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 1.

fredoniahead

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.

I already received one answer, but haven’t analyzed it for correctness yet. Shown below.

Xantos C. Guin

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.

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About misha

Imagine a story that one can't believe. Hi. Life changes here. Small things only.
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