Finding an explicit formula for $a_1 := 1, a_{n+1} := 1 + \sum_{i=1}^{n} ia_i$

Question

I need to find the formula for the general term of the recurrence relation

$$a_1 := 1, a_{n+1} := 1 + \sum_{i=1}^{n} ia_i$$

I predicted the general term is $a_n = n!$, and I've tried to prove it by strong induction on $n$. The base case is true because $a_1 := 1 = 1!$. Then, I assume that $a_i = i!$ for all $i$'s less than $n+1$ with the intention of establishing that $a_{n+1} = (n+1)!$. By definition,

$$a_{n+1} := 1 + \sum_{i=1}^{n} ia_i$$

Using the inductive hypothesis, we've got

$$a_{n+1} := 1 + \sum_{i=1}^{n-2} ia_i + (n-1)(n-1)! + n.n! = 1 + \sum_{i=1}^{n-2} ia_i - (n-1)! + (n+1)!$$

And if we expand this sum,

$$a_{n+1} := 1 + 1! - 2! + ... + (n+1)! = (n+1)!$$

Therefore $a_n = n!$. Is this rigorous enough? I don't know what else to say about the expansion of the sum.

Is there any other way to solve this exercise?, i.e., any general strategy to find the general term for recurrence relations that involve summation?

Answer 1

Were you expected to do it by induction? Because we have

\begin{eqnarray} a_{n+1} &=& 1 + \sum_{i=1}^{n} ia_i\\ a_{n} &=& 1 + \sum_{i=1}^{n-1} ia_i\\ a_{n+1}-a_n&=&na_n\\ a_{n+1}&=&na_n+a_n\\ a_{n+1}&=&(n+1)a_n \end{eqnarray}

So $a_{n+1}=(n+1)!$ and $a_1=1$.

Answer 2

I do not understand some calculations in your inductive step.

I would do it like this.

Inductive hypothesis: For arbitrary but fixed $n\in\mathbb{N}$ let be $a_n=n!$.

$n\mapsto n+1$

$a_{n+2}=1+\sum_{i=1}^{n+1} ia_i=\underbrace{\left(1+\sum_{i=1}^n ia_i\right)}_{=a_{n+1}}+(n+1)a_{n+1}\stackrel{I.H.}{=}(n+1)!+(n+1)(n+1)!=(n+1)!(1+n+1)=(n+1)!(n+2)=(n+2)!$

Answer 3

The recurrence is in fact:

$$\begin{align*} a_{n+1} &=1+\sum_{i=1}^n ia_i\\ &= \underbrace{1 + \sum_{i=1}^{n-1}ia_i}_{a_n} + na_n\\ &= a_n + na_n\\ &= (n+1)a_n \end{align*}$$

Then an induction to prove that $a_n = n!$ should be easy.

Answer 4

You could use the hypothesis to get rid of the sum. Let $a_{n+1}=g(n)+\sum_{i=1}^n f(i, a_i)$ the formula defining your succession, then $a_{n+1}=g(n)+\sum_{i=1}^n f(i, a_i)=$ $=g(n)-g(n-1)+g(n-1)+\sum_{i=1}^{n-1} f(i, a_i)+f(n, a_n)=$ $=g(n)-g(n-1)+a_n+f(n, a_n)$, which is in general easier to study.