# Real Numbers, Sequences and Series: part 9

Definition.

We call a sequence $(a_{n})_{n=1}^{\infty}$ a Cauchy sequence if for all $\varepsilon >0$ there exists an $n_{0}$ such that $|a_{m}-a_{n}|<\varepsilon$ for all m, $n > n_{0}$.

Theorem:

Every Cauchy sequence is a bounded sequence and is convergent.

Proof.

By definition, for all $\varepsilon >0$ there is an $n_{0}$ such that

$|a_{m}-a_{n}|<\varepsilon$ for all m, $n>n_{0}$.

So, in particular, $|a_{n_{0}}-a_{n}|<\varepsilon$ for all $n > n_{0}$, that is,

$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.

Let $M=\max \{ a_{1}, \ldots, a_{n_{0}}, a_{n_{0}+1}+\varepsilon\}$ and $m=\min \{ a_{1}, \ldots, a_{n_{0}+1}-\varepsilon\}$.

It is clear that $m \leq a_{n} \leq M$, for all $n \geq 1$.

We now prove that such a sequence is convergent. Let $\overline {\lim} a_{n}=L$ and $\underline{\lim}a_{n}=l$. Since any Cauchy sequence is bounded,

$-\infty < l \leq L < \infty$.

But since $(a_{n})_{n=1}^{\infty}$ is Cauchy, for every $\varepsilon >0$ there is an $n_{0}=n_{0}(\varepsilon)$ such that

$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.

which implies that $a_{n_{0}+1}-\varepsilon \leq \underline{\lim}a_{n} =l \leq \overline{\lim}a_{n}=L \leq a_{n_{0}+1}+\varepsilon$. Thus, $L-l \leq 2\varepsilon$ for all $\varepsilon>0$. This is possible only if $L=l$.

QED.

Thus, we have established that the Cauchy criterion is both a necessary and sufficient criterion of convergence of a sequence. We state a few more results without proofs (exercises).

Theorem:

For sequences $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$.

(i) If $l \leq a_{n} \leq b_{n}$ and $\lim_{n \rightarrow \infty}b_{n}=l$, then $(a_{n})_{n=1}^{\infty}$ too is convergent and $\lim_{n \rightarrow \infty}a_{n}=l$.

(ii) If $a_{n} \leq b_{n}$, then $\overline{\lim}a_{n} \leq \overline{\lim}b_{n}$, $\underline{\lim}a_{n} \leq \underline{\lim}b_{n}$.

(iii) $\underline{\lim}(a_{n}+b_{n}) \geq \underline{\lim}a_{n}+\underline{\lim}b_{n}$

(iv) $\overline{\lim}(a_{n}+b_{n}) \leq \overline{\lim}{a_{n}}+ \overline{\lim}{b_{n}}$

(v) If $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$ are both convergent, then $(a_{n}+b_{n})_{n=1}^{\infty}$, $(a_{n}-b_{n})_{n=1}^{\infty}$, and $(a_{n}b_{n})_{n=1}^{\infty}$ are convergent and we have $\lim(a_{n} \pm b_{n})=\lim{(a_{n} \pm b_{n})}=\lim{a_{n}} \pm \lim{b_{n}}$, and $\lim{a_{n}b_{n}}=\lim {a_{n}}\leq \lim {b_{n}}$.

(vi) If $(a_{n})_{n=1}^{\infty}$, $(b_{n})_{n=1}^{\infty}$ are convergent and $\lim_{n \rightarrow \infty}b_{n}=l \neq 0$, then $(\frac{a_{n}}{b_{n}})_{n=1}^{\infty}$ is convergent and $\lim_{n \rightarrow \frac{a_{n}}{b_{n}}}= \frac{\lim {a_{n}}}{\lim{b_{n}}}$.

Reference: Understanding Mathematics by Sinha, Karandikar et al. I have used this reference for all the previous articles on series and sequences.

More later,

Nalin Pithwa

# Real Numbers, Sequences and Series: Part 7

Exercise.

Discover (and justify) an essential difference between the decimal expansions of rational and irrational numbers.

