Counter-intuitions of divergent series

A common mistake when it comes to understanding series, motivated by highschool-esque and contest mathematics type thought process, is to think that the behaviour of the series is dominated by the first (finitely many) terms. For example, we think of the series 1 + 2 + 3 + \cdots as ‘divergent’ because the first few terms seem to add up really quickly.

This intuition is challenged with the classic `smallest’ (note: there is no such thing as an actual smallest divergent series) divergent series: the harmonic series

\displaystyle \sum_{n=1}^\infty \frac{1}{n}.

It is challenging because it just `barely’ diverges. If we replace the exponent 1 of n by anything bigger, we will end up with a convergent series. Further, it doesn’t look divergent because the first few terms don’t add up very quickly.

The following really challenges our intuition: if we delete every positive integer n that contains the number 9 in their usual decimal representation, and then add up their reciprocals, we get something convergent! This seems ludicrous because of we try to add up the first 100 positive integers, sans those that need to be omitted, it seems to barely make a difference. However, if we think of the `later’ parts of the series, the argument becomes clear. For each positive integer n, let us consider the integers 10^n \leq k < 10^{n+1}. There are 8 admissible choices (namely 1,2,3,4,5,6,7,8) for the first digit, and 9 choices for each subsequent digit, for a total of 8 \cdot 9^n choices. Each denominator is of size at least 10^n, hence

\displaystyle \sum_{\substack{10^n \leq k < 10^{n+1} \\ k \text{ does not contain } 9 \text{ in its decimal representation}}} \frac{1}{k} < \frac{8 \cdot 9^n}{10^n}.

Let \sideset{}{'}\sum denote summation over positive integers that do not contain 9 in their decimal representation. Then, it follows that

\displaystyle \sideset{}{'} \sum_{m \geq 10} \frac{1}{m} \leq \sum_{n=1}^\infty \sideset{}{'} \sum_{10^n \leq k < 10^{n+1}} \frac{1}{k} \leq \sum_{n=1}^\infty \frac{8 \cdot 9^n}{10^n} < \infty.

Therefore, it is the tail of the series that determines its behaviour with respect to convergence, not the front part! This is a particularly vexing thing to emphasize to students who are trained to think in the fashion (and only in the fashion) “do a few computations, make a conjecture, then prove conjecture using contradiction or induction”.


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