Intellectual battle with limits begins!

$Let\quad x=\cfrac { 1 }{ t } \quad so,\\ \\ { \left\{ (x+{ a }_{ 1 })(x+{ a }_{ 2 })....(x+{ a }_{ n }) \right\} }^{ \cfrac { 1 }{ n } }=\cfrac { { { \left\{ (1+{ a }_{ 1 }t)(1+{ a }_{ 2 }t)...(1+{ a }_{ n }t) \right\} }^{ \cfrac { 1 }{ n } } } }{ t } \\ \\ Let\quad y={ \left\{ (1+{ a }_{ 1 }t)(1+{ a }_{ 2 }t)...(1+{ a }_{ n }t) \right\} }^{ \cfrac { 1 }{ n } }\\ \\ \Rightarrow L={ { lim }_{ t\rightarrow 0 }\cfrac { y-1 }{ t } =lim }_{ t\rightarrow 0 }\cfrac { { \left\{ (1+{ a }_{ 1 }t)(1+{ a }_{ 2 }t)...(1+{ a }_{ n }t) \right\} }^{ \cfrac { 1 }{ n } }-1 }{ t } \quad \quad (\cfrac { 0 }{ 0 } form)\\ \\ \Rightarrow L={ lim }_{ t\rightarrow 0 }{ y }^{ ' }={ lim }_{ t\rightarrow 0 }\cfrac { y }{ n } \left\{ \cfrac { { a }_{ 1 } }{ 1+{ a }_{ 1 }t } +\cfrac { { a }_{ 2 } }{ 1+{ a }_{ 2 }t } +...+\cfrac { { a }_{ n } }{ 1+{ a }_{ n }t } \right\} \\ \\ \Rightarrow nL={ lim }_{ t\rightarrow 0 }{ \left\{ (1+{ a }_{ 1 }t)(1+{ a }_{ 2 }t)...(1+{ a }_{ n }t) \right\} }^{ \cfrac { 1 }{ n } }\left\{ \cfrac { { a }_{ 1 } }{ 1+{ a }_{ 1 }t } +\cfrac { { a }_{ 2 } }{ 1+{ a }_{ 2 }t } +...+\cfrac { { a }_{ n } }{ 1+{ a }_{ n }t } \right\} \\ \\ \Rightarrow nL=\sum _{ r=1 }^{ n }{ { a }_{ r } } \\ \\ \Rightarrow { lim }_{ n\rightarrow \infty }nL=\sum _{ r=1 }^{ \infty }{ { a }_{ r } } =\cfrac { { a }_{ 1 } }{ 1-\cfrac { { a }_{ 2 } }{ { a }_{ 1 } } } =\cfrac { \cfrac { 1 }{ 2 } }{ 1-\cfrac { 1 }{ 2 } } =1$

In this General Case :

$L=\displaystyle \lim_{x \to \infty}((x+a_1)(x+a_2)••(x+a_n))^{\frac{1}{n}}-x= \dfrac{a_1+a_2+a_3+..+a_n}{n}$

$\therefore \displaystyle \lim_{n\to\infty} n \times \left( \dfrac{a_1+a_2+a_3+....+a_n}{n} \right)=a_1+a_2+a_3+....+a_n$

which is an infinite G.P.

$\implies Required \ \ Limit=\left(\dfrac{a_1}{1-\frac{a_2}{a_1}}\right)=\boxed{1}$

Consider $x > 0$ and $n > 0$ .

By the AM-GM-HM inequality, $\frac{(x+a_1)+ \cdots +(x+a_n)}{n}-x \geq ((x+a_1) \cdots (x+a_n))^{\frac{1}{n}}-x \geq \frac{n}{\frac{1}{x+a_1}+ \cdots + \frac{1}{x+a_n}} - x$ .

We have $\frac{(x+a_1)+ \cdots +(x+a_n)}{n}-x = \frac{a_1 + \cdots + a_n}{n}$ .

Also,

$\begin{aligned} \frac{n}{\frac{1}{x+a_1}+ \cdots + \frac{1}{x+a_n}} - x & = \frac{n - (\frac{x}{x+a_1}+ \cdots + \frac{x}{x+a_n})}{\frac{1}{x+a_1}+ \cdots + \frac{1}{x+a_n}} \\ & = \frac{\frac{a_1}{x+a_1}+ \cdots + \frac{a_n}{x+a_n}}{\frac{1}{x+a_1}+ \cdots + \frac{1}{x+a_n}} \\ &= \frac{(a_1+ \cdots + a_n)x^{n-1} + O(x^{n-2})}{nx^{n-1} + O(x^{n-2})} \end{aligned}$

where $O(x^{n-2})$ here represents a polynomial of degree at most $n-2$ . (In the last step, we multiplied both the numerator and denominator by $(x+a_1) \cdots (x+a_n)$ .)

