MHB Limits of Sequences .... Sohrab Exercise 2.2.7

[Math Amateur]

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:View attachment 7189I 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 ...

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==========================================================================================

To help readers understand Sohrab's notation I am posting his definitions of convergence of a sequence and the limit of a sequence ... as follows:View attachment 7190

 

[Evgeny.Makarov]

To show $$\text{lim } ( \frac{ 1 }{ 1 + na } ) = 0$$ where $$n \in \mathbb{N} $$ ...

We have $$| \frac{ 1 }{ 1 + na } - 0 | = \frac{ 1 }{ 1 + na } \lt \frac{ 1 }{ na }$$

To produce large left and right delimiters that correspond to the size of the formula inside, write \left before the left delimiter and \right before the right one, e.g., \left|\frac{1}{1+na }-0\right| to produce $$\left|\frac{1}{1+na }-0\right|$$ (ideally it should be \left\lvert\frac{1}{1+na }-0\right\rvert).

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

The previous line said $$\left|\frac{ 1 }{ 1 + na } - 0\right| = \frac{ 1 }{ 1 + na } \lt \frac{ 1 }{ na }$$. Implicitly, it is preceded by the universal quantifier $\forall n$, and the scope of this quantifiers ends at the end of the formula. This means that using $n$ is fine inside the formula, but outside $n$ is undefined. So saying "let $1/na<\epsilon$ produces an error message "$n$ is undefined" and a buzz
[YOUTUBE]FRpq7o1mKXY[/YOUTUBE]
after which the proof checking stops.

In fact, it is not a good idea to say things like "let $E$ equal ..." or "let $E$ be less than $\epsilon$" where $E$ is an expression and not an individual variable. What if it is impossible to satisfy this formula due to the shape of $E$? For example, it makes no sense to say "Let $1/(1+x^2)>2$" for a real $x$.

I would say, "Fix an arbitrary $\epsilon>0$. Then there exists an $n$ such that $1/na<\epsilon$".

By the way, strictly speaking it is also not good to write, "$\lim_{n\to\infty}\frac{1}{1+na}=0$ where $n\in\mathbb{N}$" because $\lim_{n\to\infty}$ is a variable-binding construction. It introduces a local variable $n$ that can be used only inside the expression that follows $\lim$ and is not defined outside of it. Other examples of variable-binding constructions are $\sum_{n=1}^\infty$ and $\int_0^1\dots\,dx$. One cannot say, "Consider $\sum_{n=1}^\infty 1/n^2$ where $n$ is even" or "$\int_0^1 x^2\,dx=1/3$ for $x=1/2$". In this example, $n\in\mathbb{N}$ is probably not necessary since $n$ is usually assumed to denote natural (rather than real) numbers (this reminds the implicit variable type convention in Fortran). But if it is not clear, one could say that $n$ ranges over natural numbers: this implies that $n$ is a placeholder rather than a fixed value.

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

Should be: "Let $$\epsilon^* = a \epsilon$$ so that $$\epsilon = \frac{\epsilon^*}{a}$$". This is because $\epsilon$ is already fixed and one cannot redefine it with "Let $\epsilon=\ldots$" (and it reads like a definition of $\epsilon$ at first glance rather than a definition of $\epsilon^*$, which it is).

Then we have :

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

This implication is in general false, for example, for $a=1$.

 

[Opalg]

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

.
.
.

(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 ...

[Edit. I wrote this before seeing Evgeny's comment above. You should probably look at both comments in parallel.]

I think you are right. There is no need to use Bernoulli for (1). Your argument is sort of correct, but you could shorten it (and perhaps clarify it) a bit like this:

Given $\varepsilon>0$, choose $N> \dfrac 1{a\varepsilon}$. Then $\dfrac 1{Na} < \varepsilon$, and $$n\geqslant N \Longrightarrow \frac1{1+na} <\frac1{na} \leqslant \frac1{Na} < \varepsilon.$$

It often happens in this sort of analysis that you need to tackle a problem by working backwards from the result (as you did in this case). But you then need to re-write the argument so as to lead forwards to the result.

For part (2) of this exercise you surely will want to use Bernoulli. I would do it like this:

Given $0<b<1$, let $a = \dfrac1b-1$. Then $a>0$, $b = \dfrac1{1+a}$, and $$b^n = \frac1{(1+a)^n} \leqslant \frac1{1+na}$$ by Bernoulli's inequality. But $\dfrac1{1+na} \to0$ as $n\to\infty$ by part (1), and it follows that $b^n\to0$.

