Show that $P_n = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots + \frac{(-1)^n}{n!}$ given $P_n = \frac{n-1}{n} P_{n-1} + \frac1n P_{n-2}$, $P_1 = 0$, $P_2 = \frac12$
I have absolutely no clue how to derive that. I tried expanding it by $P_n = \frac{n-1}{n} P_{n-1} + \frac1n P_{n-2} = \frac{n-1}{n}(\frac{n-2}{n-1}P_{n-2}+\frac{1}{n-1}P_{n-3}) + \frac1n (\frac{n-3}{n-2}P_{n-3} + \frac{1}{n-1}P_{n-4})$ but it's not helping I think. How should I turn this into a form with factorial?
On
See given that, $$ P_n = \sum_{k=2}^n \frac{(-1)^k}{k!} \tag{1} $$
we want to re-write $(1)$ in terms of $P_{n-1}$ and $P_{n-2}$ so, $$ P_n = \sum_{k=2}^{n-1}\frac{(-1)^k}{k!} + \frac{(-1)^n}{n!} $$ $$ P_n = \sum_{k=2}^{n-1}\frac{(-1)^k}{k!} - \frac{(-1)^{n-1}}{n(n-1)!} $$
Now see we're somewhat getting what we want to prove so,
$$P_n = \sum_{k=2}^{n-1}\frac{(-1)^k}{k!} - \frac{1}{n} \left( \frac{(-1)^{n-1}}{(n-1)!} \right) \tag{2}$$
See we can write
\begin{align*} \frac{(-1)^{n-1}}{(n-1)!} &= \sum_{k=2}^{n-1} \frac{(-1)^k}{k!} - \sum_{k=2}^{n-2} \frac{(-1)^k}{k!} \\ &= P_{n-1} - P_{n-2} \end{align*}
Just put it in $(2)$ and we're done.
\begin{align*} P_n &= \sum_{k=2}^{n-1}\frac{(-1)^k}{k!} - \frac{1}{n} \left( \sum_{k=2}^{n-1} \frac{(-1)^k}{k!} - \sum_{k=2}^{n-2} \frac{(-1)^k}{k!} \right) \\ &= \left(\frac{n-1}{n} \right) \sum_{k=2}^{n-1}\frac{(-1)^k}{k!} + \frac{1}{n} \sum_{k=2}^{n-2}\frac{(-1)^k}{k!} \\ P_n &= \left(\frac{n-1}{n} \right) P_{n-1} + \left( \frac{1}{n} \right) P_{n-2} \end{align*}
$$P_n = \frac{n-1}{n} P_{n-1} + \frac1n P_{n-2} \\ P_n - P_{n-1} = -\frac{1}{n}\left({P_{n-1}-P_{n-2}}\right) $$ Let : $V_n=P_n - P_{n-1}$
Therfore:
$$V_n=-\frac{1}{n}V_{n-1} \implies V_n=(-1)^n\frac{2}{n!}V_2$$ And :$V_2=P_2-P_1=\frac{1}{2}$
and therfore : $V_n=(-1)^n\frac{1}{n!}$
We have :$V_n=P_n-P_{n-1}=(-1)^n\frac{1}{n!}$
$$\sum^n_{k=2}(P_k-P_{k-1})=\sum^n_{k=2}\left({(-1)^n\frac{1}{n!}}\right)$$ this implies that:
$P_n-P_1=\sum^n_{k=2}\left({(-1)^n\frac{1}{n!}}\right)$
Q.E.D