1. estimation of series sum

I'm having problems with this question

Estimate $\displaystyle \sum_{n=1}^{\infty}n^{-\frac{3}{2}}$ to within .01

I know I can use the integral test with my remainder estimate being .01 but it would take such a large number to find the answer.

$\displaystyle -\frac{2}{\sqrt{x}} < .01$

I would have to sum up 40,000 terms to get it to that accuracy

Is there another way to do this? Thanks in advance

2. trying to think of an easier way, nothing is coming to mind... doesnt look like you can manipulate this to be a geometric series in any way.

I will tell you that my old graphic calculator would do a summation for you, up to a certain bound...

trying to think of an easier way, nothing is coming to mind... doesnt look like you can manipulate this to be a geometric series in any way.

I will tell you that my old graphic calculator would do a summation for you, up to a certain bound...
Yea i tried using my graphing caculator, poor thing is still displaying busy

4. You might be able to apply the Euler-Maclaurin Summation Formula:

Euler?Maclaurin formula - Wikipedia, the free encyclopedia

5. Originally Posted by 11rdc11
I'm having problems with this question

Estimate $\displaystyle \sum_{n=1}^{\infty}n^{-\frac{3}{2}}$ to within .01

I know I can use the integral test with my remainder estimate being .01 but it would take such a large number to find the answer.

$\displaystyle -\frac{2}{\sqrt{x}} < .01$

I would have to sum up 40,000 terms to get it to that accuracy

Is there another way to do this? Thanks in advance
$\displaystyle \sum_{n=1}^{\infty}n^{-\frac{3}{2}} \leq \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx$

$\displaystyle \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx=\lim_{\varepsilon->\infty} \int_{1}^{\varepsilon}x^{-\frac{3}{2}}\,dx$
$\displaystyle =\lim_{\varepsilon->\infty}\left(\frac{-2}{\sqrt{\varepsilon}}\right ) - \frac{-2}{\sqrt{1}}=2$

So $\displaystyle \sum_{n=1}^{\infty}n^{-\frac{3}{2}} \leq 2$

I hope that's right... I haven't done integral approximations in a long time...

6. Originally Posted by Prove It
$\displaystyle \sum_{n=1}^{\infty}n^{-\frac{3}{2}} \leq \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx$

$\displaystyle \int_{1}^{\infty}x^{-\frac{3}{2}}\,dx=\lim_{\varepsilon->\infty} \int_{1}^{\varepsilon}x^{-\frac{3}{2}}\,dx$
Sorry im confused what do you mean?

7. Originally Posted by 11rdc11
Sorry im confused what do you mean?
OK, think back to elementary calculus. A definite integral is really an infinite sum of the trapezoids made under the curve. The sum of terms of $\displaystyle n^{\frac{-3}{2}}$ therefore can not be any greater than an integral with the same limits. Evaluate the integral and you have an upper bound for the sum.