Proof by Induction for Factorial with Sum

Question

So, the problem is about solving the following statement through induction:

$$\frac{1}{n+1}\sum_{i=1}^n \left(\frac{1}{n} + i^2(i-1)!\right) = n!, n \in\mathbb{N}^+$$

What I've done so far:

Beginning (for $n=1$): $$\frac{1}{2}(1+1)=1=1!$$ (shortened it for the question)

Induction Step:

Show $\frac{1}{n+2} * \sum_{i=1}^{n+1}(\frac{1}{n+1}+i^2(i-1)!) = (n+1)!$

What I got so far:

$$ \begin{aligned} \frac{1}{n+2}\sum_{i=1}^{n+1}\left(\frac{1}{n+1}+i^2(i-1)!\right) &= \frac{1}{n+2}\left(\sum_{i=1}^{n}\left(\frac{1}{n}+i^2(i-1)!\right)+1\right) \\ &= \frac{1}{n+2} + \frac{1}{n+2}\sum_{i=1}^{n}\left(\frac{1}{n+1}+i^2(i-1)!\right) \\ &= \frac{1}{n+2} + \frac{n+1}{n+2}\frac{1}{n+1}\sum_{i=1}^{n}\left(\frac{1}{n}+i^2(i-1)!\right) \\ &= \frac{1}{n+2} + \frac{n+1}{n+2} n! \\ &= \end{aligned}$$

but from here I am kind of stumped. I tried two things:

1. $= \frac{1+n!n+n!}{n+2}$
2. $= \frac{1}{n+2} + \frac{n!(n+1)}{n+2} = \frac{1}{n+2} + \frac{(n+1)!}{n+2}$

But I don't see how I'd get to $(n+1)!$ from here. Hope I used the correct tags. Thanks anyone for help!

Answer 1

Your second step is wrong: $$\frac{1}{n+2}\sum_{i=1}^{n+1}\left(\frac{1}{n+1}+i^2*((i-1)!)\right) \neq \frac{1}{n+2} \sum_{i=1}^{n}\left(\frac{1}{n}+i^2*((i-1)!))+1\right)$$. Instead: $$\frac{1}{n+2}\sum_{i=1}^{n+1}\left(\frac{1}{n+1}+i^2((i-1)!)\right)$$ $$=\frac{1}{n+2}\left(\sum_{i=1}^{n}\left(\frac{1}{n+1}+i^2((i-1)!)\right)+\frac1{n+1}+(n+1)(n+1)!\right)$$ $$=\frac{1}{n+2}\left(\sum_{i=1}^{n}\left(\frac{1}{n}-\frac1{n(n+1)}+i^2((i-1)!)\right)+\frac1{n+1}+(n+1)(n+1)!\right)$$ $$=\frac{1}{n+2}\left(\sum_{i=1}^{n}\left(\frac{1}{n}+i^2((i-1)!)\right) -\frac{1}{n+1}+\frac1{n+1}+(n+1)(n+1)!\right)$$ using induction hypothesis, this is $$=\frac{1}{n+2}\left((n+1)!+(n+1)(n+1)!\right)$$ $$=\frac{1}{n+2}\left((n+2)!\right)$$ $$=(n+1)!$$ However, I would strongly recommend doing this without induction: $$\begin{aligned} \sum_{i=1}^n \left(\frac{1}{n} + i^2(i-1)!\right) &= \sum_{i=1}^n \left( i(i)!\right)+\sum_{i=1}^n \left(\frac{1}{n}\right) \\&= \left(\sum_{i=1}^n \left( (i+1)!-i!\right)\right)+1 \\&=((n+1)!-1)+1 \end{aligned}$$ where one observes that $(i+1)!-i!$ telescopes. Now, divide by $(n+1)$ to complete the proof.

Answer 2

The first equality is false, it should be $$ =\frac{1}{n+2}\left(\sum_{i=1}^n\left(\frac{1}{n+1}+i^2(i-1)!\right)+\frac{1}{n+1}+(n+1)^2n!\right). $$ I think it would be easier if you rephrase the equality you want to prove as $$ \sum_{i=1}^n i\times i!=(n+1)!-1. $$ You already proved the case $n=1$, the induction step is now easier.

$$\sum_{i=1}^{n+1}i\times i!=(n+1)(n+1)!+(n+1)!-1=(n+2)(n+1)!-1=(n+2)!-1 $$ or, without induction, $$ \sum_{i=1}^n i\times i!=\sum_{i=1}^n ((i+1)!-i!)=(n+1)!-1. $$