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I think it is zero; $$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)$$ we can make that steps: $$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{{n}^{m}\prod_{i=1}^{m}\left(1+\frac{a_i}n\right)}-n\right)$$ and then: $$\lim_{n \rightarrow +\infty } \left(n \cdot \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-n\right)$$ each bracket aspires to unit so we have: $$\lim_{n\rightarrow \infty} (n-n)$$ and.. $$\lim_{n\rightarrow \infty} (0) = 0$$

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  • $\begingroup$ Some of your LaTeX delimiters didn't match up. I fixed them but I hope the square roots and parentheses are in the places you wanted them to be ... $\endgroup$ Dec 3, 2013 at 23:35
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    $\begingroup$ What are the $a_i$ in your limit? $\endgroup$ Dec 3, 2013 at 23:36
  • $\begingroup$ And what is $n?$ $\endgroup$
    – Igor Rivin
    Dec 3, 2013 at 23:41
  • $\begingroup$ Not sure what the last two equalities are supposed to mean, but from $\lim_{n \rightarrow +\infty } \left(n \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-n\right)$ you can only get a $0 \cdot \infty$ type thing, after factoring $n$. But every such indeterminate form can be transformed into a more managable one, for instance by writing $n=\frac{1}{1/n}$. $\endgroup$ Dec 3, 2013 at 23:47

3 Answers 3

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$$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)=\lim_{n \rightarrow +\infty } \left(\sqrt[m]{{n}^{m}\prod_{i=1}^{m}\left(1+\frac{a_i}n\right)}-n\right)=\lim_{n \rightarrow +\infty } \left(n \cdot \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-n\right)=\lim_{n \rightarrow +\infty }\frac{ \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-1}{\frac{1}{n}}=$$$$=\lim_{n \rightarrow +\infty }\frac{ \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-1}{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)-1}\cdot\frac{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)-1}{\frac{1}{n}}=$$$$=\frac{1}{m}\cdot\lim_{n \rightarrow +\infty }\frac{1+\frac{a_1+a_2+...a_m}{n}+\frac{a_1a_2+a_2a_3+...+a_{m-1}a_m}{n^2}+...-1}{\frac{1}{n}}=\frac{a_1+a_2+...a_m}{m}.$$ We applied $$\lim_{x_n \rightarrow 0 }\frac{(1+x_n)^r-1}{x_n}=r.$$

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\color{#ff0000}{\large\mbox{With}\ n \gg 1}$: \begin{align} \ln\pars{\root[m]{\prod_{i = 1}^{m}\pars{n + a_{i}}}} &= {1 \over m}\sum_{i = 1}^{m}\ln\pars{n + a_{i}} = {1 \over m}\sum_{i = 1}^{m}\bracks{\ln\pars{n} + \ln\pars{1 + {a_{i} \over n}}} \sim \ln\pars{n} + {1 \over m}\sum_{i = 1}^{m}{a_{i} \over n} \\[3mm]&\mbox{With}\ \ol{a} \equiv {1 \over m}\sum_{i = 1}^{m}a_{i}\ \mbox{we'll get} \\ \root[m]{\prod_{i = 1}^{m}\pars{n + a_{i}}} - n &\sim \exp\pars{\ln\pars{n} + {\ol{a} \over n}} - n = n\pars{\expo{\ol{a}/n} - 1} \sim n\bracks{\pars{1 + {\ol{a} \over n}} - 1} = \ol{a} \end{align} $$\color{#0000ff}{\large% \lim_{n \to \infty}\bracks{\root[m]{\prod_{i = 1}^{m}\pars{n + a_{i}}} - n}} = \ol{a} = \color{#0000ff}{\large{1 \over m}\sum_{i = 1}^{m}a_{i}} $$

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The hard part is showing that $ P_n =\sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)} \to 1$ as $n \to \infty$.

If $n > km\max(|a_i|)$, $1+\frac{{a}_{i}}{n} < 1+\frac1{km} $ so $P_n <\sqrt[m]{\prod_{i=1}^{m}\left(1+\frac1{km}\right)} < 1+\frac1{km} $.

By choosing $k$ large enough, $P_n$ can be made as close to $1$ as wanted, showing that $\lim_{n \to \infty} P_n = 1$.

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