#### idontknow

Prove the following statements :
After the limit turns to be an equation where $$\displaystyle L=pL$$ or another transformation , then :
For p>1 , $$\displaystyle L=\infty$$ ; For p<1 $$\displaystyle L=0$$

Example: $$\displaystyle L=\lim_{n\rightarrow \infty} \dfrac{n^2 }{2^n }=\lim_{n\rightarrow \infty} \dfrac{n^2 +2n +1 }{2^{n+1} }=\dfrac{1}{2}L$$.
Since $$\displaystyle p=1/2 < 1 \; \Rightarrow L=0$$.

#### DarnItJimImAnEngineer

I don't see how that can be. If $L = pL$, then $L = p^{-1}L$, right?

$\displaystyle L=\lim_{n\rightarrow \infty} \frac{n^2 }{2^n }=\lim_{n\rightarrow \infty} \frac{n^2 -2n +1 }{2^{n-1} }=2L$
This fact doesn't change the limit, though.

Aren't $L=0$ and $L\rightarrow \infty$ both potential solutions to $L=pL$, regardless of $p$?

(Note: Probably playing a little fast and loose with the notation. I think the intended meaning is clear, though, right?)

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