# Real Numbers, Sequences and Series: part 8

Definition. A sequence is a function $f:N \rightarrow \Re$. It is usual to represent the sequence f as $(a_{n})_{n=1}^{\infty}$ where $f(n)=a_{n}$.

Definition. A sequence $(a_{n})_{n=1}^{\infty}$ is said to converge to a if for each $\varepsilon>0$ there is an $n_{0}$ such that

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

It can be shown that this number is unique and we write $\lim_{n \rightarrow \infty}a_{n}=a$.

Examples.

(a) The sequence $\{ 1, 1/2, 1/3, \ldots, 1/n \}$ converges to 0. For $\varepsilon>0$, let $n_{0}=[\frac{1}{\varepsilon}]+1$. This gives

$|\frac{1}{n}-0|=\frac{1}{n}<\varepsilon$ for all $n>n_{0}$.

(b) The sequence $\{ 1, 3/2, 7/4, 15/8, 31/16, \ldots\}$ converges to 2. The nth term of this sequence is $\frac{2^{n}-1}{2^{n-1}}=2-\frac{1}{2^{n-1}}$. So $|2-(2-\frac{1}{2^{n-1}})|=\frac{1}{2^{n-1}}$. But, $2^{n-1} \geq n$ for all $n \geq 1$. Thus, for a given $\varepsilon >0$, the choice of $n_{0}$ given in (a) above will do.

(c) The sequence $(a_{n})_{n=1}^{\infty}$ defined by $a_{n}=n^{\frac{1}{n}}$ converges to 1.

Let $n^{\frac{1}{n}}=1+\delta_{n}$ so that for $n >1$, $\delta_{n}>0$. Now, $n=(1+\delta_{n})^{n}=1+n\delta_{n}+\frac{n(n-1)}{2}(\delta_{n})^{2}+\ldots \geq 1+\frac{n(n-1)}{2}(\delta_{n})^{2}$, thus, for $n-1>0$, we have $\delta_{n} \leq \sqrt{\frac{2}{n}}$. For any $\varepsilon > 0$, we can find $n_{0}$ in N such that $n_{0}\frac{(\varepsilon)^{2}}{2}>1$. Thus, for any $n > n_{0}$, we have $0 \leq \delta_{n} \leq \sqrt{\frac{2}{n}} \leq \sqrt{\frac{2}{n_{0}}}<\varepsilon$. This is the same as writing

$|n^{\frac{1}{n}}-1|<\varepsilon$ for $n > n_{0}$ or equivalently, $\lim_{n \rightarrow \infty}n^{\frac{1}{n}}=1$

(d) Let $a_{n}=\frac{2^{n}}{n!}$. The sequence $(a_{n})_{n=1}^{\infty}$ converges to o. Note that

$a_{n}=\frac{2}{1}\frac{2}{2}\frac{2}{3}\ldots\frac{2}{n}<\frac{4}{n}$ for $n>3$.

For $\varepsilon>0$, choose $n_{1}$ such that $n_{1}\varepsilon>4$. Now, let $n_{0}=max \{ 3, n\}$ so that $|a_{n}|<\varepsilon$ for all $n > n_{0}$.

(e) For $a_{n}=\frac{n!}{n^{n}}$, the sequence $(a_{n})_{n=1}^{\infty}$ converges to 0. (Exercise!)

In all the above examples, we somehow guessed in advance what a sequence converges to. But, suppose we are not able to do that and one asks whether it is possible to decide if the sequence converges to some real number. This can be helped by the following analogy: Suppose there are many people coming to Delhi to attend a conference. They might be taking different routes. But, as soon as they come closer and closer to Delhi the distance between the participants is getting smaller.

This can be paraphrased in mathematical language as:

Theorem. If $(a_{n})_{n=1}^{\infty}$ is a convergent sequence, then for every $\varepsilon>0$, we can find an $n_{0}$ such that

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

Proof. Suppose $(a_{n})_{n=1}^{\infty}$ converges to a. Then, for every $\varepsilon$ we can find an $n_{0}$ such that

$|a_{n}-a|<\varepsilon/2$ for all $n > n_{0}$.

So, for $m, n > n_{0}$, we have

$|a_{m}-a_{n}|=|a_{n}-a+a-a_{m}|\leq |a_{n}-a|+|a-a_{m}|<\varepsilon$. QED.

The proof is rather simple. This very useful idea was conceived by Cauchy and the above theorem is called Cauchy criterion for convergence. What we have proved above tells us that the criterion is a necessary condition for convergence. Is it also sufficient? That is, given a sequence $(a_{n})_{n=1}^{\infty}$ which satisfies the Cauchy criterion, can we assert that there is a real number to which it converges? The answer is yes!

We call a sequence $(a_{n})_{n=1}^{\infty}$ monotonically non-decreasing if $a_{n+1} \geq a_{n}$ for all n.

Theorem. A monotonically non-decreasing sequence which is bounded above converges.