Giving a decimal expansion of a real number means that given $n \in N$, we can find $a_{0} \in Z$ and $0 \leq a_{1}, \ldots, a_{n} \leq 9$ such that

$|x-\sum_{k=0}^{n}\frac{a_{k}}{10^{k}}|< \frac{1}{10^{n}}$

In other words if we write

$x_{n}=a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots +\frac{a_{n}}{10^{n}}$

then $x_{1}, x_{2}, x_{3}, \ldots, x_{n}, \ldots$ are approximate values of x correct up to the first, second, third, …, nth place of decimal respectively. So when we write a real number by a non-terminating decimal expansion, we mean that we have a scheme of approximation of the real numbers by terminating decimals in such a way that if we stop after the nth place of decimal expansion, then the maximum error committed by us is $10^{-n}$.

This brings us to the question of successive approximations of a number. It is obvious that when we have some approximation we ought to have some notion of the error committed. Often we try to reach a number through its approximate values, and the context determines the maximum error admissible. Now, if the error admissible is $\varepsilon >0$, and $x_{1}, x_{2}, x_{3}, \ldots$ is a scheme of successive is approximation of a number x, then we should be able to tell at which stage the desired accuracy is achieved. In fact, we should find an n such that $|x-x_{n}|<\varepsilon$. But this could be a chance event. If the error exceeds $\varepsilon$ at a later stage, then the scheme cannot be a good approximation as it is not “stable”. Instead, it would be desirable that accuracy is achieved at a certain stage and it should not get worse after that stage. This can be realized by demanding that there is a natural number $n_{0}$ such that $|x-x_{n}|<\varepsilon$ for all $n > n_{0}$. It is clear that $n_{0}$ will depend on $varepsilon$. This leads to the notion of convergence, which is the subject of a later blog.

More later,

Nalin Pithwa

# What is analysis and why do analysis — part 2 of 2

We had discussed this on Nov 17 2015 blog. We finish the article with more examples from the work of Prof. Terence Tao. (If you like it, please send a thanks to him :-))

Example 1. (Interchanging limits and integrals).

For any real number y, we have

$\int_{-\infty}^{\infty}\frac {dx}{1+(x-y)^{2}}=\arctan (x-y)\mid_{x=-\infty}^{\infty}$ which equals

$(\pi/2)-(-\pi/2)=\pi$.

Taking limits as $y \rightarrow \infty$, we should obtain

$\int_{-\infty}^{\infty}\lim_{y \rightarrow \infty}\frac {dx}{1+(x-y)^{2}}=\lim_{y \rightarrow \infty} \int_{-\infty}^{\infty}\frac {dx}{1+(x-y)^{2}}=\pi$

But, for every x, have $\lim_{y \rightarrow \infty} \frac {1}{1+(x-y)^{2}}=0$. So, we seem to have concluded that $0=\pi$. What was the problem with the above argument? Should one abandon the (very useful) technique of interchanging limits and integrals?

Example 2. Interchanging limits and derivatives.

Observe that if $\in > 0$, then

$\frac {d}{dx}\frac {x^{3}}{\in^{2}+x^{2}}=\frac {3x^{2}(\in^{2}+x^{2})+x^{2}-2x^{4}}{(\in^{2}+x^{2})^{2}}$,

and in particular that

$\frac {d}{dx}\frac {x^{3}}{\in^{2}+x^{2}}\mid_{x=0}=0$.

Taking limits as $\in \rightarrow 0$, one might then expect that

$\frac {d}{dx}\frac {x^{3}}{0+x^{2}}\mid_{x=0}=0$.

But, the right hand side is $\frac {dx}{dx}=1$. Does this mean that it is always illegitimate to interchange limits and derivatives?

Example 3. Interchanging derivatives.

$Let^{1}$ $f(x,y)$ be the function $f(x,y)=\frac {xy^{3}}{x^{2}+y^{2}}$. A common manoeuvre in analysis is to interchange two partial derivatives, thus one expects

$\frac {\partial^{2}f(0,0)}{\partial x \partial y}=\frac {\partial^{2}f(0,0)}{\partial y \partial x}$.

