The most direct way of computing the sum of an alternating series, simply computing the partial sums in the terms get small enough, may not be the most efficient.

Euler figured this out in the 18th century.

For our demo we’ll evaluate the Struve function defined by the seriesNote that the the terms in the series are relatively expensive to evaluate since each requires evaluating a gamma function.

Euler’s acceleration method will be inexpensive relative to computing the terms it takes as input.

Here’s what we get by evaluating the first partial sums for H1.

2(3.

4):2.

34748 0.

67236 1.

19572 1.

10378 1.

11414 1.

11332 So if we were to stop here, we’d report 1.

11332 as the value of H1.

2(3.

4).

Next let’s see what we’d get using Euler’s transformation for alternating series.

I’ll full precision in my calculations internally but only displaying four digits to save horizontal space that we’ll need shortly.

2.

3474 1.

5099 0.

6723 0.

9340 1.

1957 1.

1497 1.

1037 1.

1089 1.

1141 1.

1137 1.

1133 Now we repeat the process again, taking averages of consecutive terms in each column to produce the next column.

2.

3474 1.

5099 1.

2219 1.

1319 1.

1087 1.

1058 0.

6723 0.

9340 1.

0418 1.

0856 1.

1029 1.

1957 1.

1497 1.

1293 1.

1203 1.

1037 1.

1089 1.

1113 1.

1141 1.

1137 1.

1133 The terms across the top are the Euler approximations to the series.

The final term 1.

1058 is the Euler approximation based on all six terms.

And it is worse than what we get from simply taking partial sums!The exact value is 1.

11337… and so the sixth partial sum was accurate to four decimal places, but our clever method was only good to one decimal place.

What went wrong?!.Euler’s method is designed to speed up the convergence of slowly converging alternating series.

But our series converges pretty quickly because it has two terms in the denominator that grow like factorials.

When you apply acceleration when you don’t need to, you can make things worse.

But Euler’s method works well when it’s needed.

For example, let’s look at the slowly converging seriesπ = 4 – 4/3 + 4/5 – 4/7 + 4/9 – …Then we get the following array.

4.

0000 3.

3333 3.

200o 3.

1619 3.

1492 3.

1445 2.

6666 3.

0666 3.

1238 3.

1365 3.

1399 3.

4666 3.

1809 3.

1492 3.

1434 2.

8952 3.

1174 3.

1376 3.

3396 3.

1578 2.

9760 The sequence of partial sums is along the left side, and the series of Euler terms is across the top row.

Neither gives a great approximation to π, but the approximation using Euler’s acceleration method is significantly better.

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