Posted by Jason Polak on 24. December 2017 · Write a comment · Categories: analysis

Series hold endless fascination. To converge or not to converge? That is the question.

Let’s take the series $1 + 1/2 + 1/3 + \cdots$. It’s called the harmonic series, and it diverges. That’s because it is greater than the series
$$1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + \cdots = 1 + 1/2 + \cdots$$

The harmonic series diverges rather slowly, however. In fact, by comparing with the integral of $1/x$, we see that $1 + 1/2 + \cdots + 1/N$ can never be more than $\log(N) + 1$. For example, the sum of the first two hundred million terms is about 19.691044.

On the other hand, the sum of the reciprocals of the squares $1 + 1/4 + 1/9 + 1/16 + \cdots$ converges, which can be seen by comparing it to the integral of $1/x^2$ from one to infinity. In fact, $1 + 1/4 + 1/9 + \cdots = \pi^2/6$ as proved by Leonhard Euler. Here’s a question for you: does $\sum_{n=1}^\infty 1/n^s$ converge or diverge for $1 < s < 2$?

Even though the sum of reciprocals of squares converges, the sum of reciprocals of primes $1/2 + 1/3 + 1/5 + 1/7 + 1/11 + \cdots$ diverges. One could say by the convergence-divergence metric that the primes are more numerous than the squares. Also, in a similar vein to the harmonic series, the sum $1/2 + 1/3 + 1/5 + \cdots + 1/p – \log\log p$ is bounded.

Let’s go back to that harmonic series: $1 + 1/2 + 1/3 + 1/4 + \cdots$. Take this series, and delete every term whose denominator has the digit “9” somewhere in its decimal expansion. The resulting series converges! Can you prove it?

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