I'm trying to prove this statement:

Let $ \alpha, t >0 $, then there exists a positive constant $ c $ such that $$ \left( 1+\frac{t}{2}\right)^{-\alpha} \leq c \left( 1+t\right)^{-\alpha}. $$ Is this true? How could i proceed?


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  • $\begingroup$ Are $\alpha,t>0$ fixed? $\endgroup$ – Dbchatto67 Apr 18 at 16:45
  • $\begingroup$ Yes they are fixed. $\endgroup$ – C. Bishop Apr 18 at 16:47
  • 2
    $\begingroup$ Then you can simply use Archimedean property of real numbers to prove that inequality. $\endgroup$ – Dbchatto67 Apr 18 at 16:49

Hint: use the Archimedean property, and the fact that N has no upper bound.

  • $\begingroup$ Archimidean property: Given any positive $x $ and $y$ in A, there is an integer $ n $ such that $ nx> y.$ In this case: $ \left( 1+\frac{t}{2}\right)^{-a} $ and $ \left( 1+t\right)^{-a} $ are fixed, then exists an integer $ c $ such that $$ \left( 1+\frac{t}{2}\right)^{-a} \leq c\left(1 +t\right)^{-a}. $$ Isn't it? $\endgroup$ – C. Bishop Apr 18 at 16:59
  • $\begingroup$ Yes. Just substitute the value for x , and y. $\endgroup$ – yousef magableh Apr 18 at 17:02

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