This gives us $\frac{a_1 + \cdots + a_n}{n} \geq ((x+a_1) \cdots (x+a_n))^{\frac{1}{n}}-x \geq \frac{(a_1+ \cdots + a_n)x^{n-1} + O(x^{n-2})}{nx^{n-1} + O(x^{n-2})}$ for all $x > 0$ and $n > 0$ . But as $x \to \infty$ , $\frac{(a_1+ \cdots + a_n)x^{n-1} + O(x^{n-2})}{nx^{n-1} + O(x^{n-2})} \to \frac{a_1 + \cdots + a_n}{n}$ . So by sandwiching, $((x+a_1) \cdots (x+a_n))^{\frac{1}{n}}-x \to \frac{a_1 + \cdots + a_n}{n}$ as $x \to \infty$ , i.e. $L$ exists and equals $\frac{a_1 + \cdots + a_n}{n}$ .

Then $nL = a_1 + \cdots + a_n$ , and since the sequence $(a_n)$ is a geometric progression with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$ , we know that $\lim_{n\to\infty} (nL)$ exists and equals $\frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$ .

Let's look at some element

$x+a_i$

You can write it as

$x (1+\frac{1}{x 2^i})$

Using Bernoulli inequality that in limes becomes equality you have it's equal to

$x (1+\frac{1}{x})^{\frac{1}{2^i}}$

Taking product of all of these we have it's equal to

$x^n (1+\frac{1}{x})^{\sum\limits_{i=1}^n \frac{1}{2^i}}$

The following sum is equal to $1$

And now

$nL=n((x^n(1+\frac{1}{x}))^{\frac{1}{n}}-x)$

$nL=n (x(1+\frac{1}{nx})-x)=1$

The solution can be done without de l'Hospital's formula for "0/0"-fractions, using a variant of the geometric sum. For simplicity, define

$\begin{aligned} P_n(x)&:=\prod_{k=1}^n(x+a_k)=x^n+x^{n-1}\left(\sum_{k=1}^na_k\right)+\ldots=:x^n+Q_{n-1}(x)\\\\ \tilde{L}_n&:=P_n(x)^{1/n}-x \end{aligned}$

with polynomial degrees given by the index. Now to the variant of the geometric sum. Assuming we don't divide by zero, we have

$\begin{aligned} a-b &= \frac{a^n-b^n}{ \sum_{k=0}^{n-1}a^kb^{n-1-k} },\qquad n\in\mathbb{N},\quad a,b\in\mathbb{R} \end{aligned}$

Now consider the problem. We get rid of the n-th root and the subtraction by applying the above to $\tilde{L}_n$ :

$\begin{aligned} \tilde{L}_n&=\frac{P_n(x)-x^n}{ \sum_{k=0}^{n-1}P_n(x)^{\frac{k}{n}}x^{n-1-k} }=\frac{ Q_{n-1}(x) }{ \sum_{k=0}^{n-1}P_n(x)^\frac{k}{n}x^{n-1-k} }=\frac{ Q_{n-1}(x)/x^{n-1} }{ \sum_{k=0}^{n-1}\left( P_n(x)/x^n \right)^{\frac{k}{n}} }\qquad\left|\cdot\frac{x^{-n+1}}{x^{-n+1}}\right.\\\\ \Rightarrow L_n&=\lim_{x\rightarrow\infty}\tilde{L}_n=\frac{\sum_{k=1}^na_k}{\sum_{k=0}^{n-1}1^{\frac{k}{n}}}=\frac{1}{n}\sum_{k=1}^na_k,\\\\ \lim_{n\rightarrow\infty}(nL_n)&=\lim_{n\rightarrow\infty}\sum_{k=1}^n a_k=\sum_{k=1}^\infty2^{-k}=1\qquad\text{q.e.d.} \end{aligned}$

During the first limit $x\rightarrow\infty$ the first coefficient of $Q_{n-1}(x)$ remains in the numerator, while the denominator was normalized to 1.

We can rewrite the limit as: $L_n=\lim_{x\rightarrow\infty} x \left[\left(\dfrac{x^n+O(x^{n-1})}{x^n}\right)^{\frac{1}{n}}-1\right]=\lim_{x\rightarrow\infty} \dfrac{\left[1+O(\frac{1}{x})\right]^{\frac{1}{n}}-1}{\frac{1}{x}}$

Where $O(x^{n-1})$ contains the rest of the terms of the polynomial and $O(\frac{1}{x})$ contains these terms divided by $x^n$

Since it is of the $\frac{0}{0}$ form we can apply L'Hopital's Rule. $L_n=\lim_{x\rightarrow\infty} \dfrac{\frac{1}{n}\left[1+O(\frac{1}{x})\right]^{\frac{1}{n}-1}\left[O(\frac{1}{x^2})\right]}{-\frac{1}{x^2}}$

Now we can multiply the denominator and numerator by $-x^2$ which will perfectly cancel the $-\frac{1}{x^2}$ in $O(\frac{1}{x^2})$ . The coefficient of the $-\dfrac{1}{x^2}$ term is $\displaystyle\sum_{j=1}^{n}\dfrac{1}{2^i}$ and when multiplying $-x^2$ by the rest of the terms we are left with a bunch of $x$ terms in the denominators:

$\implies L_n=\dfrac{1}{n}\sum_{j=1}^{n}\dfrac{1}{2^j}$ $\lim_{n\rightarrow\infty} nL_n=\sum_{j=1}^{\infty}\dfrac{1}{2^j}=\boxed{1}$