
There are particularly simple criteria for alternating series, which are those which terms alternate in sign.
First the terms must themselves converge to zero. This is a necessary condition for convergence of the partial sums, which is what convergence of the series means, for any series.
For alternating series for which each term is smaller in magnitude than its predecessor, this condition (that terms converge to 0) is sufficient for convergence as well.
How come?
The idea is, you can pair up consecutive terms. In any pair the first, by definition dominates over the second, for such series, so the sign is that of the first term of the pair.
This means that the pairs starting with odd entries are all the same sign as the first entry, while the pairs starting with even entries all have the opposite sign.
Suppose the first term is positive. That means that the even partial sums are each a positive number more than the previous even partial sums. So they will always increase. Meanwhile the odd partial sums will be a negative number plus the previous one, and so will start at the first term and then constantly decrease.
The sum is then trapped between the rising tide of the even partial sums and the falling odd partial sums. And the difference between consecutive even and odd sums is just a term of the series, which, by definition goes to 0. This implies that the difference of partial sums converges to 0, which in fact implies that the series converges to some real number.
If you look at the partial sums of an alternating series, they zigzag up and down. This means that the average of two successive partial sums is a better estimate of the total sum than the first partial sum, and often is better than the second one as well.
Here is something you can do to improve your estimate of the sum of an alternating series. First write down the sequence of partial sums up to some point. Then average pairs of adjacent partial sums. If these still zigzag, average adjacent pairs again, and so on.
This is an extra easy thing to do. If you put the terms themselves in column B starting at B10, say, and the partial sums in column C starting at C10 , you can enter =(C10+C11)/2 in D11, and copy this last instruction down and to the right as many columns as you want. This will do averaging over and over again. If you put the letter x in E11, that will kill off those entries in which you averaged nothing along with some partial sums.
With the alternating harmonic series, I find that not only is each column still zigzagging, each row is as well. This means we can try the same averaging of successive terms in any row, and this ought to improve the answer.
What i then find is that at some point in each row, you get a zigzag failure; which means three in a row that are increasing or decreasing. If we estimate the complete answer by the middle entry of these three in the first zigzag failure we find, we get amazing results.
If we stop at the 13^{th} term of our series, which is , the answer we get as the middle entry in 3 in the 13^{th} row is accurate to 8 decimal places. If we go as far as the 20^{th} row, which is computed using only the first 20 entries, the last of which is .05, we get ten place accuracy in our answer. (which turns out to be ln(2), which you can deduce by summing the geometric series, and integrating it term by term, and evaluating as x approaches 1).
Exercises:
15.1 Perform the spreadsheet computation indicated above for the alternating harmonic sequence. And verify or disprove the claims above.
15.2 Use the fact that the geometric series converges uniformly and absolutely to (1x)^{1} for x < a < 1 to deduce that the value of this series is ln(2), by integrating both sides of the identity stating this fact, integrating the series term by term.
