The integral test

  • #1
How many terms of the series [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} [/tex] would you need to add to find its sum to within 0.01?

Here's what i got:

let [tex] f(n) = \frac{1}{n(ln\;n)^{2}} [/tex]. Since [tex] f(n) [/tex] is continuous, positive and decreasing for all n over the interval [tex] [2,\infty] [/tex], we can use the integral test to evaluate the series.

[tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{2}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]

thus,

[tex] R_{n} \leq \frac{1}{ln\;n} [/tex]

since we want [tex] R_{n} \leq 0.01 [/tex],

[tex] \frac{1}{ln\;n} \leq 0.01 [/tex]

implying [tex] n = e^{100}[/tex].


but that can't be right...e^100 is way too big, isn't it? thanks in advance.
 

Answers and Replies

  • #2
mathman
Science Advisor
7,970
508
It may not be. The series you have is just barely convergent. Without the (ln n)2 term, it would diverge.
 
  • #3
So it's right?
 
  • #4
NateTG
Science Advisor
Homework Helper
2,450
6
3.14159265358979 said:
but that can't be right...e^100 is way too big, isn't it? thanks in advance.

I haven't followed the math, but I don't see why it should be. You can have sequences that converge *very* slowly like, apparently, the one you cite, and the series definitely looks like a slowly converging one. Consider the sequence:
[tex]\sum_{n=2}^{\infty}\frac{1}{n \ln n}[/tex]
which converges significantly more slowly. And that makes sense considering that [itex]\ln x[/itex] grows very slowly, and
[tex]\sum_{n=2}^{\infty}\frac{1}{n}[/tex]
is divergent.
 
  • #5
shmoe
Science Advisor
Homework Helper
1,992
1
NateTG said:
I haven't followed the math, but I don't see why it should be. You can have sequences that converge *very* slowly like, apparently, the one you cite, and the series definitely looks like a slowly converging one. Consider the sequence:
[tex]\sum_{n=2}^{\infty}\frac{1}{n \ln n}[/tex]
which converges significantly more slowly.

This actually diverges by the integral test, [tex]\int_{2}^{b}\frac{1}{t \ln t}dt=\ln\ln b-\ln\ln 2[/tex], which goes to infinity as b does (though obviously very slowly).


For the OP,the result is fine, but the process seems garbled:

[tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{2}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]

This isn't correct, the sum and the integral are not equal, and the evaluation of the limit and/or integral is off. This may be a translation error from paper to screen though?
 
  • #6
it should be [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{n}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]
right?
 
  • #7
(noting that the lower limit is n now, instead of 2)
 
  • #8
3.14159265358979 said:
it should be [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{n}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]
right?

what you have above is wrong.



[tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} \neq \lim_{b \to \infty} \int_{2}^{b} \frac{1}{n(ln\;n)^{2}} dn= \frac{1}{ln\;2} [/tex]

if you want it to be equal to [tex]\frac{1}{ln\;n} [/tex]
then write


[tex]\lim_{b \to \infty} \int_{n}^{b} \frac{1}{t(ln\;t)^{2}} dt= \frac{1}{ln\;n} [/tex]
 
Last edited:
  • #9
shmoe
Science Advisor
Homework Helper
1,992
1
3.14..., do you know how to use Zone Ranger's correction to get an estimate for [tex]R_n[/tex]? Remember that [tex]R_n[/tex] is the 'tail' of your series.

edit-What I'm hoping is that you can justify why [tex]R_n[/tex] can be bounded using that integral.
 
Last edited:

Related Threads on The integral test

Replies
1
Views
443
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
5
Views
3K
Replies
7
Views
2K
Top