Evaluating π and e with infinite series | Approximations of π | Approximations of e
There are many infinite series that can help us evaluate important mathematical constants. For example, consider the series
\sum_{k=1}^{\infty}\frac{1}{(k-1)!}Written out term by term, this series is
\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...=1+1+\frac{1}{2}+\frac{1}{6}+...If we use a calculator to work out the first few partial sums of this series, we get
(1, 2, 2·5, 2·66667, 2·70833, 2·71667, 2·71806, . . .),
where we have written down some of the terms to five decimal places.
Now you might have noticed that this sequence of partial sums seems to be getting closer and closer to the number e, which is 2·71828 to five decimal places. In fact it can be shown that the partial sums do tend to e. So working out the partial sums of this series is a useful way of calculating e to a large number of decimal places
Now let us look at the infinite series
\sum_{k=1}^{\infty}(-1)^{k+1}\frac{4}{(2k-1)!}
For this series, we need to recall the meaning of the power (−1)^{k+1}. If k is odd then k+1 is even, and so (−1)^{k+1} = 1. On the other hand, if k is even then k + 1 is odd, and so (−1)^{k+1} = −1. We can now write out the series term by term as
\sum_{k=1}^{\infty}(-1)^{k+1}\frac{4}{(2k-1)!}=4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-...
Again we can use a calculator to work out the first few partial sums of this series. We get
(4, 2·6667, 3·4667, 2·8952, 3·3397, 2·9760, 3·2837, . . .)
where we have written down some of the terms to four decimal places.
This sequence of partial sums looks like it might be getting close to some number just greater
than 3. In fact it can be shown that the partial sums tend to π, which is 3·1416 to four decimal
places.
If we keep on calculating the partial sums for this series, we will eventually get
The value of π to several decimal places.