Question

Find the value of $\frac{1}{1+}\frac{2}{2+}\frac{3}{3+}\cdots$.

Answer 1

The law of formation holds for $p_n$ and $q_n$; let us take $u_n$ to denote either of them; then $u_n=n{u}_{n-1}+n{u}_{n-2}$,
or $u_n-(n+1)u_{n-1}=-(u_{n-1}-n{u}_{n-2})$.
Similarly, $u_{n-1}-n{u}_{n-2}=-(u_{n-2}-\overline{n-1}{u}_{n-3})$.
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$u_3-4u_2=-(u_2-3u_1)$;
whence by multiplication, we obtain
$u_n-(n+1)u_{n-1}={(-1)}^{n-2}(u_2-3u_1)$.
The first two convergents are $\frac{1}{1}$, $\frac{2}{4}$; hence
$p_n-(n+1)p_{n-1}={(-1)}^{n-1}$, $q_n-(n+1)q_{n-1}={(-1)}^{n-2}$.
Thus $\frac{p_n}{(n+1)!}-\frac{p_{n-1}}{n!}=\frac{{(-1)}^{n-1}}{(n+1)!}$, $\frac{q_n}{(n+1)!}-\frac{q_{n-1}}{n!}=\frac{{(-1)}^{n-2}}{(n+1)!}$,
$\frac{p_{n-1}}{n!}-\frac{p_{n-2}}{(n-1)!}=\frac{{(-1)}^{n-2}}{n!}$, $\frac{q_{n-1}}{n!}-\frac{q_{n-2}}{(n-1)!}=\frac{{(-1)}^{n-3}}{n!}$,
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$\frac{p_2}{3!}-\frac{p_1}{2!}=-\frac{1}{3!}$, $\frac{q_2}{3!}-\frac{q_1}{2!}=\frac{1}{3!}$,
$\frac{p_1}{2!}=\frac{1}{2!}$, $\frac{q_1}{2!}=\frac{1}{2}=1-\frac{1}{2!}$;
whence, by addition
$\frac{p_n}{(n+1)!}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots +\frac{{(-1)}^{n-1}}{(n+1)!}$;
$\frac{q_n}{(n+1)!}=1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\cdots +\frac{{(-1)}^{n-2}}{(n+1)!}$.
By making $n$ infinite, we obtain
$\underset{}{lim}\frac{p_n}{q_n}=\frac{1}{e}÷(1-\frac{1}{e})=\frac{1}{e-1}$,
which is therefore the value of the given expression.