Inspired by Jason Dyer

Calculus Level 4

It is known that the harmonic series

$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +\cdots$

diverges. What about if we only took those terms without a 1 in the denominator; does it still diverge?

$\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{20} + \frac{1}{22} + \frac{1}{23} + \cdots$

Inspiration .

It converges It still diverges

1 solution

Brian Charlesworth
Jan 30, 2017

This series is $K_{1}$ of the set of Kempner series , all of which converge.

While a somewhat surprising result, intuitively, as the number of digits in an integer increases, the greater the likelihood that the digit $1$ will be present, so the number of elements in $K_{1}$ "thins out" in comparison to the harmonic series, enough so to achieve convergence. From a probability standpoint, as the the number of digits in an integer goes to infinity, the probability that such a number contains a $1$ goes to $1$ , and thus the probability that its reciprocal is included in $K_{1}$ goes to $0$ .

Using that idea, can you provide an upper bound for the sum?

Chung Kevin - 4 years, 4 months ago

Inspired by Jason Dyer

1 solution

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