Showing Convergence: 0<β<1, n*β^n -> 0

[St41n]

Let 0<β<1

So, β^n -> 0 as n -> infty
Also, we can find γ>1 so that
{γ^n}*{β^n} -> 0 as n -> infty
e.g. γ = β^{-(1/2)}

My question is how can i show that:
n*β^n -> 0 as n -> infty
and there exists γ>1 so that:
{γ^n}*{n*β^n} -> 0 as n -> infty

I appreciate any help

 

[Petr Mugver]

Try showing that for any a&gt;1 there exists an integer \bar n such that

n&lt;a^n\qquad\textrm{for all}\qquad n&gt;\bar n

To prove this you can use induction or differential calculus. After you have done this, you should be able to prove easily your claim.