Here's what I have so far:
Estimate the number of terms required to approximate to within .
Solving gives:
Therefore so terms.
This seems to be a really big answer. I might just be confusing myself but I was also wondering if there was an easier way of doing it (if this is the right way).
awww, I thought I had the answer in the bag! =(
I should have spotted there was an error there since I did know that any alternating decreasing sequence, when summed to infinity, can be rearranged to sum to any real number.
The hardest part here is making the series into a geometric series so it can be summed. I've just got no idea what to turn it into.
But you cannot really make it into a geometric series. Instead, consider the series given by . Now how could you get your sum from this? Hint in white: consider the integral from zero to one, consider it when the function is as it is and when it is in series notation.
Numerical experiment is always worth a look:
at which point brute force runs out of umph!Code:log_10(N) log_10(|error|) 2 -2.0043 3 -3.0003 4 -4.00004 5 -5.00000 6 -6.00000 7 -7.00000
But it looks like a tight bound on the number of terms needed should be ~=10^10
.
An alternative approach to this problem is to group the terms in pairs to turn the series into a series of decreasing positive terms, then to get bounds for the sum of the tail by constructing integrals that over and under estimate the sum of the tail.
We start by writting:
(though if we need terms this series to get the error under we need terms of the original series)
.
That's a very clever method!
So is this what I need to work out? if k is even, but can I use this fact at all? (or is this what you used to get the sum to 2N?).
I have another query: Since this series is a sum to infinity, can the individual terms be grouped together in this way? I was wondering this because if then is a decreasing null sequence of non-negative terms. Therefore can be used to sum the series to any real number.
The idea here is that if is a positive decreasing function for then
So finding an such that
will give us a number of terms that will suffice to give the required error limit on the sum.
Doing the sums (at this point I resort to the use of Maxima for doing the integral. and the algebra) gives (in fact slightly less if we use double precision floating point, but with higher precision we find that any will do) but this is half the number of terms needed for the original series so for the original series you need no more than terms
.