Precise calculation of infinite sum

Question

How to precisely calculate $\sum_{k = 0}^\infty \frac{(-1)^k}{2^k}$ ? And which algorithm can be used for the related problems?

Answer 1

This is a geometric series.

Use the formula:

$$\sum_{k=0}^{\infty} r^k =\frac{1}{1-r}$$

where $|r|<1$.

Answer 2

There is a faster way.

Call your sum as $S$:

$$S = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \ldots$$

Collecting the second term:

$$S = 1 - \frac{1}{2}\left(1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \ldots \right)$$

What you have inside the bracket is nothing but $S$:

$$S = 1 - \frac{1}{2}S$$ hence

$$S + \frac{1}{2}S = 1$$

$$\frac{3}{2}S = 1$$

And finally

$$\boxed{{S} = \frac{2}{3}}$$

Answer 3

For related problems, where the series does not have a closed form, one can take into account that the series is alternating (its terms alternate the sign from $+$ to $-$) in order to produce an approximate result.

For a general alternating series $$ \sum_{k=0}^{\infty}(-1)^{k}a_{k}% $$ where $a_{k}>0$ for all $k$ and the sequence $(a_{k})_{k\geq0}$ is decreasing, denote by $S_{n}$ the sum of the first $n+1$ terms ($n\geq0$) and by $S$ the sum of the series $$ S_{n}=\sum_{k=0}^{n}(-1)^{k}a_{k} $$ $$ S=\sum_{k=0}^{\infty}(-1)^{k}a_{k}. $$ It is easy to check that $$ S_{0}>S_{2}>S_{4}>\ldots>S_{2n}>\ldots>S>\ldots>S_{2n+1}>\ldots>S_{5} >S_{3}>S_{1}. $$

This means that $S$ can be approximated with desired precision by increasing $n$ accordingly.