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Let $a_n$ be a recursive sequence with initial values $a_1=2$ and $a_2=\frac{3}{2}$ such that

$$a_{n+2}=a_n-na_{n+1}+\frac{1}{(n+2)!}$$

Find the explicit formula of $a_n$.

I observed five first terms : $ a_1=2 ,a_2=\frac{3}{2},a_3=\frac{4}{3!}, a_4=\frac{5}{4!},a_5=\frac{6}{5!},...$

So I guessed the explicit sequence is $a_n=\frac{n+1}{n!}$, then prove it by mathematical induction.

Basis step : $a_1=\frac{2}{1!}=2$. true

Inductive step : Assume for some $n$ i.e. $a_n=\frac{n+1}{n!}$ is true, then prove $a_{n+1}=\frac{n+2}{(n+1)!}$ is also true.

$$\begin{align}a_{n+1}&=a_{n-1}-(n-1)a_n+\frac{1}{(n+1)!} \\&=\frac{n}{(n-1)!}-\frac{(n-1)(n+1)}{(n)!}+\frac{1}{(n+1)!}\\&=\frac{1}{(n+1)!}\cdot (n^2(n+1)-(n-1)(n+1)^2+1) \\&=\frac{n+2}{(n+1)!} \end{align}$$

I could prove the explicit formula using induction, but I would like to prove using another technique. All help would be appreciated.

  • 0
    again: generating functions – 2017-02-13

1 Answers 1

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You can show that the sequence $a_n = \frac{n+1}{n!}$ satisfies the equation $$a_{n+2}=a_n-na_{n+1} + \frac{1}{(n+2)!}$$

i.e. you show that $$\frac{n+3}{(n+2)!} = \frac{n+1}{n!} - n\cdot \frac{n+2}{(n+1)!} + \frac{1}{(n+2)!}$$ is always true.

It's simple to show that the sequence satisfies the two initial values, so $a_n$ clearly satisfies all the required equations.

Together, this means that $a_n$ is one of the possible sequences, defined by the recursive relation. However, since the recursive relation is limiting enough so that only one possible sequence can satisfy it (technically, this claim can be proved by induction, but it's more or less obvious), $a_n$ must be the solution.