Stolz Theorem
To find limits of indeterminate expressions $$$\frac{{{x}_{{n}}}}{{{y}_{{n}}}}$$$ of type $$$\frac{{\infty}}{{\infty}}$$$ often can be useful following theorem.
Stolz Theorem. Suppose that sequence $$${y}_{{n}}\to+\infty$$$ and starting from some number with increasing of $$${n}$$$ also increases $$${y}_{{n}}$$$ (in other words if $$${m}>{n}$$$ then $$${y}_{{m}}>{y}_{{n}}$$$). Then $$$\lim\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\lim\frac{{{x}_{{n}}-{x}_{{{n}-{1}}}}}{{{y}_{{n}}-{y}_{{{n}-{1}}}}}$$$ if limit of the expression on the right side exists (finite or infinite).
Example 1. Find $$$\lim\frac{{{{a}}^{{n}}}}{{n}}$$$ where $$${a}>{1}$$$.
Let $$${x}_{{n}}={{a}}^{{n}}$$$ and $$${y}_{{n}}={n}$$$ then $$${y}_{{n}}$$$ is increasing and $$${y}_{{n}}\to\infty$$$ therefore we can apply Stolz Theorem.
$$$\lim\frac{{{{a}}^{{n}}}}{{n}}=\lim\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\lim\frac{{{x}_{{n}}-{x}_{{{n}-{1}}}}}{{{y}_{{n}}-{y}_{{{n}-{1}}}}}=\lim\frac{{{{a}}^{{n}}-{{a}}^{{{n}-{1}}}}}{{{n}-{\left({n}-{1}\right)}}}=\lim{\left({{a}}^{{n}}-{{a}}^{{{n}-{1}}}\right)}=$$$
$$$=\lim{{a}}^{{n}}{\left({1}-\frac{{1}}{{a}}\right)}=+\infty$$$.
Example 2. If $$$\lim{a}_{{n}}={A}$$$ find $$$\lim\frac{{{a}_{{1}}+{a}_{{2}}+..+{a}_{{n}}}}{{n}}$$$ (arithmetic mean of first $$${n}$$$ values of sequence $$${a}_{{n}}$$$).
Let $$${x}_{{n}}={a}_{{1}}+{a}_{{2}}+\ldots+{a}_{{n}}$$$ and $$${y}_{{n}}={n}$$$ then $$${y}_{{n}}$$$ is increasing and $$${y}_{{n}}\to\infty$$$ therefore we can apply Stolz Theorem.
$$$\lim\frac{{{a}_{{1}}+{a}_{{2}}+\ldots+{a}_{{n}}}}{{n}}=\lim\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\lim\frac{{{x}_{{n}}-{x}_{{{n}-{1}}}}}{{{y}_{{n}}-{y}_{{{n}-{1}}}}}=$$$
$$$=\lim\frac{{{\left({a}_{{1}}+{a}_{{2}}+\ldots+{a}_{{n}}\right)}-{\left({a}_{{1}}+{a}_{{2}}+\ldots.+{a}_{{{n}-{1}}}\right)}}}{{{n}-{\left({n}-{1}\right)}}}=\lim\frac{{{a}_{{n}}}}{{1}}=\lim{a}_{{n}}={A}$$$.
For example, since $$${\sqrt[{{n}}]{{{n}}}}\to{1}$$$ then $$$\frac{{{1}+\sqrt{{{2}}}+{\sqrt[{{3}}]{{{3}}}}+\ldots+{\sqrt[{{n}}]{{{n}}}}}}{{n}}\to{1}$$$.