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

In summary, for 0<β<1, we can show that β^n -> 0 as n -> ∞ and there exists a γ>1 such that {γ^n}*{β^n} -> 0 as n -> ∞. To prove that n*β^n -> 0 as n -> ∞, we can use the fact that for any a>1, there exists an integer \bar n such that n<a^n for all n>\bar n. Using this, we can also show that there exists a γ>1 such that {γ^n}*{n*β^n} -> 0 as n -> ∞.
  • #1
St41n
32
0
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
 
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  • #2
Try showing that for any [tex]a>1[/tex] there exists an integer [tex]\bar n[/tex] such that

[tex]n<a^n\qquad\textrm{for all}\qquad n>\bar n[/tex]

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

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

1. What does the notation "n*β^n" mean in the context of convergence?

The notation "n*β^n" represents the product of n and β^n, where β is a constant and n is a variable. This is commonly used in mathematics to represent exponential growth or decay.

2. How does "0<β<1" impact the convergence of n*β^n?

The inequality "0<β<1" indicates that β is a fraction or decimal less than 1. This means that as n increases, the value of n*β^n will approach 0, indicating convergence towards 0.

3. What is the significance of showing convergence of n*β^n?

Showing convergence of n*β^n is important in mathematical analysis to determine the behavior of a sequence or series. It can help predict the long-term trend or limit of the sequence as n approaches infinity.

4. Can you provide an example of a sequence that demonstrates "0<β<1, n*β^n -> 0"?

One example of a sequence that satisfies "0<β<1, n*β^n -> 0" is the sequence 0.5^n, where β = 0.5. As n increases, the value of n*0.5^n approaches 0, demonstrating convergence towards 0.

5. Is it possible for n*β^n to converge to a value other than 0?

No, if the inequality "0<β<1" is satisfied, n*β^n will always converge to 0. This is because as n increases, the value of β^n decreases and approaches 0, causing the product n*β^n to also approach 0.

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