Is $\lim_{n\to\infty} S_n$ irrational?

Question

Q: Is $$\lim_{n\to\infty} S_n$$ irrational?

Here $p_n$ is the nth prime.

$$S_1= \frac{.9}{p_1}+\frac{.8}{p_2}+\frac{.7}{p_3}+...+\frac{.1}{p_9} $$

$$ S_2=\frac{.99}{p_1}+\frac{.98}{p_2}+\frac{.97}{p_3}+...+\frac{.01}{p_{99}} $$

$$ S_3= \frac{.999}{p_1}+\frac{.998}{p_2}+\frac{.997}{p_3}+...+\frac{.001}{p_{999}} $$

$$ ... $$

I think it is, because if you keep adding zeros to make the numerators smaller while retaining equal space among successive numerators, the period of the resultant rational sum extends to infinity, which would mean in limit form, $S$ would approach an irrational number. How can I formally prove this?

Answer 1

The sum $\sum_{k=1}^\infty {1\over p_k}$ of the reciprocals of the primes diverges, see here. Since $$S_n>{1\over2}\sum_{k=1}^{10^n/2}{1\over p_k}\qquad(n\geq1)$$ it follows that $\lim_{n\to\infty}S_n=\infty$.

Answer 2

Essentially you're averaging $j/p_j$ for $j$ from $1$ to $10^n$. Since $j/p_j \to 0$ as $j \to \infty$, your limit will be $0$.

EDIT: For the new sequence, by the monotone convergence theorem for sums your limit is $\sum_{n=1}^\infty 1/p_n$, which is $\infty$.

Answer 3

Not an answer, but a strong suggestion

Suppose you instead look at your series not just for powers of 10, but others as well, so that your first sum gets renamed $$ S_{10} = \frac{1}{10} \frac{1}{p_1} + \frac{2}{10} \frac{1}{p_2} + \ldots + \frac{10-1}{10} \frac{1}{p_{10-1}}. $$ Then the general form is $$ S_k = \frac{1}{k} \frac{1}{p_1} + \ldots + \frac{k-1}{k} \frac{1}{p_{k-1}} = \frac{1}{k} \left(1 \cdot \frac{1}{p_1} + \ldots + (k-1) \cdot \frac{1}{p_{k-1}} \right ). $$ The density of primes is roughly $\frac{\log n}{n}$, so the $n$th prime is roughly $n \log n$. So let's replace all those terms with estimates to get $$ S_k \approx \frac{1}{k} \left(1 \cdot \frac{1}{p_1} + 2 \cdot \frac{1}{2 \ln 2}\ldots + (k-1) \cdot \frac{1}{(k-1) \ln (k-1)} \right ). $$ Lots of terms cancel, and we get

$$ S_k \approx \frac{1}{k} \left(1 \cdot \frac{1}{p_1} + \frac{1}{ \ln 2}\ldots + \frac{1}{\ln (k-1)} \right ). $$ Now the inside sum consists of about $k$ terms whose average value is about $\frac{1}{ln \frac{k}{2}}$ (I admit this is pretty crude as an estimate!), and when we multiply by $1/k$, we find that $S_k$ is approximately (up to a multiplicative constant) $\frac{1}{\ln k}$, which tends to zero as $k \to \infty$.

In short: the sequence appears to approach a very rational number, namely zero.