Prove that $$\lim_{n\rightarrow\infty}\frac{f(n+1)}{n!}=e\tag{1}$$where $$f(n+1)=n(1+f(n))$$
The recurrence relation of $n!$ is $a_n=na_{n-1}$ or $a_{n+1}=(n+1)a_n$. I thought of making a new recurrence relation but we swap the $n$ and the $a_n$. Then, I noticed that although $f(n)<n!$, $f(n)>(n-1)!$ when $n>2$. At first, I thought the limit of their quotient would be $2$, but I found that $\frac{f(8)}{7!}\approx2.718$, so there is a very small chance that $(1)$ is not true.
According to Wolfram Alpha, $f(n+1)=c\Gamma(n+1)+en\Gamma(n,1)$, where $$\Gamma(s,x)=\int_x^\infty t^{s-1}e^{-t}dt$$is the incomplete gamma function. Setting $c=0$, our limit becomes $$e\lim_{n\rightarrow\infty}\frac{\Gamma(n,1)}{\Gamma(n)}=e\lim_{n\rightarrow\infty}\left(1-\frac{\gamma(n,1)}{\Gamma(n)}\right)=e$$Where $\Gamma(n)=\Gamma(n,x)+\gamma(n,x)$. The last equality is obtained by L'hopital's rule. My question here is either proving $(1)$ in a completely different way or proving the closed form that Wolfram Alpha gives (the proof I give of $(1)$ is technically not valid unless there is proof for that closed form).
Apparently, we have $$\frac{f(n+1)}{n!}=\sum_{k=0}^{n-1}\frac1{k!}$$So proving this obviously proves $(1)$
Best Answer
Note that, according to the given functional equation, $$\dfrac{f(n+1)}{n!}-\dfrac{f(n)}{(n-1)!}=\dfrac{1}{(n-1)!}.$$ Now telescope both sides and solve for $f(n).$