How to find $\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)$? 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.. 
$$\lim_{n\rightarrow \infty} (0) = 0$$
 A: $$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)=\lim_{n \rightarrow +\infty } \left(\sqrt[m]{{n}^{m}\prod_{i=1}^{m}\left(1+\frac{a_i}n\right)}-n\right)=\lim_{n \rightarrow +\infty } \left(n \cdot \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-n\right)=\lim_{n \rightarrow +\infty }\frac{ \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-1}{\frac{1}{n}}=$$$$=\lim_{n \rightarrow +\infty }\frac{ \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-1}{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)-1}\cdot\frac{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)-1}{\frac{1}{n}}=$$$$=\frac{1}{m}\cdot\lim_{n \rightarrow +\infty }\frac{1+\frac{a_1+a_2+...a_m}{n}+\frac{a_1a_2+a_2a_3+...+a_{m-1}a_m}{n^2}+...-1}{\frac{1}{n}}=\frac{a_1+a_2+...a_m}{m}.$$
We applied $$\lim_{x_n \rightarrow 0 }\frac{(1+x_n)^r-1}{x_n}=r.$$
A: $\newcommand{\+}{^{\dagger}}%
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$\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}
$$\color{#0000ff}{\large%
\lim_{n \to \infty}\bracks{\root[m]{\prod_{i = 1}^{m}\pars{n + a_{i}}} - n}}
=
\ol{a} = \color{#0000ff}{\large{1 \over m}\sum_{i = 1}^{m}a_{i}}
$$
A: 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
$\lim_{n \to \infty} P_n = 1$.
