# Proof of Convergence

• August 17th 2009, 01:35 AM
olekaiwalker
Proof of Convergence
I'm Kinda Stuck With An Assignment:

For $y>1$ Prove That $\lim_{n\rightarrow\infty}\frac{n^\alpha}{y^n}=0, y,\alpha\in\mathbb{R}$ By Applying The Binomial Expansion of $\left(1+x\right)^n, y=1+x$

Can Someone Please Figure That Out?
• August 17th 2009, 02:46 AM
Plato
Quote:

Originally Posted by olekaiwalker
For $y>0$ Prove That $\lim_{n\rightarrow\infty}\frac{n^\alpha}{y^n}=0, y,\alpha\in\mathbb{R}$ By Applying The Binomial Expansion of $\left(1+x\right)^n, y=1+x$

This not true as written. For example let $y=\frac{1}{2}$.

Do you mean $y>1?$
• August 17th 2009, 03:57 AM
olekaiwalker
You're Right. It Should Be

$y>1$
• August 17th 2009, 02:09 PM
Plato
Quote:

Originally Posted by olekaiwalker
I'm Kinda Stuck With An Assignment:

For $y>1$ Prove That $\lim_{n\rightarrow\infty}\frac{n^\alpha}{y^n}=0, y,\alpha\in\mathbb{R}$ By Applying The Binomial Expansion of $\left(1+x\right)^n, y=1+x$

I was hoping someone else might help you. I hate this problem.
Let say that $n>2k~\&~k>\alpha$
$n>n-1>n-2>\cdots>n-k+1>n-k>\frac{n}{2}$
$x+1=y~\&~\left( {1 + x} \right)^n > \frac{{n(n - 1) \cdots (n - k + 1)}}{{k!}}x^k > \left( {\frac{n}{2}} \right)^k \frac{{x^k }}{{k!}}$

From that it follows $0 < \frac{{n^\alpha }}{{\left( {1 + x} \right)^n }} < \frac{{k!}}{{x^k }}\frac{{2^k }}{{n^k }}n^\alpha = \frac{{2^k k!}}{{x^k n^{k - \alpha } }}~~(k>\alpha)$.

It should be clear that $\left( {\frac{{2^k k!}}{{x^k n^{k - \alpha } }}} \right) \to 0$.