Integral Test for \sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} - 0.01

[3.14159265358979]

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

Here's what i got:

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

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

thus,

$$R_{n} \leq \frac{1}{ln\;n}$$

since we want $$R_{n} \leq 0.01$$,

$$\frac{1}{ln\;n} \leq 0.01$$

implying $$n = e^{100}$$.


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

 

[mathman]

It may not be. The series you have is just barely convergent. Without the (ln n)² term, it would diverge.

 

[3.14159265358979]

So it's right?

 

[NateTG]

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:
$$\sum_{n=2}^{\infty}\frac{1}{n \ln n}$$
which converges significantly more slowly. And that makes sense considering that $\ln x$ grows very slowly, and
$$\sum_{n=2}^{\infty}\frac{1}{n}$$
is divergent.

 

[shmoe]

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:
$$\sum_{n=2}^{\infty}\frac{1}{n \ln n}$$
which converges significantly more slowly.

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


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

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

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?

 

[3.14159265358979]

it should be $$\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{n}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n}$$
right?

 

[3.14159265358979]

(noting that the lower limit is n now, instead of 2)

 

[Zone Ranger]

it should be $$\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{n}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n}$$
right?

what you have above is wrong.



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

if you want it to be equal to $$\frac{1}{ln\;n}$$
then write


$$\lim_{b \to \infty} \int_{n}^{b} \frac{1}{t(ln\;t)^{2}} dt= \frac{1}{ln\;n}$$

 

[shmoe]

3.14..., do you know how to use Zone Ranger's correction to get an estimate for $$R_n$$? Remember that $$R_n$$ is the 'tail' of your series.

edit-What I'm hoping is that you can justify why $$R_n$$ can be bounded using that integral.