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 $\frac{{{1}+\sqrt{{{2}}}+{\sqrt[{{3}}]{{{3}}}}+\ldots+{\sqrt[{{n}}]{{{n}}}}}}{{n}}\to{1}$ .