Let $(a_n)_{n \geq 1}$ be an arithmetic progression of positive numbers. Compute $$ \Lambda =\lim_{n \to \infty} \frac{n (a_1 \cdots a_n)^{1/n}}{a_1 + \cdots + a_n}.$$
so first by AM-GM inequality I got $\Lambda \leq 1$ and equality holds when it is a constant progression, next I don't know how to proceed.
You can set $a_1=a,a_2-a_1=d$ so that $a_i=a+(i-1)d$ and then $$a_1+a_2+\dots+a_n=\frac{n(2a+(n-1)d) }{2}$$ and hence $$\lim_{n\to\infty} \frac{n(a_1a_2\dots a_n) ^{1/n}}{a_1+a_2+\dots+a_n}=2\lim_{n\to\infty}\frac{(a_1\dots a_n) ^{1/n}}{2a+(n-1)d}=\frac{2}{d}\lim_{n\to\infty}\frac{(a_1a_2\dots a_n) ^{1/n}}{n}\cdot\frac{n} {(2a/d)+n-1}$$ Now we can observe that if $b_n=(a_1\dots a_n) /n^n$ then $$\frac{b_{n+1}}{b_n}=\frac{a_{n+1}}{n+1}\left(1+\frac{1}{n}\right)^{-n}$$ and this tends to $d/e$. It follows that $b_n^{1/n}=(a_1\dots a_n) ^{1/n}/n\to d/e$ and hence desired limit is $2/e$.
The above works when $d>0$. For $d=0$ the limit is trivially equal to $1$. The case $d<0 $ is not allowed as $a_n>0$ for all $n$.
The general form of an arithmetic progression is $$ a_n = a_1 + (n-1) d $$ where $d$ is the difference between successive terms. By assumption, $a_1$ and $d$ are positive.
Substituting that into the limit, you obtain $$ \Lambda = \lim\limits_{n\rightarrow\infty} \frac{n\left(\prod\limits_{k=1}^n (a_1 + (k-1)d)\right)^{\frac1n}}{n a_1 + \sum\limits_{k=1}^n (k-1)d} $$ The sum $\sum\limits_{k=1}^n (k-1)$ is simply the $(n-1)$th triangular number, hence it equals $\frac{n(n-1)}2$. Therefore we have $$ \lim\limits_{n\rightarrow\infty} \frac{n\left(\prod\limits_{k=1}^n (a_1 + (k-1)d)\right)^{\frac1n}}{n a_1 + \sum\limits_{k=1}^n (k-1)d} =\lim\limits_{n\rightarrow\infty} \frac{n\left(\prod\limits_{k=1}^n (a_1 + (k-1)d)\right)^{\frac1n}}{n a_1 + d\frac{n(n-1)}2} = \lim\limits_{n\rightarrow\infty} \frac{\left(\prod\limits_{k=1}^n (a_1 + (k-1)d)\right)^{\frac1n}}{a_1 + \frac d 2 (n-1)} $$ The denominator is asymptotically $\frac d 2 n$, hence we have $$ \Lambda = \lim\limits_{n\rightarrow\infty} \frac{\left(\prod\limits_{k=1}^n (a_1 + (k-1)d)\right)^{\frac1n}}{\frac d 2 n} $$ The numerator looks a lot like $(n!)^{\frac1n}$, so if you're familiar with Stirling's approximation, you can probably see where to go next right away. Note that, by Stirling's approximation, for any $\alpha\in \mathbb{R}$ and $m\in\mathbb{Z}$, we have $$ \lim\limits_{n\rightarrow\infty} \frac{(\alpha \cdot (n-m)!)^{\frac1n}}{n} = \frac1e $$ Let's manipulate our equation so that it looks similar: \begin{eqnarray} \Lambda&=&\lim\limits_{n\rightarrow\infty} \frac{\left(\prod\limits_{k=1}^n (a_1 + (k-1)d)\right)^{\frac1n}}{\frac d 2 n} = 2\lim\limits_{n\rightarrow\infty} \left(\frac{\prod\limits_{k=1}^n (a_1 + (k-1)d)}{d^n n^n} \right)^{\frac1n} \\&=& 2\lim\limits_{n\rightarrow\infty} \prod\limits_{k=1}^n \left(\frac{\frac {a_1}d + (k-1)}{ n} \right)^{\frac1n} = 2\lim\limits_{n\rightarrow\infty} \prod\limits_{k=0}^{n-1} \left(\frac{\frac {a_1}d + k}{ n} \right)^{\frac1n} \end{eqnarray} Now the numerator looks even more like a factorial than before. Note that $$ k \le \frac{a_1}d + k \le \lceil\frac{a_1}d\rceil + k $$ Therefore $$ \frac{a_1}d (n-1)!\le\prod\limits_{k=0}^{n-1}(\frac {a_1}d + k)\le(\lceil\frac{a_1}d\rceil+n-1)!$$ so we only need to apply the squeeze theorem and we're done.
