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Let $a_1=0,a_2=1,$ and $a_{n+2}=\dfrac{(n+2) a_{n+1}-a_n}{n+1}$. Prove that $\lim_{n\to \infty}a_n=e$.

I know that $\lim_{n\to\infty}\left(1+\frac1{2!}+\frac1{3!}+...+\frac1{n!}\right)=e$ and $a_n = \dfrac{(n) a_{n-1}-a_{n-2}}{n-1}$ from $a_{n+2}=\dfrac{(n+2) a_{n+1}-a_n}{n+1}$. However, I don't know how to link them together.

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    $\begingroup$ What do you see when you write out several terms of the sequence? Are there any clear patterns? $\endgroup$ – abiessu Sep 18 '13 at 17:16
  • $\begingroup$ It actually converges to $e-1$ $\endgroup$ – user84413 Jul 26 '16 at 23:23
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$$a_{n+2}-a_{n+1}=\dfrac{1}{n+1}(a_{n+1}-a_{n})=\cdots=\dfrac{1}{(n+1)!}(a_{2}-a_{1})=\dfrac{1}{(n+1)!}$$ so $$a_{n}=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots+\dfrac{1}{(n-1)!}(n\ge 2)$$

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