Sum of digits of a prime number in base 10

Question

Define $S(i)$ as the sum of the digits of the $i$-th prime number $p_i$. Furthermore, define $$T(i) := \frac{S(i)}{p_i}.$$

Is it true that $T(i) < T(i+1)$ infinitely many times?

Response to comment: I don't have much progression other than assuming by contradiction that there is a number $N$ for which $T(i) > T(i+1)$ for all $i>N$. Then I want to show that this isn't true, my first idea was using Bertrand & Dirichlet (Bertrand for bounding $p_i$ and $p_{i+1}$ to eachother, and Dirichlet for constructing a prime of the form ending on $00001$ or something to lower its digit sum).

I am fairly confident this statement is true, as surely you have primes of the form $10111101013201$ with a very low digit sum, and the next prime being of the form $10111101099999$ for example, with a much higher digit sum and therefore making our statement very probable.

I created this problem myself, and context was purely random. I wanted to create an olympiad style problem with digit sum of a prime, but this problem seems too hard for an olympiad. I am assuming we need strong tools.

Answer 1

I agree that strong tools are needed for this problem, but I think that such tools are available.

Katai proved in 1986 that the sum of digits of primes acts like a normal random variable in the limit with mean $\frac92\log N$ and variance $\frac{33}4\log N$ (a central limit theorem), which is what we'd expect if the digits of primes were distributed randomly. More precisely, Katai proved that for every real number $y$, $$ \lim_{N\to\infty} \frac1N \# \biggl\{ i\le N\colon T(i) \le \frac92\log N + y \sqrt{\frac{33}4\log N} \biggr\} = \frac1{\sqrt{2\pi}} \int_{-\infty}^y e^{-t^2/2}\,dt. $$

It's not hard to check that this is incompatible with a world where $T(i) \ge T(i+1)$ always past some point (it implies, for example, that when $N$ is large, there are always primes $p_i$ with $N\le i\le 2N$ for which $T(i) < \frac92\log N$ and also primes in that range for which $T(i) > \frac92\log N$).

One moral of this argument is that detecting inequalities between consecutive elements of a sequence seems like a delicate local problem, but often they can be deduced from coarser global results like average values, which are much more likely to be achievable.