# Limit proofs by definition

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.

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$N$ 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
@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

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 $n \to \infty$.

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