$a_n$ is a sequence where $a_1=0$ and $a_2=100$, and for $n \geq 2$:
$$ a_{n+1}=a_n+\frac{a_n-1}{(n)^2-1} $$
I have a basic understanding of sequences. I wasn't sure how to deal with this recurrence relation since there is $n$ in the equation.
By using an excel sheet, I know the limit is 199. And I confirmed this with Wolfram Alpha, which showed that the "Recurrence equation solution" is: $f(x)=199-\frac{198}{x}$
My question: Is it possible to find the limit of this sequence or even the "recurrence equation solution" without using an excel sheet or Wolfram Alpha? If so, can you clearly explain how this is done?
Best Answer
You have: $$(n^2-1)\,a_{n+1} = n^2 a_n - 1,$$ that by putting $b_n = n a_n$ becomes: $$(n-1) b_{n+1} = n b_n - 1,$$ or: $$\frac{b_{n+1}}{n}-\frac{b_n}{n-1}=-\frac{1}{n(n-1)}=\frac{1}{n}-\frac{1}{n-1},$$ so if we set $c_n=\frac{b_n}{n-1}=\frac{n}{n-1}a_n$, we end with: $$ c_{n+1}-c_{n} = \frac{1}{n}-\frac{1}{n-1}.\tag{1}$$ If $c_2=2a_2=200$ (notice that only one starting value is needed), by summing both sides of $(1)$ with $n$ going from $2$ to $N-1$ you get: $$ c_{N}-c_2 = \sum_{n=2}^{N-1}\left(\frac{1}{n}-\frac{1}{n-1}\right)=\frac{1}{N-1}-1,$$ then: $$ c_{N} = \frac{1}{N-1}+199$$ and: $$ a_{N} = \frac{1}{N}+199\cdot\frac{N-1}{N} = 199 - \frac{198}{N}$$ as claimed by Wolfram Alpha.