Limit $\lim\limits_{x \to +\infty} \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ =?

Question

What is the limit of $ f(x) = \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ when $x$ goes to plus infinity? The number $n$ is fixed and $a_i$ are some constants.

All I know is the answer: $(a_1 + a_2 + ... + a_n) / n$

I figured out that I can limit the root from above with the arithmetic mean. And this gives me some nice expression as an upper limit. But OK... this only proves the limit is $\leq (a_1 + a_2 + ... + a_n) / n$.

But how do I limit it from below with the same value. I tried using the harmonic mean but it seems it does not lead me to anything nice. So... any ideas?

Answer 1

Notice that $$(x+a_1)(x+a_2)\cdot (x+a_n)=x^n+ b x^{n-1}+\mathcal{O}(x^{n-2})$$ Where $b=a_1+a_2+\dots +a_n$, and $$\sqrt[n]{x^n+bx^{n-1}+\mathcal{O}(x^{n-2})}=\left(x^n+bx^{n-1}+\mathcal{O}(x^{n-2})\right)^{1/n}=x \left(1+bx^{-1} +\mathcal{O}(x^{-2})\right)^{1/n}$$ Now, we can use the first order Taylor approximation of $(1+x)^q$ around $x=0$ (which is $1+q x+\mathcal{O}(x^2)$) to get

$$x \left(1+b +\mathcal{O}(x^{2-n})\right)^{1/n}= x\left(1+\frac{b}{n x} +\mathcal{O}(x^{-2})\right)=x-\frac{b}{n } +\mathcal{O}(x^{-1}). $$

From this, we can see that the limit is $\frac{b}{n}$.

A more elementary solution: Let $f(x)=\sqrt[n]{(x+a_1)(x+a_2)\cdot (x+a_n)}$, note that

$$\lim_{x\to \infty} \frac{f(x)}{x}=\lim_{x\to \infty} \sqrt[n]{(1+a_1x^{-1})(1+a_2x^{-1})\cdot (1+a_nx^{-1})}=1\tag{1}$$

Note that

$$f(x)-x=\frac{f(x)^n-x^n}{\sum_{k=0}^{n-1} f(x)^{n-1-k} x^{k}}= \frac{f(x)^n-x^n}{f(x)^{n-1}+f(x)^{n-2}x+\dots+f(x)x^{n-2}+x^{n-1}}$$

The numerator is a polynomial of degree $n-1$ with a leading coefficient of $a_1+a_2+\dots+a_n$, the denominator is composed of the sum of $n$ terms $f(x)^{n-1-k}x^k$. Divide both by $x^{n-1}$ to get

\begin{align*}\lim_{x\to \infty}f(x)-x&=\lim_{x\to \infty} \frac{f(x)^n-x^n}{x^{n-1}} \cdot \frac{1}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1}\\ &=(a_1+a_2+\dots+a_n)\cdot \frac{1}{1+1+\dots+1} \\ & =\frac{a_1+a_2+\dots+a_n}{n}\end{align*}

Answer 2

Using $$ a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ ab^{n-2}+b^{n-1})$$ one has \begin{eqnarray} &&\lim\limits_{x \to +\infty} \sqrt[n]{(x+a_1)(x+a_2)\cdots(x+a_n)} - x\\ &=&\lim\limits_{x \to +\infty} \frac{(x+a_1)(x+a_2)\cdots(x+a_n) - x^n}{\sqrt[n]{(x+a_1)(x+a_2)\cdots(x+a_n)}^{n-1} + \sqrt[n]{(x+a_1)(x+a_2)\cdots(x+a_n)}^{n-2}x+\cdots+\sqrt[n]{(x+a_1)(x+a_2)\cdots(x+a_n)}x^{n-2}+x^{n-1}}\\ &=&\lim\limits_{x \to +\infty} \frac{(a_1+a_2+\cdots+a_n)+o(\frac1x)}{n+o(\frac1x)}\\ &=&\frac{a_1+a_2+\cdots+ a_n}{n}. \end{eqnarray}

Answer 3

Note that for any $a,b \in \mathbb{C}$, we have $$a^n-b^n = (a-b)\sum_{i=0}^{n-1} a^i b^{n-1-i}$$

Hence, we can rewrite the expression as follows \begin{align*} \sqrt[n]{(x+a_1)\cdots(x+a_n)} - x &= \frac{(x+a_1)\cdots(x+a_n)-x^n}{\sum_{i=0}^{n-1} (\sqrt[n]{(x+a_1)\cdots(x+a_n)})^i x^{n-1-i}} \\ \text{divide by } x^{n-1} \text{ above and below} \\ &= \frac{a_1 + a_2 + \ldots + a_n + \mathcal{O}(x^-1)}{\sum_{i=0}^{n-1} (\sqrt[n]{(x+a_1)\cdots(x+a_n)})^i x^{-i}} \\ &= \frac{a_1 + a_2 + \ldots + a_n + \mathcal{O}(x^-1)}{\sum_{i=0}^{n-1} (\sqrt[n]{\frac{(x+a_1)\cdots(x+a_n)}{x^n}})^i} \\ &= \frac{a_1 + a_2 + \ldots + a_n + \mathcal{O}(x^-1)}{\sum_{i=0}^{n-1} (\sqrt[n]{1+\mathcal{O}(x^{-1})})^i} \end{align*}

Now, taking a limit as $x \to \infty$, we get the desired result.