# Calculus 2 - Infinite Series Question - Estimating Series with Positive Terms

1. Oct 21, 2011

### GreenPrint

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

Consider the following convergent series. Then complete parts a throw d below.

sum[k=1,inf] 5/k^7

a. Find an upper bound for the remainder in terms of n

2. Relevant equations

Estimating Series with Positive Terms
Let f be a continuous, positive, decreasing function for x >= 1 and let a_k = f(k) for k = 1,2,3,.... Let S = sum[k=1,inf] a_k be a convergent series and let S_n = sum[k=1,n] a_k be the sum of the first n terms of the series. The remainder R_n = S - S_n satisfies

R_n <= integral[n,inf] f(x)dx.

Furthermore, the exact value of the series is bounded as follows:

S_n + integral[n+1,inf] f(x)dx <= sum[k=1,inf] a_k <= S_n + integral[n,inf] f(X)dx

3. The attempt at a solution

I'm unsure how to do this problem. I believe that I'm trying to evaluate

S_n + integral[n,inf] f(X)dx

I have no problem find the value of integral[n,inf] f(X)dx
but am not sure how to find the value of S_n. I would now how to find the value of this if I was asked to find upper bound for the error for the first 50 terms, I could then find S_50 by just finding the sum which would be a finite number, but I am unsure how to find the upper bound in this case were I guess I'm trying to find the value of S_n in this case would be S_inf which I'm not sure how to do. Thank's for any help which you can provide me with.

2. Oct 21, 2011

### Staff: Mentor

The remainder (or error) is
$$\int_n^{\infty}f(x)dx = \int_n^{\infty}\frac{dx}{x^7}$$

3. Oct 22, 2011

### GreenPrint

Well this problem was one of my homework questions which I do online in this program in which I input my answer and it told me I was wrong when I entered 5/6. Have I done something wrong?

5*integral[1,inf] dk/k^7 = 5/6

4. Oct 22, 2011

### Staff: Mentor

Yes. If you want to estimate the series by using the first 50 terms of the series, the error is
$$R_{50} = \int_{50}^{\infty}5x^{-7}dx$$

You have this in your relevant equations, but you must not have thought it to be relevant...

Last edited: Oct 22, 2011
5. Oct 22, 2011

### GreenPrint

Alright well I entered 5/(6n) and it still told me I was wrong.

6. Oct 22, 2011

### HallsofIvy

Staff Emeritus
You have been told that an upper bound for the error is
$$\int_n^\infty \frac{5}{x^7}dx= 5\int_n^\infty x^{-7}dx$$

What is that? (It is NOT 5/(6n)!)