Limits of Sequences .... Sohrab Exercise 2.2.7 ....

[Math Amateur]

Homework Statement

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with a part of Exercise 2.2.7 Part (1) ... ...

Exercise 2.2.7 Part (1) reads as follows:

I have managed a solution to this exercise and am posting it because

(1) I am unsure that the proof is valid/correct ... particularly given my arithmetic on $\epsilon$ ...

(2) I did not use Sohrab's hint ... and that concerns me ... but I am also interested in how Bernoulli's Inequality is used in this proof ... I can see how to simply involve Bernoulli's Inequality, but not how to use it profitably ...


2. Homework Equations

The concepts of convergence of a sequence and the limit of a sequence are relevant ... so I am posting Sohrab's text corresponding to these notions ...

3. The Attempt at a Solution

My proof is as follows:

To show $\text{lim } ( \frac{ 1 }{ 1 + na } ) = 0$ where $n \in \mathbb{N}$We have $| \frac{ 1 }{ 1 + na } - 0 | = \frac{ 1 }{ 1 + na } \lt \frac{ 1 }{ na }$

Now for $\epsilon \gt 0$ let $\frac{ 1 }{ na } \lt \epsilon$

Then let $\epsilon = \frac{\epsilon^*}{a}$ so that $\epsilon^* = a \epsilon$ ...

Then we have :

$\frac{ 1 }{ 1 + na } \lt \frac{\epsilon^*}{a} \Longrightarrow \frac{ 1 }{ n } \lt \epsilon^*$ ...So ... choose $N$ so that $\frac{ 1 }{ N } \lt \epsilon^*$ ...

So that then we have ... for $n \gt N$ ...

$| \frac{ 1 }{ 1 + na } - 0 | = \frac{ 1 }{ 1 + na } \lt \frac{ 1 }{ na } = ( \frac{ 1 }{ n } ) ( \frac{ 1 }{ a } ) \lt \epsilon^* ( \frac{ 1 }{ a } ) = \epsilon a ( \frac{ 1 }{ a } ) = \epsilon$Is that correct?

and further ...

Can someone please indicate how Bernoulli's Inequality is used in a proof ...

Peter

 

[andrewkirk]

Then we have :

$\frac{ 1 }{ 1 + na } \lt \frac{\epsilon^*}{a} \Longrightarrow \frac{ 1 }{ n } \lt \epsilon^*$ ...

This step is not justified. Consider $\epsilon=\frac58, n=1, a=1, n=1$. The first inequality is satisfied but not the second.

I can't see any use for Bernoulli's inequality, but perhaps the author has in mind a different version of it from the one on wikipedia.

For (1) I'd be inclined to try using Corollary 2.1.32(a) with $\frac1\epsilon$ for $x$ and $a$ for $y$.

 

[Math Amateur]

Hi Andrew,

I was guided to the following solution by Opalg from the Math Help Boards ...

" ... ... Given $\varepsilon>0$, choose $N> \dfrac 1{a\varepsilon}$. ... ... Then $\dfrac 1{Na} < \varepsilon$, and ...

$n\geqslant N \Longrightarrow \frac1{1+na} <\frac1{na} \leqslant \frac1{Na} < \varepsilon.$ ... ... "Thanks for all your help ... I got a number of pointers from your guidance and help ... which is helping me to grasp the basics of analysis ...

Most grateful for all your help ...

Peter

 

[member 587159]

This would be my proof:

Let $\epsilon > 0$ and let $N \geq \frac{1}{a\epsilon} - \frac{1}{a}$. Then:

$n \geq N \implies n \geq \frac{1}{a\epsilon} - \frac{1}{a}$
$\implies na \geq \frac{1}{\epsilon} -1$
$\implies 1 + na \geq \frac{1}{\epsilon}$
$\implies \epsilon \geq \frac{1}{1+na} = \vert \frac{1}{1+na}\vert$

so we have $\frac{1}{1+na} \to 0$ if $n \to \infty$