Motivation
Let me ask you a question:
If infinite number of positive numbers are added, what is the answer expected?
`Infinity' is the instant reply, I get in the class.
``Must it be infinity", I repeat!
Even the considered reply I get is infinity.
There are genuine objections to my question of the kind: an infinity of numbers cannot be described or written in a finite time space; so no question arises that of adding them up!
Leaving the question unanswered and unattended let me ask a second question:
Can you in some way distribute a unit piece into infinite number of people?
And the reply I often get is “very very very small bits”.
“How small ..... can we numerically quantify it?”
To this, we don't find an answer.
In the classroom, a chalk-piece I find to be readily available object to be asked to be distributed among infinite number of people. And I get the suggestion to powder it fine, presumably to distribute it to an infinite number of people; each one getting a particle of the powder like the pious powder (Bhabhoot), the medicine for all ailments. Interesting, isn't it! Let me insist \emph{that by distribution we mean to know the exact fraction an individual would get} as his or her share. And then a final hint: I suggest that in an iniquitous world, equal distribution should not be expected to yield result. If the answer is not coming yet, offering half the chalk-piece to the front-corner student, the attempt for the distribution I start. How much to offer to the next is the biggest question? Not to frustrate the third and the subsequent ones in the row leaving nothing for them to be shared, the second should not claim the entire of the remaining half, a quarter looks reasonable for the second. That sets the trend, and a consensus voice announces the share of the remaining in the queue (the queue is much more important mathematically than culturally or civilizationally, a non-queue crowd cannot be obliged, non-queue is a symbol of barbarism) as $1/4, 1/8, 1/16, \ldots$ and so on. It is important to emphasize that in the proposal each person's share in the queue is exactly determined. We may further observe that at each stage of the distribution, the left over is as big as the fraction last given away. And the proposed distribution process can be continued endlessly. It is great discovery to know our ability to distribute one unit or one-tenth, or one-hundredth of a unit to an infinite number of queueable persons. It is important to emphasize that in school-days an infinite geometric series like $1/2+1/2^2+1/2^3+ \cdots$ is not treated with the respect it deserves. Summing infinite number of members is a GREAT STEP -- forward, worthy of jubilations and celebrations. The equality sign used below:
$1/2+1/2^2+ 1/2^3 \ldots = 1$
has a meaning (to be assigned) different from the usually known meaning of the equality sign used below:
$ 5+ 4+1 = 10, 6+5+4 = 15, 5-1 = 4.$
Giving a meaning to this equality is the beginning of interpreting infinite processes (you may note that at no stage of the summing of $1/2+ 1/2^2 + 1/2^3+ \dots$, the sum reaches 1; as is in the cases like $1/3 + 2/3$ or $1/10 + 1/10 + 1/5 + 3/5$. This entry into the study of infinite processes takes us into a new realm of Mathematics - Analysis.
Sequences : Fast and Slow
Observe the following sequences:
Clearly each of the above sequences are approaching to infinity. What is the difference? The successive following ones were in greater hurry to go to infinity. Can we get sequences going to infinity yet faster? Yes, for example
- $1, 2, 3, 4, 5, 6, \ldots, n, \ldots ;$
- $1^2, 2^2, 3^2, 4^2, 5^2, 6^2, \ldots, n^2, \ldots ;$
- $1^3, 2^3, 3^3, 4^3, 5^3, 6^3, \ldots, n^3, \ldots$.
$ 1^{10}, 2^{10}, 3^{10}, 4^{10}, \ldots, n^{10}, \ldots ;$
and more faster is
$1^{100}, 2^{100}, 3^{100}, \ldots,n^{100}, \ldots$.
Can we think of sequences faster than all of the above $ \{ n^p\} $ types? Yes we can, for example:
$ 2^1, 2^2, 2^3, \ldots, 2^n, \ldots$
and yet faster ones are
$ 5^1, 5^2, 5^3, \ldots, 5^n, \ldots ;$
$ 23^1, 23^2, 23^3, \ldots, 23^n, \ldots$.
We also find that faster than $\{20^n\}$ is the sequence:
$ 1^1, 2^2, 3^3, \ldots, n^n, \ldots $.
And rocketing yet faster is $\{n^{n^n}\}$.
You can guess, no fastest can be there.
