Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions

Solution 1:

Exponential generating function of sequence $\{a_n\}$ is $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$. Rewriting the recurrence equation as $$ \frac{a_{n+1}}{n+1} = \frac{2}{3} a_n + \frac{5}{3} n! $$ Now multiplying both sides by $\frac{x^n}{n!}$ and using recurrence relation for factorial: $$ a_{n+1} \frac{x^n}{(n+1)!} = \frac{2}{3} a_n \frac{x^n}{n!} + \frac{5}{3} x^n $$ Summing from $n=0$ to infinity: $$ \frac{1}{x} \sum_{n=1}^\infty a_n \frac{x^n}{n!} = \frac{2}{3} \sum_{n=0}^\infty a_n \frac{x_n}{n!} + \frac{5}{3} \sum_{n=0}^\infty x^n $$ or $$ \frac{1}{x} \left( f(x) - a_0 \right) = \frac{2}{3} f(x) + \frac{5}{3} \frac{1}{1-x} $$ Solving for $f(x)$ we readily get: $$ f(x) = \frac{5}{1-x} = \sum_{n=0}^\infty (5 \cdot n!) \frac{x^n}{n!} $$ Thus the solution is $a_n = 5 \cdot n!$.

Solution 2:

Generating functions are definitely overkill here: considering $$b_n=\frac{a_n}{n!},$$ one sees that the recursion on $(a_n)_n$ translates as $$b_{n+1}=\frac23b_n+\frac53.$$ This is an affine recursion hence one knows that to center the recursion at its fixed point, if such a fixed point exists, will make it linear. Here the fixed point solves $$b=\frac23b+\frac53,$$ that is, $b=5$. And, surprise, one gets the linear relation $$b_{n+1}-5=\frac23(b_n-5),$$ for every $n\geqslant0$.

Iterating this is trivial and yields $$b_n-5=\left(\frac23\right)^n(b_0-5),$$ that is, $$\frac{a_n}{n!}-5=\left(\frac23\right)^n(a_0-5).$$ Since one assumes that $a_0=5$, the RHS is zero hence, for every $n\geqslant0$, $$a_n=5\cdot n!.$$