## PROBLEM 3

Let the sequence $\left(x_{n}\right)_{n \geq 1}$ be defined by $\ln \left(\frac{x_{n-1}-1}{x_{n-1}}+\frac{1}{x_{n}}\right)=\frac{1}{n}, n \geq 2, x_{1}=1$. Calculate $\lim _{n \rightarrow \infty} \frac{1}{x_{n} \ln (n+1)}$.
Solution: The given relation is equivalent to $1-\frac{1}{x_{n-1}}+\frac{1}{x_{n}}=e^{\frac{1}{n}}$, from which

$$
\frac{1}{x_{n}}-\frac{1}{x_{n-1}}=e^{\frac{1}{n}}-1, \quad n \geq 2
$$

Summing the first $n-1$ equalities, we obtain:

$$
\begin{aligned}
& \frac{1}{x_{n}}-\frac{1}{x_{n-1}}+\frac{1}{x_{n-1}}-\frac{1}{x_{n-2}}+\ldots+\frac{1}{x_{2}}-\frac{1}{x_{1}}=e^{\frac{1}{n}}-1+e^{\frac{1}{n-1}}-1+\ldots+e^{\frac{1}{2}}-1 \\
& \frac{1}{x_{n}}-\frac{1}{x_{1}}=e^{\frac{1}{2}}+e^{\frac{1}{3}} \ldots+e^{\frac{1}{n}}-(n-1), \text{ hence } \frac{1}{x_{n}}=e^{\frac{1}{2}}+e^{\frac{1}{3}} \ldots+e^{\frac{1}{n}}-n+2
\end{aligned}
$$

Then $\lim _{n \rightarrow \infty} \frac{1}{x_{n} \ln (n+1)}=\lim _{n \rightarrow \infty} \frac{e^{\frac{1}{2}}+e^{\frac{1}{3}} \ldots+e^{\frac{1}{n}}-n+2}{\ln (n+1)}=\mathrm{L}$.

The sequence $a_{n}=\ln (n+1), n \geq 1$ is strictly increasing and unbounded above, so the Stolz - Cesàro theorem can be applied: $\mathrm{L}=\lim _{n \rightarrow \infty} \frac{e^{\frac{1}{n+1}}-1}{\ln (n+2)-\ln (n+1)}=$.

$$
=\lim _{n \rightarrow \infty} \frac{e^{\frac{1}{n+1}}-1}{\frac{1}{n+1}} \cdot \frac{1}{(n+1) \ln \frac{n+2}{n+1}}=\lim _{n \rightarrow \infty} \frac{1}{\ln \left(1+\frac{1}{n+1}\right)^{n+1}}=\frac{1}{e} \ldots \ldots \ldots \ldots \ldots . . . . . . .2 \mathrm{p}
$$