# Math Help - Infinite Series

1. ## Infinite Series

I'm having trouble with the following problem:

Use the integral test to find upper and lower bounds on the limit of the series.

$\sum_{k=1}^{\infty} \frac {1}{k^3}$

2. Originally Posted by larson
I'm having trouble with the following problem:

Use the integral test to find upper and lower bounds on the limit of the series.

$\sum_{k=1}^{\infty} \frac {1}{k^3}$
Drawing a graph and shading rectangles that represent the sum helps:

$\int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx$

3. Originally Posted by mr fantastic
Drawing a graph and shading rectangles that represent the sum helps:

$\int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx$
I get what your saying, but how would I go about graphing this... just graph $\frac {1}{x^3}$ and then do like a trapezoid sum or something? If so, what does this even mean? I'm pretty confused in this whole chapter.

4. Originally Posted by larson
I get what your saying, but how would I go about graphing this... just graph $\frac {1}{x^3}$ and then do like a trapezoid sum or something? If so, what does this even mean? I'm pretty confused in this whole chapter.
The integral can be evaluated easily.
$\int \frac {1}{x^3} \, dx = -\frac{1}{2x^2} + C$

Thus

$\int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx$

$\frac12 < \sum_{k=1}^{\infty} \frac{1}{k^3} < \frac32$

5. Originally Posted by Isomorphism
The integral can be evaluated easily.
$\int \frac {1}{x^3} \, dx = -\frac{1}{2x^2} + C$

Thus

$\int_1^{\infty} \frac{1}{x^3} \, dx < \sum_{k=1}^{\infty} \frac{1}{k^3} < 1 + \int_1^{\infty} \frac{1}{x^3} \, dx$

$\frac12 < \sum_{k=1}^{\infty} \frac{1}{k^3} < \frac32$
I get everything but the very last part... how do you get the 1/2 and the 3/2 in the end?

6. $\int \frac {1}{x^3} \, dx = -\frac{1}{2x^2} + C$

So, $\displaystyle \int_{1}^{\infty} \frac {1}{x^3} \, dx = -\frac{1}{2x^2} \mid^{\infty}_{1} = 0 - \left(-\frac{1}{2}\right) = \frac12$

7. Originally Posted by larson
I get everything but the very last part... how do you get the 1/2 and the 3/2 in the end?
Do you know how to evaluate improper integrals?

8. Originally Posted by mr fantastic
Do you know how to evaluate improper integrals?
Yes, I understand how to get the 1/2, but how do you find the upper bound (the 3/4)?

9. Originally Posted by larson
Yes, I understand how to get the 1/2, but how do you find the upper bound (the 3/4)?
Please read the replies more carefully.

The upper bound is $1 + \int_1^{\infty} \frac{1}{x^3} \, dx$.

And you've already been shown that the integral is equal to 1/2 ......

10. Originally Posted by mr fantastic
Please read the replies more carefully.

The upper bound is $1 + \int_1^{\infty} \frac{1}{x^3} \, dx$.

And you've already been shown that the integral is equal to 1/2 ......
Wowww... okay, thanks, sorry for that...