Theorem.
Assume that $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are two sequences of real numbers such that $y_n$ is strictly increasing and $\lim_{n\to\infty}y_n=+\infty$. Then,
- (a) the following three inequalities hold:
\[
\liminf \frac{x_{n+1}-x_n}{y_{n+1}-y_n}\leq \liminf\frac{x_n}{y_n}\leq \limsup\frac{x_n}{y_n}\leq \limsup \frac{x_{n+1}-x_n}{y_{n+1}-y_n}.
\] - (b) Assume that $\displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=L$. Then the limit $\displaystyle \lim_{n\to\infty}\frac{x_n}{y_n}$ exists and is equal to $L$.
(b) follows from (a) so it is enough to prove (a).
The middle inequality is trivial because $\liminf$ is always not bigger than $\limsup$.
The rightmost inequality is analogous to the leftmost one, so we prove only that.
Choose the real number $r$ such that \[ \limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}\lt r. \] Then there exists $N$ such that for all $n\gt N$ we have \[ \frac{x_{n+1}-x_n}{y_{n+1}-y_n}\lt r. \] Let $m\in\mathbf{N}$ such that $m\gt N$. Then we have \begin{align*} x_{N+1}-x_N &\lt r(y_{N+1}-y_N)\\ x_{N+2}-x_{N+1} &\lt r(y_{N+2}-y_{N+1})\\ &\vdots\\ x_{m}-x_{m-1} &\lt r(y_{m}-y_{m-1}). \end{align*} Summing all of the previous inequalities we obtain \[ x_m-x_N\lt r(y_m-y_N), \] from which it follows \[ \frac{x_m}{y_m}-\frac{x_N}{y_m}\lt r-r\frac{y_N}{y_m}. \] Since $y_m\to\infty$ as $m\to\infty$ it follows that \[ \limsup\frac{x_m}{y_m}\lt r. \] Since this inequality holds for every $r\lt\displaystyle{\limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}}$ thus we obtain \[ \limsup\frac{x_m}{y_m}\leq\limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}. \qquad\blacksquare \]
The middle inequality is trivial because $\liminf$ is always not bigger than $\limsup$.
The rightmost inequality is analogous to the leftmost one, so we prove only that.
Choose the real number $r$ such that \[ \limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}\lt r. \] Then there exists $N$ such that for all $n\gt N$ we have \[ \frac{x_{n+1}-x_n}{y_{n+1}-y_n}\lt r. \] Let $m\in\mathbf{N}$ such that $m\gt N$. Then we have \begin{align*} x_{N+1}-x_N &\lt r(y_{N+1}-y_N)\\ x_{N+2}-x_{N+1} &\lt r(y_{N+2}-y_{N+1})\\ &\vdots\\ x_{m}-x_{m-1} &\lt r(y_{m}-y_{m-1}). \end{align*} Summing all of the previous inequalities we obtain \[ x_m-x_N\lt r(y_m-y_N), \] from which it follows \[ \frac{x_m}{y_m}-\frac{x_N}{y_m}\lt r-r\frac{y_N}{y_m}. \] Since $y_m\to\infty$ as $m\to\infty$ it follows that \[ \limsup\frac{x_m}{y_m}\lt r. \] Since this inequality holds for every $r\lt\displaystyle{\limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}}$ thus we obtain \[ \limsup\frac{x_m}{y_m}\leq\limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}. \qquad\blacksquare \]
Example 1.
\[
\lim_{n\to\infty}\frac{\displaystyle{\sum_{k=1}^n \frac{1}{k}}}{\log n}=1.
\]
\[
x_n:=\sum_{k=1}^n \frac{1}{k},
\]
\[
y_n:=\log n.
\]
Then
\[
\begin{align*}
\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n} &=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\log(n+1)-\log n} \\
&=\lim_{n\to\infty} \frac{1}{(n+1)\log\left(1+\frac{1}{n} \right)}\\
&=\lim_{n\to\infty} \frac{1}{\log\left(1+\frac{1}{n} \right)^{n+1}}\\
&=1,
\end{align*}
\]
where we used A remarkable sequence Theorem 3. $\quad\blacksquare$
Example 2.
Let $\alpha\gt -1$. Then
\[
\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^n k^{\alpha}}{n^{\alpha+1}}=\frac{1}{\alpha+1}.
\]
\[
x_n:=\sum_{k=1}^n k^{\alpha},
\]
\[
y_n:=n^{\alpha+1}.
\]
Then
\[
\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n} =\lim_{n\to\infty}\frac{(n+1)^{\alpha}}{(n+1)^{\alpha+1}-n^{\alpha+1}}.
\]
By Lagrange's mean value theorem we have
\[
(n+1)^{\alpha+1}-n^{\alpha+1}=(\alpha+1)(n+c_n)^{\alpha},
\]
where $0\lt c_n\lt 1$.
So it follows
\[
\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\frac{1}{\alpha+1}.\qquad\blacksquare
\]