But, from the quotient rule, we have

$\frac {\partial f(x,y)}{\partial y}=\frac {3xy^{2}}{x^{2}+y^{2}}=\frac {2xy^{4}}{(x^{2}+y^{2})^{2}}$

and in particular,

$\frac {\partial f(x,0)}{\partial y}=\frac {0}{x^{2}}-\frac{0}{x^{4}}=0$.

Thus, $\frac {\partial^{2}f(0,0)}{\partial x \partial y}=0$.

On the other hand, from the quotient rule again, we have

$\frac {\partial f(x,y)}{\partial x}=\frac {y^{3}}{x^{2}+y^{2}} - \frac {2x^{2}y^{3}}{(x^{2}+y^{2})^{2}}$ and hence,

$\frac {\partial f(0,y)}{\partial x}=\frac {y^{3}}{y^{2}}-\frac {0}{y^{4}}=y$.

Thus, $\frac {\partial^{2}f(0,0)}{\partial y \partial x}=1$.

Since $1 \neq 0$, we thus seem to have shown that interchange of two derivatives is untrustworthy. But, are there any other circumstances in which the interchange of derivatives is legitimate?

Example 4.$L^{'} H\hat {o}pital's Rule$

We are familiar with the beautifully simple $L^{'}H \hat{0}pital's$ rule

$\lim_{ x \rightarrow x_{0}} \frac {f(x)}{g(x)}=\lim_{x \rightarrow x_{0}}\frac {f^{'}(x)}{g^{'}(x)}$

but one can still get led to incorrect conclusions if one applies it incorrectly. For instance, applying it to $f(x)=x$, $g(x)=1+x$ and $x_{0}=0$ we would obtain

$\lim_{x \rightarrow 0}\frac {x}{1+x}=\lim_{x \rightarrow 0} \frac {1}{1}=1$.

But this is an incorrect answer since $\lim_{x \rightarrow 0}\frac {x}{1+x}=\frac {0}{1+0}=0$.

Of course, all that is going on here is that $L^{'}H \hat{o}pital's rule$ is only applicable when both $f(x), g(x)$ go to zero as $x \rightarrow x_{0}$, a condition which was violated in the previous example. But, even when $f(x)$ and $g(x)$ do go to zero as $x \rightarrow x_{0}$, there is still a possibility for an incorrect conclusion. For instance, consider the limit

$\lim_{x \rightarrow 0} \frac {x^{2} \sin (x^{-4})}{x}$.

Both numerator and denominator go to zero as $x \rightarrow 0$, so it seems pretty safe to apply the rule, to obtain

$\lim_{x \rightarrow 0} \frac {x^{2}\sin (x^{-4})}{x}=\lim_{x \rightarrow 0} \frac {2x \sin (x^{-4})-4x^{-3}\cos (x^{-4})}{1}$ which equals

$\lim_{x \rightarrow 0}2x \sin (x^{-4})-\lim_{x \rightarrow 0}4x^{-3}\cos (x^{-4})$.

The first limit converges to zero by the Sandwich theorem (since the function $2xsin(x^{-4})$ is bounded above by $2|x|$ and below by $-2|x|$, both of which go to zero at 0). But the second limit is divergent (because $x^{-3}$ goes to infinity as $x \rightarrow 0$, and $\cos (x^{-4})$ does not go to zero.) So the limit $\lim_{x \rightarrow 0} \frac {2x \sin(x^{-4})-4x^{-2}\cos (x^{-4})}{1}$ diverges. One might then conclude using $L^{'}H\hat{o}pital's Rule$ that $\lim_{x \rightarrow 0}\frac {x^{2}\sin (x^{-4})}{x}$ also diverges; however, we can clearly rewrite this limit as $\lim_{x \rightarrow 0}x\sin(x^{-4})$, which goes to zero when $x \rightarrow 0$ by the Sandwich Theorem again. This does not show that $L^{"}H\hat opital's Rule$ is untrustworthy. Indeed, it is quite rigorous, but it still requires some care when applied.

That is all, once again, if you like this, please send a thanks note to Prof. Terence Tao.

More later,

Nalin Pithwa