Incorrect proposal to dealing with nested radicals?

Question

Lets try to calculate this set of nested radicals:

$$f(2)=\sqrt{2\sqrt{3\cdots}}$$

So If we call: $$f(n)=\sqrt{n\sqrt{(n+1)\sqrt{(n+2)\cdots}}}$$

Of course, this is for: $f(2)$ right, so logically here we must solve for $n=2$

So we square both sides:$$(f(n))^2=n\sqrt{(n+1)\sqrt{(n+2)\cdots}}$$

You might notice something tricky about that nested radical on the right. $$(f(n))^2=nf(n+1)$$ Now we try to plug $(n-1)$ in on both sides for $n$. $$(f(n-1))^2=nf(n)$$ And we square! We are trying to create $(f(n))^2$ again so we can substitute back with earlier steps. $$(f(n-1))^4=n^2(f(n))^2$$ And we divide to create the term that we need. $$\frac{(f(n-1))^4}{n^2}=(f(n))^2$$ And we replace: $$\frac{(f(n-1))^4}{n^2}=nf(n+1)$$ Our last step is to divide by $n$ and plug in $1$. We get a rather auspicious answer that I really didn't quite understand. We are trying to create $f(2)$, right? $$\frac{(f(n-1))^4}{n^3}=f(n+1)$$ So...we just plug in $1$ for $n.$ $$\frac{(f(0))^4}{1^3}=f(2)$$ And solve: $$f(2)=0$$ But it is obvious that $f(2)\not=0$ Can someone please where I went wrong in this proof?

Answer 1

If you have reccurence $$ (f(n))^2=nf(n+1) $$ then you will have $$ (f(n-1))^2=(n-1)f(n) $$ instead of $$ (f(n-1))^2=nf(n). $$