Home  18.013A  Chapter 30 


With a spreadsheet or program it is quite easy to compute values of the partial sums of a series.
When a series has a ratio of successive terms $\frac{{a}_{j+1}}{{a}_{j}}$ that is less than some $r$ that is less than 1, the terms decrease exponentially in $r$ as $j$ increases, and there is rarely a problem in computing value of the series to any desired accuracy.
When the ratio of successive terms becomes close to 1 convergence becomes slower, and one can hope to improve the accuracy of computations by fiddling with them.
When the rate of convergence of the jth partial sum behaves as ${j}^{k}$ for some power $k$ , you can improve convergence by extrapolating.
One way to do this, as already noted in a number of contexts, is to replace the partial sum ${s}_{2j}$ by $\frac{2k{s}_{2j}{s}_{j}}{{2}^{k}1}$ which can also be written as ${s}_{2j}+\frac{{s}_{2j}{s}_{j}}{{2}^{k}1}$ .
When the coefficients of the series are standard functions of $j$ this often happens for integer values of $k$ .
You can easily examine ratios of the differences of terms whose indices differ by a factor of 2 in order to see how the partial sums are converging, and to choose a suitable $k$ to extrapolate with. Doing this successively allows computations to increase very greatly in accuracy.
You can even determine the rate of growth of the partial sums of a divergent series by similar extrapolation.
For example, consider the sum of ${j}^{2}$ , which obviously diverges.
You can make the following deductions by looking at the partial sums ${s}_{j}$ for $j$ of the form ${2}^{k}$ :
1. The terms increase by a factor of roughly 8 as $j$ doubles. This indicates that the leading term is proportional to ${n}^{3}$ . (Which you probably should have known.)
2. By looking at the ratio of partial sums to ${n}^{3}$ and extrapolating you can see that the coefficient of ${n}^{3}$ is $\frac{1}{3}$ .
3. By taking the differences between the partial sums ${s}_{n}$ and $\frac{{n}^{3}}{3}$ you can produce another sequence whose ratios increase by roughly 4 as $n$ doubles.
4. Looking at the ratios of these to ${n}^{2}$ you can find the coefficient of ${n}^{2}$ which is $\frac{1}{2}$ .
5. By looking at the differences of partial sums with $\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2}$ you can find that it is $\frac{n}{6}$ .
Thus you can find the formula
for this series, up to the nth term by numerical spreadsheet manipulations alone.
There are, no doubt easier ways to get this result, but this one is really quite painless if you are used to spreadsheet manipulations.
When the series does converge you can extrapolate to get accurate answers with very little effort this way.
By looking at powers of 2 you can first get rid of the difference between the partial sum and the error that decreases by the leading power of 2, then the next and the next etc. until the answer is accurate to within your standard of same.
Exercises:
30.12 Do the steps above for the series starting with 1 whose $(n+1)$ th term is ${n}^{2}$ explicitly. Then do the same for $(n+1)$ th term ${n}^{3}$ .
30.13 Find the sum of ${n}^{3}$ (starting at $n=1$ ) to ten place accuracy by appropriate extrapolation.