Proof. Suppose $(a_{n})_{n=1}^{\infty}$ is monotonic and non-decreasing. We have $a_{1} \leq a_{2} \leq \ldots \leq a_{n} \leq a_{n+1} \leq \ldots$. Since the sequence is bounded above, $\{ a_{k}: k=1, \ldots\}$ has a least upper bound. Let it be a. By definition, $a_{n} \leq a$ for all n, but for $\varepsilon>0$ there is at least one $n_{0}$ such that $a_{n_{0}}+\varepsilon>a$. Therefore, for $n > n_{0}$, $a_{n}+\varepsilon \geq a_{n_{0}}+\varepsilon>a\geq a_{n}$. This gives us $|a_{n}-a|<\varepsilon$ for all $n \geq n_{0}$. QED.

We can similarly prove that: A monotonically non-increasing sequence which is bounded below is convergent.

Suppose we did not have the condition of boundedness below or above for a monotonically non-increasing or non-decreasing sequence respectively, then what would happen? If a sequence is monotonically non-decreasing and is not bounded above, then given any real number $M>0$ there exists at least one $n_{0}$ such that $a_{n_{0}}>M$, and hence $a_{n}>M$ for all $n > n_{0}$. In such a case, we say that $(a_{n})_{n=1}^{\infty}$ diverges to $\infty$. We write $\lim_{n \rightarrow \infty}a_{n}=\infty$. More generally, (that is, even when the sequence is not monotone), the same criterion above allows us to say that $\{ a_{n}\}_{n=1}^{\infty}$ diverges to $\infty$ and we write $\lim_{n \rightarrow \infty}a_{n}=\infty$. We can similarly define divergence to $-\infty$.

We make a digression here: In case a set is bounded above, then we have the concept of least upper bound. For any set A or real numbers, we define supremum of A as

$sup \{ x: x \in A\}=sup A = least \hspace{0.1in} upper \hspace{0.1in}bound \hspace{0.1in} of \hspace{0.1in} A$, if A is bounded above and is equal to $\infty$ if A is not bounded above.

Similarly, we define infimum of a set A as

$inf \{ x: x \in A\}= inf A= greatest \hspace{0.1in} lower \hspace{0.1in} bound \hspace{0.1in} of \hspace{0.1in}A$, if A is bounded below, and is equal to $-\infty$, if A is not bounded below.

This would help us to look for other criteria for convergence. For any bounded sequence (a_{n})_{n=1}^{\infty} of real numbers, let

$b_{n} = \sup \{ a_{n}, a_{n+1}, a_{n+2}, \ldots\}=\sup \{ a_{k}: k \geq n\}$.

It is clear that $(b_{n})_{n=1}^{\infty}$ is now a non-increasing sequence. So, it converges. We set

$\lim_{n \rightarrow \infty}b_{n}=limit \hspace{0.1in} superior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \sup a_{n}= \overline{\lim}_{n}a_{n}$,

Similarly, if we write

$a_{n}=\inf \{ a_{n}, a_{n+1}, \ldots\}=\inf \{ a_{k}: k \geq n\}$,

then $(a_{n})_{n=1}^{\infty}$ is a monotonically non-decreasing sequence. So we write

$\lim_{n \rightarrow \infty}c_{n}=limit \hspace{0.1in} inferior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \inf a_{n}= \underline{\lim}_{n}a_{n}$,

We may not know a sequence to be convergent or divergent, yet we can find its limit superior and limit inferior.

We have in fact the following result:

Theorem.

For a sequence $(a_{n})_{n=1}^{\infty}$ of real numbers, $\underline{\lim}a_{n} \leq \overline{\lim}a_{n}$.

Further, if $l=\underline{\lim}a_{n}$, and $L=\overline{\lim}a_{n}$ are finite, then $l=L$ if and only if the sequence is convergent.

Proof.

It is easy to see that $l \leq L$. Now, suppose l, L are finite.If $l=L$ are finite. If $l=L$, then for all $\varepsilon >0$ there exists $n_{1}, n_{2}$ such that

$l - \varepsilon< \sup_{k \geq n} a_{k}< l + \varepsilon$ for all $n \geq n_{1}$ and $l - \varepsilon < \inf_{k \geq n} < l + \varepsilon$ for all $n > n_{2}$.

Thus, with $n_{0}=max{n_{1},n_{2}}$ we have

$l - \varepsilon for all $n > n_{0}$.

This proves that equality holds only when it is convergent. Conversely, suppose $(a_{n})_{n=1}^{\infty}$ converges to a. For every $\varepsilon > 0$, we have $n_{0}$ such that $a-\varepsilon for all $n > n_{0}$. Therefore, $\sup_{k \geq n}a_{k} \leq a+\varepsilon$ and

$a-\varepsilon \leq \inf a_{n}$ for all $n \geq n_{0}$. Hence, $a-\varepsilon \leq < L \leq a+\varepsilon$. Since this is true for every $\varepsilon > 0$, we have $L=l=a$.

QED.

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