# The integral test

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.

So it's right?

NateTG
Homework Helper
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:
$$\sum_{n=2}^{\infty}\frac{1}{n \ln n}$$
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
Homework Helper
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:
$$\sum_{n=2}^{\infty}\frac{1}{n \ln n}$$
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?

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?

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

3.14159265358979 said:
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

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

Last edited:
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.