Solving $\begin{cases} T_{0}=5 \\ 2T_{n}=nT_{n-1} + 3n! \end{cases}$ using method of summation factors

Question

I need to solve the following recurrence relation using "method of summation factors".

$\begin{cases} T_{0}=5 \quad \quad \quad \quad \quad\quad\quad \iff n=0\\ 2T_{n}=nT_{n-1} + 3n! \quad \quad \iff n > 0\end{cases}$

I found this describing what method of summation factor is:

Let there be given recursive formula:

$$a_{n}T_{n} = b_{n}T_{n-1} + c_{n}$$

where:

$$a_{n} \neq 0 \iff n \geq 0$$ $$b_{n} \neq 0 \iff n > 0$$

then the solution $T_{n}$ of the recursive formula is defined as:

$$T_{n} = \frac{1}{S_{n}a_{n}}(S_{1}b_{1}T_{0} + \sum_{k = 1}^{n} S_{k}c_{k})$$

where:

$$S_{n} = \frac{a_{1}a_{2}...a_{n-2}a_{n-1}}{b_{2}b_{3}...b_{n-1}b_{n}} \quad \quad \land \quad \quad S_{1} = 1$$

---

My attempt at $\begin{cases} T_{0}=5 \quad \quad \quad \quad \quad\quad\quad \iff n=0\\ 2T_{n}=nT_{n-1} + 3n! \quad \quad \iff n > 0\end{cases}$

$S_{n} = \frac{\overbrace{2*2*2*2*...*2}^{n - 1 \text{ times}}}{\underbrace{2*3*4*...*n}^{n - 1 \text{ times}}} = \frac{2^{n-1}}{n!}$

$a_{n} = 2, \quad \quad b_{n} = n, \quad \quad c_{n} = 3n!, \quad \quad S_{1} = 1$

$$T_{n} = \frac{1}{S_{n}a_{n}}(S_{1}b_{1}T_{0} + \sum_{k = 1}^{n} S_{k}c_{k})$$

$$T_{n} = \frac{1}{\frac{2^{n-1}}{n!} 2} \Bigg( 1*n*5 + \sum_{k = 1}^{n} \frac{2^{k-1}}{k!} * 3k!\Bigg)$$

$$T_{n} = \frac{n!}{2^n}\Bigg( 5n + \sum_{k = 1}^{n} \frac{2^{k-1}}{k!} * 3k! \Bigg)$$

$$T_{n} = \frac{n!}{2^n}\Bigg( 5n + 3 \sum_{k = 1}^{n} 2^{k-1} \Bigg)$$

---

Not sure how to proceed. My questions:

1. Does the $S_{1} = 1$ and $S_{n} = \frac{a_{1}a_{2}...a_{n-2}a_{n-1}}{b_{2}b_{3}...b_{n-1}b_{n}}$ always apply? Is this somewhat considered "a constant", a rule?

2. Until that point, does my solution look good?

3. Lastly, perhaps you could recommend me some good learning materials? I need to know how to solve such kind of recurrence relation, and nothing beyond that really. I have just started learning series.

Answer 1

Just for your curiosity.

Looking at your work and related comments, without using the method of summation factors, I worked the more general problem of $$ T_n=a\, n \,T_{n-1}+b\, n!\qquad \text{with} \qquad T_0=c$$ The problem can simplify using Pochhammer symbols and making $$T_n=U_n\, (1)_n\implies U_n=a\, U_{n-1}+b\qquad \text{with} \qquad U_0=c$$ Now, using $$U_n=V_n+k \implies V_n=a\, V_{n-1}+b+(a-1)k$$ Let $k=\frac b {1-a}$ to get a simple geometric progression.

Back to $T_n$ $$T_n=\frac{a^n ((a-1) c+b)-b}{a-1}\, (1)_n $$