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.