A generalisation of $\frac{(n!)^{\frac 1n}} n$?

Question

Limit of the sequence $\{ a_n \} $ where $a_n$ is defined as

$ a_n= \frac { \big( \Pi _{i=1}^ n i^{i^p}\big)^{\frac 1{n^{p+1} }}} {n^{\frac 1{p+1} }}$ where $p\ge 0$ is

(a) $ e^{ -\frac 1{(p+1)^2} } $

(b) $e^{-\frac 1{p^2} }$

(c) $0$

(d) Not defined.

My attempt :-

For $p=0$ we have $a_n=\frac {(n!)^{\frac 1n} }n$ and taking $b_n= \frac{n!}{n^n} $ , we can show $\frac{b_{n+1}}{b_n} \rightarrow \frac 1e $ and thus $b_n ^{1/n} =a_n \rightarrow \frac 1e$ as $n\rightarrow \infty $

This suggest option (a) is correct.

An attempt to prove (a)

$\ln a_n = \frac 1{n^{p+1}}\big( \sum_{i=1}^n i^p \ln i \big) -\frac {\ln n}{p+1} $

$\implies\ln a_n = \frac 1{n}\big( \sum_{i=1}^n \big(\frac{i}{n}\big)^p \{ \ln \frac in +\ln n \}\big)- \frac {\ln n}{p+1} $

$\implies\ln a_n = \frac 1{n}\sum_{i=1}^n \big(\frac{i}{n}\big)^p \ln \frac in +\frac {\ln n}{n} \sum_{i=1}^n \big(\frac{i}{n}\big)^p-\frac {\ln n}{p+1} $

Multiplying both sides by $\frac 1n$ and taking limit , we get .

$\lim_{n\rightarrow \infty} \ln (a_n ^{\frac 1n}) =0 $

Or $\lim_{n\rightarrow \infty} a_n^{\frac 1n} =1 $

I am stuck at here. How may I proceed?

Answer 1

The term $$\frac 1n\sum_{i=1}^n\left(\frac in\right)^p\ln\left(\frac in\right)$$ is a Riemann sum, which converges to $\int_0^1 t^p\ln tdt$ and can be computed via an integration by part. The term $$ \frac {\ln n}{n} \sum_{i=1}^n \left(\frac{i}{n}\right)^p-\frac {\ln n}{p+1} $$ goes to $0$, which can be seen by noticing that $\sum_{i=1}^n i^p=(p+1)^{-1}n^{p+1}+P(n)$, where $P(\cdot)$ is a polynomial of degree $p$.