## Adding next integers vs adding next primes, and the AIR project

$\lim\limits_{n \rightarrow \infty}{{(1+{{1}\over{n}})(2+{{1}\over{n}})(3+{{1}\over{n}})..(n+{{1}\over{n}})}\over{n!}}$

Solutions:

dr Sonnhard Graubner

$\frac{\prod_{i=1}^n\left(i+\frac{1}{n}\right)}{n!}=\frac{\left(\frac{n^2+1}{n}\right)!}{n!\left(\frac{1}{n}!\right)}$. The searched limit is .

Look at the identity:

$\Gamma(z) = \frac{\Gamma(z+n)}{z(z+1)(z+2)\cdot\cdot\cdot (z+n-1)}$, remembering that $\Gamma(n+1)=n!$. Let $z=1+\frac{1}{n}$

JoeBlow

It’s equal to:

$\prod_{k=1}^n \left(1+\frac{1}{kn}\right)=1+\frac{1}{n}\sum_{k=1}^n \frac{1}{k}+O(n^{-1})=1+\frac{\log n}{n}+O(n^{-1})$.

WWW

Another approach: Apply $\ln$ to get $\sum_{k=1}^{n}\ln(k+1/n)-\sum_{k=1}^{n}\ln k =\sum_{k=1}^{n}\ln(1+1/(kn))$. Now $0<\ln (1+h) < h$ for $h>0$, so the last sum lies between $0$ and $\frac {\sum_{k=1}^{n}1/k}{n}.$ This $\to 0$ (Cesaro-Stolz or the well known fact that $\sum_{k=1}^{n}1/k \sim \ln n$). Thus the original limit is $1.$

Indeed,from the very definition, as JoeBlow mentioned. What about the following?

R-limit = $\lim\limits_{n \rightarrow \infty}{{(p_1+{{1}\over{n}})(p_2+{{1}\over{n}})(p_3+{{1}\over{n}})..(p_n+{{1}\over{n}})}\over{n!}}$, where $p_i$ is prime for $i \in N$.

Let $A_n = \dfrac{1}{n!}\prod_{k=1}^{n}(p_k + \tfrac{1}{n})$ and $B_n = \dfrac{1}{n!}\prod_{k=1}^{n}p_k$.
Clearly, $0 < B_n < A_n$ for all $n \in \mathbb{N}$. Also, $\dfrac{B_{n}}{B_{n-1}} = \dfrac{p_n}{n} \ge 2$ for all integers $n \ge 5$.
Hence, $\lim_{n \to \infty} B_n = \infty$, and thus, $\lim_{n \to \infty} A_n = \infty$ as well.