The Harmonic Series

Here's an interesting problem: an ant traverses a circle with a circumference of one metre at a rate of one centimetre per second. After each second, the circumference of the circle increases by one metre. Will the ant ever return to its starting point?

Let's consider the matter. In the first second, the ant traverses 1% of the circumference; in the second second, it traverses 1/2%; in the third second, 1/3% and so on. The cumulative distance covered is given by the sum of:

1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 

This is the harmonic series and it is divergent, meaning that the sum is continuously increasing and never reaches any upper limit. Contrast this to a convergent geometric series, the sum of whose terms approaches ever closer to 2 as more terms are added:

1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

In the case of the harmonic series, the sum will certainly surpass the 100% required for the ant to return to its starting point. It may take a long time but it will happen. In fact OEIS A004080 tells us that it would take the following number of seconds:

15092688622113788323693563264538101449859497

This of course represents more than 4.78 x 10^23 years which is far in excess of the age of the universe! The following screenshot is taken from the WolframAlpha article about the harmonic series.

Taken from http://mathworld.wolfram.com/HarmonicSeries.html

This Numberphile YouTube video was the inspiration for this post.

The Riemann Prime Counting Function

I've been looking back over some of the entries in my Evernote notebook titled "Your Days Are Numbered" that used to be automatically posted to a blog that wanted to charge for the privilege of accessing my posts. However, the entries in the notebook remain and one that caught my attention related to the prime number 24121. OEIS A226473 states:

a(n) is the first prime index where the gap between R(n), Riemann's prime counting function, and Pi(n), the exact prime counting function, is greater than n.

The first terms in the sequence are:

109, 556, 1327, 3296, 5380, 10343, 11767, 19202, 19361, 19371, 24121, 42863 

The example given for 109 is that RiemannR(109) = 27.4664... and PrimePi(109) = 29 give the first gap greater than 1, hence a(1) = 109. 

24121 is the 11th member of the sequence. Pi(24121) = 2686 and RiemannR(24121) = 2674.9424 ... and 11 subtracted from 2686 is 2675. Thus the gap between PrimePi and RiemannR is more than 11 and so a(11) = 24121.

There is a discussion of the Riemann prime counting function on WolframAlpha.