 

[Math Amateur]

To produce large left and right delimiters that correspond to the size of the formula inside, write \left before the left delimiter and \right before the right one, e.g., \left|\frac{1}{1+na }-0\right| to produce $$\left|\frac{1}{1+na }-0\right|$$ (ideally it should be \left\lvert\frac{1}{1+na }-0\right\rvert).

The previous line said $$\left|\frac{ 1 }{ 1 + na } - 0\right| = \frac{ 1 }{ 1 + na } \lt \frac{ 1 }{ na }$$. Implicitly, it is preceded by the universal quantifier $\forall n$, and the scope of this quantifiers ends at the end of the formula. This means that using $n$ is fine inside the formula, but outside $n$ is undefined. So saying "let $1/na<\epsilon$ produces an error message "$n$ is undefined" and a buzz

after which the proof checking stops.

In fact, it is not a good idea to say things like "let $E$ equal ..." or "let $E$ be less than $\epsilon$" where $E$ is an expression and not an individual variable. What if it is impossible to satisfy this formula due to the shape of $E$? For example, it makes no sense to say "Let $1/(1+x^2)>2$" for a real $x$.

I would say, "Fix an arbitrary $\epsilon>0$. Then there exists an $n$ such that $1/na<\epsilon$".

By the way, strictly speaking it is also not good to write, "$\lim_{n\to\infty}\frac{1}{1+na}=0$ where $n\in\mathbb{N}$" because $\lim_{n\to\infty}$ is a variable-binding construction. It introduces a local variable $n$ that can be used only inside the expression that follows $\lim$ and is not defined outside of it. Other examples of variable-binding constructions are $\sum_{n=1}^\infty$ and $\int_0^1\dots\,dx$. One cannot say, "Consider $\sum_{n=1}^\infty 1/n^2$ where $n$ is even" or "$\int_0^1 x^2\,dx=1/3$ for $x=1/2$". In this example, $n\in\mathbb{N}$ is probably not necessary since $n$ is usually assumed to denote natural (rather than real) numbers (this reminds the implicit variable type convention in Fortran). But if it is not clear, one could say that $n$ ranges over natural numbers: this implies that $n$ is a placeholder rather than a fixed value.

Should be: "Let $$\epsilon^* = a \epsilon$$ so that $$\epsilon = \frac{\epsilon^*}{a}$$". This is because $\epsilon$ is already fixed and one cannot redefine it with "Let $\epsilon=\ldots$" (and it reads like a definition of $\epsilon$ at first glance rather than a definition of $\epsilon^*$, which it is).

This implication is in general false, for example, for $a=1$.

Hi Evgeny ...

Well! Thank you for all the guidance and support ... not only with the logic of proofs ... but also with Latex ...

I appreciate your support ... might even get me to the point of understanding analysis ... :)

Peter

- - - Updated - - -

[Edit. I wrote this before seeing Evgeny's comment above. You should probably look at both comments in parallel.]

I think you are right. There is no need to use Bernoulli for (1). Your argument is sort of correct, but you could shorten it (and perhaps clarify it) a bit like this:

Given $\varepsilon>0$, choose $N> \dfrac 1{a\varepsilon}$. Then $\dfrac 1{Na} < \varepsilon$, and $$n\geqslant N \Longrightarrow \frac1{1+na} <\frac1{na} \leqslant \frac1{Na} < \varepsilon.$$

It often happens in this sort of analysis that you need to tackle a problem by working backwards from the result (as you did in this case). But you then need to re-write the argument so as to lead forwards to the result.

For part (2) of this exercise you surely will want to use Bernoulli. I would do it like this:

Given $0<b<1$, let $a = \dfrac1b-1$. Then $a>0$, $b = \dfrac1{1+a}$, and $$b^n = \frac1{(1+a)^n} \leqslant \frac1{1+na}$$ by Bernoulli's inequality. But $\dfrac1{1+na} \to0$ as $n\to\infty$ by part (1), and it follows that $b^n\to0$.

Thanks Opalg ... appreciate the help ...

Great to see a solution that stands up to scrutiny ...

Peter