:

1/(n+1)⁶ ≤ 10^-4

(n+1)⁶ ≥ 10⁴

n ≥ ⁶√(10⁴) - 1 = 3.64159

So you need to sum 4 or more terms.

 

2. ∑_k=1^∞ 6(-1)^k/k!

E = 6/k!

 

3. ∑_k=1^∞ 9(-1)^k+1/k

E = 9/k

 

You should be able to solve the last two from there. Please comment if you're still confused.