Let us think of slower sequences than all those discussed above and yet going to infinity. A simple answer is:
$1^{1/2}, 2^{1/2}, 3^{1/2}, \ldots, n^{1/2}, \ldots; $
and yet slower shall be:
$1^{1/3}, 2^{1/3}, 3^{1/3}, \ldots ,n^{1/3},\ldots;$
$1^{1/10},2^{1/10}, 3^{1/10}, \ldots ,n^{1/10},\ldots $.
Shall we look for sequences slower than all of these $\{n^{1/p}\}$ type sequences ? Yes thinking the reverse of exponentials, we have the sequence
$ \log2, \log3, \log4, \ldots, \log n, \ldots$.
It is really worthwhile to compare the growth of sequences $\{n\}$ and $\{\log n\}$. Logarithm considered on base 10, the sequence has to take eight steps to attain the value 2, and 90 steps more to attain the value 3 and 900 more steps to reach the value 4 and so on. Let us explore for even more slower sequences and here they are:
$ \log \log100, \log \log 101, \log \log 102, \ldots, \log \log n,\ldots;$
$\{ \log \log \log n\} ,\, n\geq 10^{100} \, \text{(to avoid negative terms)}.$
We observe that there is no slowest growing sequence approaching to infinity.
Having seen that $\{ n^2\} $ goes towards infinity faster than $\{n\}$, without much effort we observe that $\{ 1/n^2\} $ approaches to zero faster than $\{1/n\}$; in general if $ \{x_n\}$ goes to infinity faster than $\{y_n\}$; then $\{1/x_n\} $ goes to zero faster than $\{1/y_n\}$. We may further expect that $\{x_n\} \to 0$ faster than $\{y_n\} \to 0$ shall imply $\{x_n/y_n\} \to 0$, or equivalently, $\{y_n/x_n\} \to \infty$. We avoid to exactly define `faster' or `slower' sequences used above. We may instead assume or define $\{x_n\} \to 0$ faster than $\{y_n\} \to 0$ as $\{x_n / y_n\} \to 0$. It should be clear that $\{1/n^2\} \rightarrow 0$ neither faster nor slower than $\left\{\dfrac{1}{n^2+5n+7} \right\} $ as
$\left\{\frac{1/n^2}{1/(n^2+5n+7)}\right\} = \left\{ \frac{n^2+5n+7}{n^2}\right\} = \left\{1+\frac{5}{n}+\frac{7}{n^2}\right\} \to1.$
Attending public functions, you might have found people looking towards M.P.'s and M.L.A.'s and on the arrival of the Chief Minister ignoring them altogether. I have seen heads of people and cameras of media turning away from the Cabinet Minister on the arrival of Madam Sonia Gandhi. Same is our approach towards infinity, you will soon learn to ignore $5n$ and $7$ in the presence of $n^2$; and shall ignore $n^5+3n^3-40$ in the presence of $3^n$. The dominance of a particular term is well-recognized in Mathematics as in everyday life.
Infinite Series
Deferring the exploration of the meaning and definition of convergence of series yet allows me to put a small weight on your head, very small not to offend you in the least. We may assume its weight to be 0.001 grams. But if allowed to put another weight of 0.001 grams and then one more.... can I go indefinitely? Obviously not, this add on will collapse you at some stage and even a stronger person than you may be at a later stage. An endless add on, even with an arbitrary small number, cannot result in a situation like
$1/2+1/2^2+1/2^3+\ldots=1.$
So to obtain a non-collapsing convergent situation, beginning with a small number is not enough. The `fixed' shall have to be done away with. Small or big, a subsequent squeezing is unavoidable. Even a subsequent squeezing, if stopped, will lead to the collapsing situation again and so endless squeezing of terms is a must for convergence of series. So we find that for the series
$ a_1 + a_2 + a_3 + \dots+ a_n+ \ldots,$
with $ a_n > 0$ for all $n$, to converge, the necessity is that $a_1\geq a_2\geq a_3 \geq a_4 \geq \ldots a_n \geq \ldots $. Shall I tell you, squeezing or giving away does not always lead to bankruptcy. To demonstrate this let us start with $1/100$. Break
$ 1/100 = 1/200 +1/200 > 1/200+1/201 > 1/200+1/202 \cdots >$
$ 1/200+ 1/300 \cdots > 1/200+ 1/3000 \cdots$.
