A few days ago, the John D. Cook’s excellent [@CompSciFact](https://twitter.com/CompSciFact) put the following tweet in my Twitter feed.

This is one of my favorite data structure facts, which at first appears quite surprising, but after a bit of thought seems quite reasonable. The naive approach to building a binary heap from a list of elements, inserting each element and reheaping every time requires \(O(n \log n)\) time. With this complexity, we may as well just sort the list. Fortunately, as per the tweet, an \(O(n)\) algorithm for contructing the heap exists.

This post won’t discuss the algorithm directly (Wikipedia does a fairly good job). Rather, we’ll explore see how to evaluate the series \[ \begin{align*} \sum_{k = 1}^\infty \frac{k}{2^k}, \end{align*} \] which appears in the proof that the algorithm has linear time complexity (again, consult Wikipedia for details). We evaluate these series by clever reindexing.

The first step rather innocently factors out a factor of \(\frac{1}{2}\) as follows:

\[ \begin{align*} \sum_{k = 1}^\infty \frac{k}{2^k} & = \frac{1}{2} \sum_{k = 1}^\infty \frac{k}{2^{k - 1}} \end{align*} \]

We then reindex the right hand sum, so that \[ \begin{align*} \sum_{k = 1}^\infty \frac{k}{2^k} & = \frac{1}{2} \sum_{k = 1}^\infty \frac{k}{2^{k - 1}} = \frac{1}{2} \sum_{k = 0}^\infty \frac{k + 1}{2^k} = \frac{1}{2} \left( \sum_{k = 0}^\infty \frac{k}{2^k} + \sum_{k = 0}^\infty \frac{1}{2^k} \right). \end{align*} \]

When \(k = 0\),

\[ \begin{align*} \frac{k}{2^k} & = \frac{0}{2^0} = \frac{0}{1} = 0, \end{align*} \]

so the first \(k = 0\) term of the series is zero, and

\[ \begin{align*} \sum_{k = 0}^\infty \frac{k}{2^k} & = \sum_{k = 1}^\infty \frac{k}{2^k}. \end{align*} \]

Therefore, after reindexing, we get that

\[ \begin{align*} \sum_{k = 1}^\infty \frac{k}{2^k} & = \frac{1}{2} \sum_{k = 1}^\infty \frac{k}{2^{k - 1}} = \frac{1}{2} \sum_{k = 0}^\infty \frac{k + 1}{2^k} = \frac{1}{2} \left( \sum_{k = 1}^\infty \frac{k}{2^k} + \sum_{k = 0}^\infty \frac{1}{2^k} \right). \end{align*} \]

solving for the series in question, we get that

\[ \begin{align*} \sum_{k = 1}^\infty \frac{k}{2^k} & = 2 \sum_{k = 0}^\infty \frac{1}{2^k} = 2 \end{align*} \]

by the formula for summing a geometric series.

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