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<a_{n}<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<a_{n}<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

 

everywhere differentiable function — an interesting problem

Question: Determine all the functions f, which are everywhere differentiable and satisfy 

f(x)+f(y)=f((x+y)/(1-xy)) for all real x and y with xy \neq 1. —- Equation I

Solution:

Let f(x) satisfy Equation I above. Differentiating Equation I partially with respect to each of x and y, we obtain

f^{'}(x)=((1+y^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy)) Equation II

f^{'}(y)=((1+x^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy)) Equation III

Eliminating common  terms in Equation II and Equation III, we deduce that

(1+x^{2})f^{'}(x)=(1+y^{2})f^{'}(y)

As the left side of Equation III depends only on x and the  right side only on y, each side of Equation III must be equal to a constant c. Thus, we have

f^{'}(x)=\frac {c}{1+x^{2}}

and so, f(x)=c\arctan x +d,

for some constant d. However, taking y=0 in Equation III, we obtain f(x)+f(0)=f(x), so that f(0)=0 and

d=0. Clearly, f(x)=c\arctan x satisfies Equation I and so all solutions of Equation I are given by

f(x)=c\arctan x where c is a constant.

Note: In Equation II, the f^{'} is w.r.t. x and in Equation III, the f^{'} is w.r.t. y, yet in equation II 

f^{'}(x+y)/(1-xy) is the same as the corresponding f^{'} in Equation III because the argument (x+y)/(1-xy) is symmetric w.r.t. x and y.