A calculus of the absurd

3.2 "Telescoping" series

I think this is a further maths topic.

Sometimes, we have sums where everything cancels out nicely. A specific form of this is called a "telescoping" series, and the term refers to anything in the form

\begin{equation} \sum _{k=1}^{n} \sbrackets {f(k+a) - f(k)} \end{equation}

Here, \(a\) is a constant natural number (e.g. \(1, 2, 3, ...\)) and \(k\) is the index of summation.

The key idea is that eventually the \(-f(k)\) will "catch up" to the \(+f(k+a)\). If \(a\) were equal to \(1\), for example, we would have that

\begin{equation} \sum _{k=1}^{n} \sbrackets {f(k+1) - f(k)} = f(2) - f(1) + f(3) - f(2) + f(4) - f(3) + ... \end{equation}

A lot of terms are going to cancel here! When we add \(f(2)\) and \(-f(2)\) we get \(0\). The same thing happens for \(f(3)\) and \(-f(3)\), and so on. Every \(-f(k)\) term in our series cancels out the \(f(k)\) term from the previous term. This means that all the terms, except the first and the last terms are going to cancel each other out, leaving us with just

\begin{equation} - f(1) + f(n) = f(n) - f(1) \end{equation}

One way to visualise this it to rewrite the sum.

\begin{align} \sum _{k=1}^{n} \sbrackets {f(k+a) - f(k)} &= \sum _{k=1}^{n} \sbrackets {f(k+a)} - \sum _{k=1}^{n} \sbrackets {f(k)} \\ &= \sum _{k=a}^{n+a} \sbrackets {f(k)} - \sum _{k=1}^{n} \sbrackets {f(k)} \end{align}

We can then draw this geometrically

(-tikz- diagram)

The orange section in the middle represents all the common terms. When we subtract the two series from each other, we are left with the last \(a\) terms from the sum of all the \(+f(k+a)\) sum and the first \(a\) terms from the \(-f(k)\) sum.