Whats the minimum number of terms of the series
infinity (1) / ((n)Ln^2(n))
E
n = 2
..that must be added to find the sum with error at most 0.1?
Not sure about this one. where to start, and how to calculate.
Thanks for any help with this one.
Whats the minimum number of terms of the series
infinity (1) / ((n)Ln^2(n))
E
n = 2
..that must be added to find the sum with error at most 0.1?
Not sure about this one. where to start, and how to calculate.
Thanks for any help with this one.
You need to determine the convergence of,
$\displaystyle \sum_{k=2}^{\infty} \frac{1}{n\ln^2 n}$
This sequence can be represented by this function,
$\displaystyle f(x)=\frac{1}{x\ln^2 x} $
This function, is continous, positive and decreasing.
Thus,
$\displaystyle \int_2^{\infty} \frac{1}{x\ln^2 x} dx$
Let, $\displaystyle u=\ln x$.
Thus,
$\displaystyle \int \frac{1}{u^2} du=-\frac{1}{u}$.
Thus, subsitution for $\displaystyle u$ and the limits,
$\displaystyle -\frac{1}{\ln x}\big|^{\infty}_{2}$
This clearly converges.
okay understood as far as the integral goes. However, knowing that it converges makes sense. But where do i go from here to find the minimum number of terms in the series?
also, whats it mean to have no more error than 0.1?
thanks for sticking with me so far!
We are interested in the sum:
$\displaystyle \sum_2^{\infty}\frac{1}{n[\ln(n)]^2}$
This may be written:
$\displaystyle \sum_2^{\infty}\frac{1}{n[\ln(n)]^2} =\sum_2^{N}\frac{1}{n[\ln(n)]^2} + \sum_{N+1}^{\infty}\frac{1}{n[\ln(n)]^2}$
Now look at the tail:
$\displaystyle \sum_{N+1}^{\infty}\frac{1}{n[\ln(n)]^2} < \int_{N}^{\infty} \frac{1}{x[\ln(x)]^2} dx = \frac{1}{\ln(N)}$
(for the explanation of the last step we need to observe that the sequence of terms is strictly decreasing so the sum is bounded above by the given integral, then ImPerfectHackers fine post tells us what the integral is)
Now we want the error in the truncated sum to be $\displaystyle <0.1$, so if we find an $\displaystyle N$ such that:
$\displaystyle \frac{1}{\ln(N)}<0.1$
we will have found how many terms we need to sum to get errors less than this. Now solving
$\displaystyle \frac{1}{\ln(N)}=0.1$
gives $\displaystyle N=22026.5..$, so we need $\displaystyle N \ge 22027$, which is $\displaystyle 22026$ terms in the truncated series.
(comment: we can in fact trap the remainder between a pair of bounds
so providing an estimate of the remainder rather than just a bound
Interestingly constructing a pair of bounds on the remainder gives with N=10
that the remainder is between 0.417 and 0.434, so the sum can be determined
to better precisions with 10 terms plus an estimate of the remainder than
by summing ~22000 terms!).
RonL