All we did was to add a zero at the start so I hope we can agree that we haven't changed the sum at all. If we now write $S$ twice $$S=1-1+1-1+1-1+1-1+...$$ $$S=0+1-1+1-1+1-1+1-...$$ and add them together we get $$S+S=2S=(1+0)+(-1+1)+(1-1)+(-1+1)+...$$ Hence $2S$ should be equal to $1,$ which gives that $S$ should be equal to $\frac{1}{2}.$

Infinite but tame

As you can see, sums containing an infinite number of terms, known as infinite series, can challenge our understanding of very basic mathematical concepts such as addition and subtraction. Our next stop in our exploration of infinite series is the following geometric series:

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

There's a clever way of figuring out the value of this sum, using the diagram below

Take a square of side length $1$ and it divide it in half to get two rectangles, which both have area $\frac{1}{2}.$ Now divide one of those rectangles in half as shown, to get two squares of area $\frac{1}{4}.$ Divide one of those squares in half to get two rectangles of area $\frac{1}{8},$ and so on, ad infinitum. The total area (of the squares and rectangles we left undivided) is the same as the sum of all terms of the geometric series. Since that area is just the area of the large square, the geometric series appears to be equal to $1.$ And indeed, mathematicians agree with this statement. They say that the series converges to $1.$ Formally convergence is defined by looking at the sequence of partial sums: $$\begin{array}{rcl}{S}_1& =& \frac{1}{2}\\ S_2& =& \frac{1}{2}+\frac{1}{4}=\frac{3}{4}\\ S_3& =& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}\\ S_4& =& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{15}{16}\end{array}$$ and so on. The partial sums gives us a sequence of numbers, $\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},...$ which get closer and closer to $1;$ in fact they get arbitrarily close to $1$ as we include more and more terms. In general, when the sequence of partial sums of an infinite series converges on some limit number $a$ in this way, then we say that the infinite series converges to $a.$ Notice that this doesn't happen for the Grandi's series above. The partial sums are $$\begin{array}{rcl}{S}_1& =& 1\\ S_2& =& 1-1=0\\ S_3& =& 1-1+1=1\\ S_4& =& 1-1+1-1=0\end{array}$$ and so on. The partial sums eternally skip from 1 to 0 and back again. They don't converge to a limiting value, so there's no obvious way of assigning a value to the series.

Infinite and divergent

If this sounds hard to believe, here is a proof by contradiction of the divergence of the harmonic series. In a proof by contradiction we start by assuming that the opposite of what we are trying to prove is true and then show that this leads to a contradiction. If we are trying to prove statement $A$ is true, then we start by assuming the opposite: that not $A$ is true. If this assumption leads to a contradiction, then this implies that the not $A$ must be false, so our original statement $A$ must be true. In this case, we want to show that the harmonic series diverges, so we assume that the harmonic series converges to some positive value $H.$ We know that $$\frac{1}{3}>\frac{1}{4},$$ $$\frac{1}{5}>\frac{1}{6},$$ $$\frac{1}{7}>\frac{1}{8},$$ and so on. So if we replace the fractions with odd denominators by their consecutive even ones we find the following inequality $$H=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\frac{1}{8}+\frac{1}{8}+...$$ If we now combine the fractions with the same denominator we get $$H=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...>1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$$ On the right-hand side the harmonic series appears again, plus an extra term of $\frac{1}{2}.$ This proves that $$H>H+\frac{1}{2}!$$ As this contradiction followed from our assumption that the harmonic series converges, we can conclude that this assumption is false: the harmonic series does not converge. As the partial sums get larger and larger by smaller and smaller increments (we add a term of the form $\frac{1}{n}$ at every step) the only other possibility is that the sequence of partial sums grows beyond all bounds (rather than skipping around akin to Grandi's series). This proves that $H$ diverges.

Infinite and bizarre

Things get seriously bizarre when we examine the alternating harmonic series. Built from the harmonic series but with every other term negative, the alternating harmonic series is defined as follows:

$$S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...$$
If we now simplify the fractions so that all fractions with the even denominators are reduced and then combine the terms with the same denominator – and you can see that only fractions with odd denominators will be combined – we get $$2\ln 2=(2-1)-\frac{1}{2}+(\frac{2}{3}-\frac{1}{3})-\frac{1}{4}+(\frac{2}{5}-\frac{1}{5})-\frac{1}{6}...$$ Removing the brackets we have $$2\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...$$ The right-hand side is the alternating harmonic series, so we seem to have proved that $$2\ln 2=\ln 2$$ But, hang on, what? There must be a mistake in our working. I find it hard to believe, but in fact there is nothing wrong with our working. The paradox is explained by the fact that, when we rearrange the terms of the alternating harmonic series, the series converges to a different value. The right-hand side of expression (1) above can be written as $$2\ln 2=(2-1)-\frac{1}{2}+(\frac{2}{3}-\frac{1}{3})-\frac{1}{4}+(\frac{2}{5}-\frac{1}{5})-\frac{1}{6}...=2(1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}...).$$ The problem comes from the fact that while $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...$$ converges to $\ln 2,$ the rearranged series $$1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}...$$ converges to $$\ln 2+\frac{1}{2}\ln \left(\frac{1}{2}\right),$$ despite the fact the two series have exactly the same terms! Let’s see if we can make sense of this. In the classic formulation of the alternating harmonic series we start with $1$ and subtract $\frac{1}{2}$, then add $\frac{1}{3}$ and subtract $\frac{1}{4}$ and so on. For each added term we subtract exactly one term. In the second formulation of the alternating harmonic series, we subtract two terms for each term that we add. What do you think will happen? The graph below shows what happens to the partial sums as we add terms one at a time. It shows the first 25 partial sums. The green dots are the partial sums for the classic alternating harmonic series and the purple dots are the partial sums for our rearrangement of the series. The graph shows that the two series will converge to different values. It turns out that by rearranging the alternating harmonic series we can make it converge to any value we like. Try to work out for yourself what value $$S=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$$

converges to. See here for an answer.

You will be glad to know that, for a series with only positive terms, altering the order of the terms does not affect the sum.

To read more about infinite series see An infinite series of surprises.

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About the author

Luciano Rila works in the Department of Mathematics at University College London and is also an Area Coordinator for the Further Maths Support Programme. He is keen to promote the beauty of mathematics to young mathematicians and the general public. He also teaches static trapeze. You can follow him on Twitter @DrTrapezio