2007-10-31 - Alternating series remainder theorem

Today's nonsense

Alternating Series Remainder

In a convergent alternating series, the absolute value of the remainder, R_{_N}, involved in approximating the sum , s, by S_{_N} is less than (or equal to) the first neglected term. That is, such that, just thought I'd throw in the such that, for a reminder, so Bill- this series goes on forever, we will do partial sums, the first partial sum is the first term, the fourth partial sum is the sum of the first four terms, okay, so partial sums, whatever numbers partial sums, that's just how many of the terms you've added up, and you can't add them up to infinity, and at some point the partial sums will stop, so like at 112, and the first neglected one is 113. So you're just approximating the sum. So the question is: what kind of approximation do you have? How accurate is it? Your approximation is within an accuracy of whatever your first neglected term is, in the case of alternating series.

|S - S_{_N}| = R_{_N} (the Remainder) < = a_{_(N+1)}
Now, this remainder, this difference in your approximation and the actual answer, is going to be < = the next term, which has a_{_N+1}.

So, say the next term was 7, and the next one is subtract 7, so that's why it can't be any bigger than the next term, so ..

ex 1) Approximate the sum of the following series by its first six terms.
Sigma from n=1 to infinity of (-1)^(n+1) * (1/n!)
-- Converges by AST.
91/144 - 1/5040 < = S < = 91/144 + 1/5040

ex) Approximate the sum of the series with an error of less than .001 ...
a) Series from n=1 to infinity of (-1)^{^(n+1)} / 4^{^n}
So we're looking for the first term less than 0.001, and then we add up all the terms prior to it, in this case that will be a total of four terms, and what do you get when you add up the four terms? And this sum is approximately .199 (becuase we are estimating to the nearest thousandths).

b) Series from n=1 to infinity of (-1)^{^n} * e^{^(-n)}
-- Have to add up seven terms. 0.732.

ex) Determine how many terms are needed to approximation the sum to within 0.001.

a) Series from n=0 to infinity of (-1)^n /(n+3)^2
- Just scroll through the tableset on the calculator in order to get to that point where you see the "within 0.001" and find that next neglected term. The first neglected term is term #30 where n=29, term #30 is the first term less than 0.001. So find the partial sum up to n=29.
b) Series from n=1 to infinity of 2(-1)^(n+1) / ( sqrt(x) * x^3 )