Problem 1. Let $\left(a_{n}\right)_{n \geq 1}$ be a sequence of strictly positive real numbers, with the property that the sequence $\left(a_{n+1}-a_{n}\right)_{n \geq 1}$ is convergent, with a non-zero limit. Calculate the limit

$$
\lim _{n \rightarrow \infty}\left(\frac{a_{n+1}}{a_{n}}\right)^{n}
$$

Mathematical Gazette
## Solution and Scoring:

Let $L=\lim _{n \rightarrow \infty}\left(a_{n+1}-a_{n}\right)$. If $L>0$, then there exists $n_{0} \in \mathbb{N}^{*}$ such that $a_{n+1}-a_{n}>\frac{L}{2}, \forall n>n_{0}$. Then, for $n>n_{0}-\frac{2 a_{n_{0}}}{L}$, we obtain $a_{n}>0$..................................................................................................................... According to the Stolz-Cesàro theorem, $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=L \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$......................................... Since $L>0$, there exists $n_{0} \in \mathbb{N}^{*}$, with the property that $a_{n+1}-a_{n}>\frac{L}{2}, \forall n>n_{0}$. It follows that $a_{n}>a_{n_{0}}+\left(n-n_{0}\right) \cdot \frac{L}{2}, \forall n>n_{0}$, from which $\lim _{n \rightarrow \infty} a_{n}=\infty \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ For $n>n_{0}$ we have $a_{n+1}-a_{n}>0$ and

$$
\left(\frac{a_{n+1}}{a_{n}}\right)^{n}=\left[\left(1+\frac{a_{n+1}-a_{n}}{a_{n}}\right)^{\frac{a_{n}}{a_{n+1}-a_{n}}}\right]^{\frac{n}{a_{n}} \cdot\left(a_{n+1}-a_{n}\right)}
$$

$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$
Since $\lim _{n \rightarrow \infty} \frac{a_{n+1}-a_{n}}{a_{n}}=0$, we obtain

$$
\lim _{n \rightarrow \infty}\left(\frac{a_{n+1}}{a_{n}}\right)^{n}=e^{\frac{1}{L} \cdot L}=e
$$

$2 \mathrm{p}$ $\qquad$