# Proof of Product Rule for Sequences using definition of infinitesimal and properties of infinitesimal sequences.

I have been trying to understand this proof for the product rule of sequences, where the author makes use of some properties for infinitesimals, to prove this theorem. This is quite a long question, but please answer it as explicitly as you possibly can.

"A sequence ($y_nz_n$) is convergent to $ab$ if sequences ($y_n$) and($z_n$) are convergent to $a$ and $b$, respectively."

• First of all how would you prove this.
• The author uses an important property described earlier in the book: That for any convergent sequence ($y_n$) there corresponds an infinitesimal sequence ($\alpha_n$) where $\alpha_n$ = $y_n$- $a$. Why is this true, is there any intuition/ a precise reason behind this? Explain this property please.
• Lastly after initial steps are taken we get:
($y_nz_n$) = ab + $\gamma_n$ where $\gamma_n$ = $b\alpha_n$+$a\beta_n$+ $\alpha_n\beta_n$

The author then states:

the sequences ($b\alpha_n$) , ($a\beta_n$) , ($\alpha_n\beta_n$) are infinitesimal as well.

• Why? Is it true that if we multiply a limit with a infinitesimal, we get another infinitesimal as $n\to\infty$? Explain please.

Your textbook has a typically roundabout way of proving simple things (typically I strongly dislike this kind of books). But anyway let's proceed with the same.

A sequence $\alpha_{n}$ of real numbers is said to be an infinitesimal sequence if $\lim_{n \to \infty} \alpha_{n} = 0$.

The correct/better term is null sequence. The meaning of the above definition is that if $\alpha_{n}$ is a null sequence then for any given number $\epsilon > 0$ it is possible to find a positive integer $m$ such that $|\alpha_{n}| < \epsilon$ whenever $n \geq m$.

It is easy to prove (and you should try to prove it, if you face a problem you can post that here) that if $\alpha_{n}, \beta_{n}$ are null sequences and $a, b$ are any real numbers then $a\alpha_{n} + b\beta_{n}, \alpha_{n}\beta_{n}$ are also null sequences.

Further it is almost obvious (your question suggests that it is not obvious to you, but its hard to believe unless you are lost in symbols and jargon which is typical of such bad textbooks) that $\lim_{n \to \infty}x_{n} = a$ if and only if $\alpha_{n} = x_{n} - a$ is a null sequence. Similary if $y_{n} \to b$ then $\beta_{n} = y_{n} - b$ is also a null sequence.

Now consider the sequence \begin{align} \gamma_{n} &= x_{n}y_{n} - ab\notag\\ &= (\alpha_{n} + a)(\beta_{n} + b) - ab\notag\\ &= \alpha_{n}\beta_{n} + a\beta_{n} + b\alpha_{n}\notag \end{align} Clearly each of the three terms in the expression for $\gamma_{n}$ is a null sequence and hence $\gamma_{n}$ is a null sequence. It follows that $x_{n}y_{n} \to ab$ as $n \to \infty$.

Let me know if you need more elaboration.

Update: Based on comments from OP I provide a short and simple proof that if $\alpha_{n},\beta_{n}$ are null sequences then so is their product $\alpha_{n}\beta_{n}$.

Note the following simple idea. If $|x| < 1, |y| < 1$ then $|x||y| = |xy| < 1$ and more strongly $|xy| < |x|, |xy| < |y|$. Thus if we we take two very small numbers (at least smaller than $1$) $x, y$ then their product is going to be much smaller (at least smaller than each of them individually). This simple fact is the key to understanding the proof below.

To prove that $\alpha_{n}\beta_{n}$ is a null sequence we need to show that values of sequence can be made arbitrarily small as $n$ increases. We are given that each sequence $\alpha_{n}, \beta_{n}$ is a null sequence. Let $\epsilon$ be an arbitrarily given positive number and we would like to ensure $$|\alpha_{n}\beta_{n}| < \epsilon$$ for large $n$. Since $\alpha_{n}, \beta_{n}$ are themselves null sequences it follows that there are positive integers $m_{1}, m_{2}$ such that $$|\alpha_{n}| < \epsilon$$ whenever $n \geq m_{1}$ and $$|\beta_{n}| < 1$$ whenever $n \geq m_{2}$. Thus if $m = \max(m_{1}, m_{2})$ then both the inequalities above will hold for all $n \geq m$. Therefore on multiplication of the inequalities we have $$|\alpha_{n}\beta_{n}| = |\alpha_{n}||\beta_{n}| < \epsilon$$ for all $n \geq m$. It follows that $\alpha_{n}\beta_{n}$ is a null sequence.

We have not used anywhere the symbol $0$ and the fact that $0 \times 0 = 0$. The fact $0 \times 0 = 0$ belongs to elementary algebra whereas analysis mostly deals with inequalities like $<, >$ instead of $=$ and the above proof is typical of arguments used in analysis. So in order to understand concepts of calculus you need to focus less on computations (dealing with $=$) and more on comparison of big and small (dealing with $<, >$).

• Would you advise me to read a different textbook (e.g. Micheal Spivak)? For a more unambiguous approach to proofs etc.
– xAly
Aug 20 '15 at 9:04
• @LostAce: For self study I have suggested Hardy's "A Course of Pure Mathematics" many times on MSE. If you want for your university then better to stick to the book suggested by your professor. Spivak's book is good but it is written in a formal style not so suitable for self study. Aug 20 '15 at 9:12
• So if $\alpha_n\beta_n$ is a null sequence then$a\alpha_n + b\beta_n, \alpha_n\beta_n$ are also null sequences because (My attempt at a proof): $\lim_{n \to \infty} \alpha_{n} = 0$ and $\lim_{n \to \infty}\beta_n = 0$ so in other words as we let n approach infinity we are able to say that $\alpha_n$ and $\beta_n$ = $0$ and we know that any real number multiplied by $0$ is also $0$. QED?
– xAly
Aug 20 '15 at 12:57
• is that correct?
– xAly
Aug 22 '15 at 4:32
• @LostAce: I am afraid your approach is not correct. Calculus/analysis is much more than elementary algebra and you are not adding any details more than algebra. Aug 22 '15 at 4:44