Ne folosim de relația de recurență de la subpunctul b)
[tex]I_{n+1}+3I_n = \dfrac{1}{n+2}\Big|\cdot n \\ \\ nI_{n+1}+3nI_n = \dfrac{n}{n+2} \\ \\ \\ \lim\limits_{n\to +\infty} \Big(nI_{n+1}+3nI_{n}\Big) = \lim\limits_{n\to +\infty} \dfrac{n}{n+2} \\ \\ \\I_{n+1}\approx I_{n},\quad n\to +\infty \\ \\\\ \lim\limits_{n\to +\infty}(nI_{n}+3nI_{n}) = 1 \\ \\ \lim\limits_{n\to +\infty}(4nI_{n}) = 1 \\ \\ 4\cdot \lim\limits_{n\to +\infty}(nI_n) = 1 \\ \\ \\\Rightarrow \lim\limits_{n\to +\infty}(nI_{n})= \dfrac{1}{4}[/tex]
