Maths on a Mug #12



One of the more common things that I’ve been asked recently. Mostly this comes from people who have watched the incredibly successful Numperphile video, but the result was well known before then. There are a number of ways to ‘prove’ this result, giving rise to much internet angst, my favourite (and simplest) is that of Ramanujan:

\(\begin{align*}
c &= 1+2+3+4+5+6+\cdots\ \\
4c &= \quad\,\,\, 4\quad\,\,\ +8\quad\,\, +12 + \cdots\ \\
-3c &= 1-2+3-4+5-6 + \cdots\ \\
-3c &= \frac{1}{(1+1)^2}, \\
-3c &= \frac{1}{4},\\
c &= \frac{-1}{12}.
\end{align*}\)

Can you spot the mistake?

Fundamentally the flaws of these argument often arise from treating infinite sums like finite sums. In general associativity and commutativity do not hold for infinite series. As an example, take the series \(1-1+1-1+1-1+1-1+\cdots=0\), and then add some brackets:

\(\begin{align*}
(1-1)+(1-1)+(1-1)+(1-1)+\cdots=0, \\
1+(-1+1)+(-1+1)+(-1+1)+(-1+\cdots = 1.
\end{align*}\)

Which give you different answers!

For an excellent description of all the errors with the \(1+2+3+\cdots = \frac{-1}{12}\) proof see the excellent Plus magazine article.

This entry was posted in Maths on a Mug. Bookmark the permalink.

Leave a Reply

Your email address will not be published.