Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

a.) Show that $\lim(1/\sqrt{n})=0$ by using the definition.

b.) Let $d\in \mathbb{R}$ satisfy $d>1$. USe Bernoulli's Inequality to show that the sequence $d^n$ is not bounded in $\mathbb{R}$, hence it is not convergent.

c.) Let $b\in \mathbb{R}$ satisfy $0<b<1$, show that $\lim(nb^n)=0$ by using the Binomial Theroem.

My attempt:

a.) Given an $\epsilon$, I need to find an $N$ that satisifies the condition of convergence. Thus I have to bound $|\frac{1}{\sqrt{n}}-0|<\epsilon$; which essentially means, when is $|\frac{1}{\sqrt{n}}|<\epsilon$, correct? So, given $\epsilon>0$ we have $|\frac{1}{\sqrt{n}}|<\epsilon$ which implies that $N=\frac{1}{\epsilon^2}$, correct? Thus $\forall$ $n>N$ then $n>\frac{1}{\epsilon^2}$, hence $\frac{1}{\sqrt{n}}<\epsilon$, so we have $|x_n-x|=|\frac{1}{\sqrt{n}}-0|=|\frac{1}{\sqrt{n}}|<\epsilon$. Q.E.D.

share|improve this question
1  
$N$ should be integer so choose $N=\lfloor \frac{1}{\epsilon^2}\rfloor+1$. –  Sami Ben Romdhane Mar 21 '13 at 7:29
    
@SamiBenRomdhane what are those brackets you have outside the fraction? And why would adding by one make $N$ an interger? –  Q.matin Mar 21 '13 at 7:42
1  
@Q.matin it's the floor function, the next smaller integer to a given number. –  Stefan Mar 21 '13 at 8:35
    
@Stefan Thanks for clarifying, Stefan. –  Q.matin Mar 22 '13 at 7:05
add comment

1 Answer

up vote 0 down vote accepted

For part c, since $0 < b < 1$, let $b = 1/(1+c)$ where $c > 0$. This is an example of the slogan "always expand around $0$."

Then $n b^n = n/(1+c)^n$.

Showing $n b^n \to 0$ is the same as showing $(1+c)^n/n \to \infty$.

Using the binomial theorem, we can show much more, namely that $(1+c)^n/n^r \to \infty$ for any positive real $r$.

To show this, let $m = \lceil r+2 \rceil$, and $n$ any integer greater than $m^2$.

By the binomial theorem, $(1+c)^n = \sum_{k=0}^n \binom{n}{k}c^k >\binom{n}{m}c^m > (n-m+1)^m \dfrac{c^m}{m!} = n^m(1-(m-1)/n)^m \dfrac{c^m}{m!} $. Since $n > m^2$, $(m-1)/n < 1/m$ so $(1-(m-1)/n)^m > (1-1/m)^m > 1/e$.

Therefore $(1+c)^n > n^m\dfrac{c^m}{e m!} > n^{r+1}\dfrac{c^m}{e m!} $, or $(1+c)^n/n^r > n \dfrac{c^m}{e m!} $. Since $c$ and $r$ are fixed, and $m$ is also, $\dfrac{c^m}{e m!}$ is fixed (once $c$ and $r$ are chosen), so $(1+c)^n/n^r \to \infty$ as $n \to \infty$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.