See how I have reserved $1/200$ and squeezed the other $1/200$. Is it not a miserly give away method? More miserly way is:
$ 1/200+ 1/300 \cdots > 1/200+ 1/3000 \cdots$.
$1/100 = 9/1000+1/1000 =a_1 >9/1000+1/1001= a_2 > 9/1000+1/1002 $
$ = a_3 \cdots 9/1000+1/2000 = a_{1001} > 9/1000+1/5000= a_{4001} \cdots $.
What needs to be observed is that a miserly squeezing$ = a_3 \cdots 9/1000+1/2000 = a_{1001} > 9/1000+1/5000= a_{4001} \cdots $.
$( a_1>a_2>a_3> \ldots a_n> a_{n+1} > \ldots)$
will give $a_1+ a_2 +a_3 + \ldots$ a collapsing situation. And so the convergence of $a_1+a_2+a_3+ \ldots $ necessitates a non--miserly (no reservation for oneself) squeezing that leads to bankruptcy. In other words the convergence of $a_1+a_2+a_3+\ldots $ necessitates $a_1 > a_2 > a_3 > \ldots > a_n > \ldots $; and $a_n \to 0$. A landmark conclusion we arrive at is that convergence of $\Sigma a_n$ implies $\{a_n\}$ decreases to $0$. However a careful consideration of the series :
$ 1+1/2+1/3+1/4+1/5+1/6+ \ldots $
$ \geq 1+1/2+1/4+1/4+1/8+1/8 + 1/8 + 1/8 + \ldots $
$ \geq 1+1/2+1/4+1/4+1/8+1/8 + 1/8 + 1/8 + \ldots $
explains $a_n \mapsto 0$ is not enough for convergence. We find that what is required is that $\{a_n\} \to 0$ faster than $\{1/n\}$ does. We find actually $a_n \to 0$ as fast as $\{1/n^2\}$, so hard squeezing of terms $a_n$ gives convergence of $\Sigma a_n$.
In reality, so hard squeezing is not required; even if $a_n \to 0$ as fast as
$\left\{\dfrac{1}{n^{1+1/2}}\right\}$, even $\left\{\dfrac{1}{n^{1+1/200}}\right\}$ or even $\left\{\dfrac{1}{n^{1+1/2000}}\right\}\to 0$
guarantees convergence of $\Sigma a_n$. But don't be tempted to squeeze $a_n$ at the rate $\left\{\dfrac{1}{n^{1+1/n}}\right\}$ or even at $\left\{\dfrac{1}{n\log n}\right\}$ as that would give divergence. However a little harder squeezing at the rate of
$\left\{\dfrac{1}{n(\log n)^2}\right\}$ or even $\left\{\dfrac{1}{n(\log n)^{200}}\right\}$
guarantees convergence of $\Sigma a_n$.
Convergence of series defined
Let us explore, the meaning of this new equality sign in $1/2 + 1/2^2 +\ldots = 1$; or the definition we wish to assign to the statement:
$a_1+ a_2 +a_3+ \ldots \text{ad inf} = 1$.
In what sense, are we inclined to say that $1/2+1/2^2+\ldots = 1$. One obvious property of the infinite addition (or infinite series) above is that at no finite stage the sum is 1; it is always less than 1. It is tending to 1; or is approaching to 1 can be seen. But we may as well say it is approaching to 2 or 1.1 or 33. (This crooked logic is of the type: suppose our car is approaching Delhi from South direction, someone may say that we are approaching Chandigarh and yet someone else may say, we are approaching Amritsar, places further north). So what is the specialty or uniqueness about 1, which 2, 1.1 or $33$ do not possess, when we say that the infinite series is approaching 1. We may for convenience denote $S_n$ to be the sum of the first $n$ terms of the series. We observe that the partial sum $S_n$ is always at a distance greater than 1 from 2, it is always at a distance greater than 0.1 from 1.1, it is always at a distance greater than 32 from 33. Symbolically, we may write
$2- S_n> 1 $
$1.1 - S_n > .1 $
$33 - S_n > 32$
for all $n$. But corresponding to 1, there is no such (as 1, .1, 32 above) positive $p$ so that we may say that $1-S_n > p$ for all $ n$. The truth is that for arbitrary positive $p$, we find a corresponding stage $N$ of the sequence where from the inequality gets reversed, that is
$1-S_n < p , \forall \,n \geq N.$$
The property referred above of 1 is geometrically much more appealing. The specialty of 1 is that the partial sums $S_n$ come near and near, arbitrarily near to 1. That is to say whatever be the standard of nearness chosen, $S_n$ is found close to 1. Let $1/10$ be a standard of nearness then we find that at the 4th stage of the series
$1/2+ 1/2^2+1/2^3+ 1/2^4 = 1- 1/2^4$.
