Prove $\displaystyle\lim_{n\to \infty}{nb^n}=0, 0
One last thing I used a certain theorem in my proof which I will give.
Theorem 3.1.10 Let (xn) be a sequence of real numbers and let x∈R If (an) is a sequence of positive real numbers with lim and if for some constant C > 0 and some m \in\mathbb{N} we have |x_n-x|\leq Ca_n for all n > m,
then it follows that lim(x_n ) = x.
Here is how my proof goes. Though I don't think its right.
Since $00 and n \in
\mathbb{N} It follows that nb^n=\frac{n}{(1+h_n)^n}. Then by Bernoulli's inequality we get \frac{n}{(1+h_n)^n}\leq \frac{n}{(1+nh_n)}\leq \frac{n}{nh_n} By the theorem above we let C=\frac{n}{h_n} and a_n=\frac{1}{n}. Thus \displaystyle\lim_{n\to \infty}{nb^n}=0$
No comments:
Post a Comment