
Suppose we have a sum of the form
or more generally
for
and want to find its sum.
How should we go about it?
Another question: using the first 5 terms in the series only, or the first 25, how accurately can we sum it?
The answer is quite amazing. You can get the answer to better than three decimal places from the first five terms alone, and to ten decimal places from the first 25! And with ridiculous ease. (And you can get further accuracy similarly with more terms.)
How?
Obviously it is not enough to compute the partial sums of the series up to the fifth or twenty fifth term and estimate the series sum as the fifth or twenty fifth partial sum. But it is enough to compute these partial sums and estimate the sum from them.
How?
You can calculate the partial sums easily with a spreadsheet. (Put A3=A2+1, and put B1=your value for x, C1= your y, B3=($B$1)^A3)/(A3)^$C$1 and C3=C2+B3, and copy A3, B3 and C3 downward and you will have your partial sums in column C.)
Once you have them, you will see that the odd partial sums form a decreasing sequence, while the even ones form form an increasing one. The sequence sum is therefore trapped between any pair of successive values of the partial sums. Thus, the odd ones partial sums give upper bounds on the sequence sum, while the even ones give lower bounds; these bounds get better and better, though rather slowly for x = 1, y = 1.
This means that if you take the average of two adjacent partial sums, you get closer to the answer than the worse of them, since one definitely lies above the answer and one below.
The curious property possessed by these particular series, and some others,
is that after averaging the adjacent partial sums, the same property: that the
odd terms are decreasing and the even terms increasing still holds true when
the terms are averages of successive partial sums (or averages of averages,
etc., of successive partial sums) just as when they are partial sums.
This means if you average these you get closer to the answer, and if you continue
to average, you get still closer, forever, or until you have the answer to the
accuracy of your machine.
How can you do all this averaging? Just enter D3=(C3+C4)/2 and copy D3 into
the rectangle with corners D3 and AE30.
You can estimate the sum from your entries in row 3. Of course the estimates
improve as you go downward, but each time you go down one you use one further
partial sum, so if you only want to use information up to the kth partial sum,
you might as well just go across the third row k  1 steps from row C. (You
actually can do a bit better, but who cares.)
You can do slightly better by averaging this result with its predecessor on the same diagonal. Thus, instead of using say G3 as estimate, you do a bit better by using (G3+F4)/2. Both of these use information in E5 and hence D6 and C7 which uses information from the first five terms in the series only.
For the original sum, x = 1, y = 1, the best estimates we get using only the first k terms in the sequence are as follows
k value 
Estimate 
Source of estimate 
True answer 
% error 
1 
1 
C3 
0.693147... 
44.27 
2 
0.625 
D3/2+C4/2 
0.693147 
9.8316 
3 
0.6825 
E3/2+D4/2 
0.693147 
0.8147 
4 
0.692708 
F3/2+E4/2 
0.693147 
0.0633 
5 
0.693229 
G3/2+F4/2 
0.693147 
0.0118 
In other words, the error in estimating this sum as noted from its first 5 terms is on the order of one hundredth of a percent.
When k is larger you are better off with a slightly different estimate (in a higher row; that is, sum several terms, then start averaging), and you can get accuracy here to one part in 10^{10} looking only at the first 20 terms of the sequence.
Roughly the same or better accuracy appears to be possible for any x and y in the indicated range.
Exercise A.1 Have a friend pick a value of x and y and compute the first
five terms in the series whose jth term is .
Use the method described above to estimate the sum of the series using only
this information. Then let the friend determine the sum to high accuracy by
using the indicated method and a k value of 50, and find the error in your estimate.
(With any kind of luck your answer will be accurate to better than one part
in a thousand, and your friend will be willing to bet that you cannot achieve
such accuracy from only five terms.)
