## PROBLEM 4

Let the sequence $\left(a_{n}\right)_{n \geq 1}$ be defined by $a_{1}=1$ and $a_{n+1}=\frac{a_{n}}{1+n a_{n}}, n \geq 1$. Calculate

$$
\lim _{n \rightarrow \infty} \frac{1}{n^{4}}\left(\frac{1}{a_{1}}+\frac{2}{a_{2}}+\ldots+\frac{n}{a_{n}}\right)
$$
The sequence $\left(a_{n}\right)_{n \geq 1}$ has strictly positive terms. $a_{n}>0 \Rightarrow\left(a_{n}\right)_{n \geq 1}$ is a sequence bounded below $\frac{a_{n+1}}{a_{n}}=\frac{1}{1+n a_{n}}0$, then $\lim _{n \rightarrow \infty}\left(1+n a_{n}\right)=\infty \Rightarrow \lim _{n \rightarrow \infty} \frac{a_{n}}{1+n a_{n}}=0 \quad \Rightarrow l=0$, false.

Therefore, $\lim _{n \rightarrow \infty} a_{n}=0$.

We apply the Stolz-Cesaro lemma successively:

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1}{n^{4}}\left(\frac{1}{a_{1}}+\frac{2}{a_{2}}+\ldots+\frac{n}{a_{n}}\right)=\lim _{n \rightarrow \infty} \frac{\frac{n+1}{a_{n+1}}}{(n+1)^{4}-n^{4}}=\lim _{n \rightarrow \infty} \frac{(n+1) \frac{1+n a_{n}}{a_{n}}}{(n+1)^{2}-n^{2}\left\lceil(n+1)^{2}+n^{2}\right]}= \\
& =\lim _{n \rightarrow \infty} \frac{n+1}{2 n+1} \cdot \lim _{n \rightarrow \infty} \frac{\frac{1}{a_{n}}+n}{2 n^{2}+2 n+1}=\frac{1}{2} \lim _{n \rightarrow \infty} \frac{n^{2}}{2 n^{2}+2 n+1} \cdot \lim _{n \rightarrow \infty} \frac{\frac{1}{a_{n}}+n}{n^{2}}=\quad \ldots \ldots \ldots \ldots . .2 \text { points } \\
& \stackrel{c-s}{=} \frac{1}{4} \lim _{n \rightarrow \infty} \frac{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}+1}{(n+1)^{2}-n^{2}}=\frac{1}{4} \lim _{n \rightarrow \infty} \frac{\frac{1+n a_{n}}{a_{n}}-\frac{1}{a_{n}}+1}{2 n+1}=\frac{1}{4} \lim _{n \rightarrow \infty} \frac{n+1}{2 n+1}=\frac{1}{8} . \quad \ldots \ldots \ldots \ldots \ldots \ldots . .2 \text { points }
\end{aligned}
$$