That is to say that the distance of the partial sum at the 4th stage from 1 is less than 1/10, that is
$1-S_4= 1-(1/2 +1/2^2+ 1/2^3 + 1/2^4) = 1/2^4 = 1/16 < 1/10.$
Also
$1-S_5= 1-(1/2 +1/2^2+ 1/2^3 + 1/2^4 + 1/2^5) = 1/2^5 = 1/32 < 1/10,$
$1-S_6= 1-(1/2 +1/2^2+\ldots+1/2^6) = 1/2^6 = 1/64 < 1/10.$
Figure 1 below shows that $ 1-S_n<1/10$ for $n \geq 4$.
 |
| Figure 1 |
$ 1-S_7 = 1- (1/2 + 1/2^2+ \ldots+ 1/2^7) = 1/128 < 1/100,$
$ 1-S_8 = 1- (1/2 +1/2^2 + \ldots+1/2^8) = 1/256 < 1/100, $
$ 1-S_9 = 1- (1/2 + 1/2^2 + \ldots+1/2^9) = 1/512 < 1/100, $
$ \text{or},\, 1-S_n < 1/100 \quad \text{for all} \, n\geq 7.$
 |
| Figure2 |
Truly, there is no smallest positive number.
And as such we are left with no choice than to take smaller and smaller positive numbers and demonstrate our ability to find a corresponding $ N$. To fulfill our objective, we may choose arbitrary positive number $p$ and show our ability to find the corresponding $N$ in terms of $p$, so that we get
$1-S_n < p \quad \text{for all} \quad n \geq N.$
For a given $p > 0$, to ensure $1- S_n =1/2^n < p $ for $n\geq N$; it is readily seen that the stage $ N$ be chosen as the integer greater than $\log_2(1/p)$.
 |
| Figure 3 |
 |
| Figure 4 |
\begin{align*}
&S_1 =1\\
&S_2 =1- 1/2\\
&S_3 =1- 1/2+ 1/3\\
&\vdots\\
&S_n = 1-1/2 + 1/3 -\ldots + (-1)^{n-1} /n;
\end{align*}
 |
| Figure 5 |
$ L-p < S_n < L + p.$
Or equivalently,
$-p< L - S_n < p \text{ for } n\geq N,$
or,
$ |S_n- L| < p \quad \text{for all } \, n\geq N.$
The modulus used here could be used to cover the earlier two cases also where the series had only positive terms or only negative terms. And without giving importance to the approach from upper (greater) or lower side; we may straight look for the stage in the sequence where onwards it falls in the strip $(L-p, L+p)$. It is not difficult to dispense away with the two--dimensional scene and look for the stage in the sequence from where onwards, the sequence falls in the given interval $(L-p, L+p)$ of the real line for an arbitrary positive $p$. We feel as much satisfied, whether this stage is 100 or $100^{100}$. Indeed, success in getting a stage ensures that except a finite number (100 or $100^{100}$ or....) of members, the entire sequence lies in $(L-p, L+p)$, and so does it, for each positive $ p$.
Sequences Revisited
\begin{align*}
&2/1, 3/2, 4/3, 5/4, 6/5, \ldots\\
&1/3, 1/3, 1/3, 1/3, 1/3, 1/3, \ldots\\
&1/5, 1/5, 1/5, 1/5, 1/5, 1/5,\ldots\\
&1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10,\ldots
\end{align*}
The sequences above are converging to $1, 1/3, 1/5$ and $0$ respectively. We adulterate and form the sequence:
$ 2, 1/3, 1/5, 1/4, 3/2, 1/3, 1/5, 1/5, 4/3, 1/3, 1/5, 1/6, 5/4,\ldots . $
Does this sequence behave like the earlier sequences? The adulteration has changed its character; the new sequence does not converge. It is natural to try to reverse the process of adulteration. Mixing of water with milk is easy - but the reverse process is not so. It is believed that a mythological bird on Himalayas called `Hans' (similar to Swan) has the strange ability to separate the two. We invoke the ability of Hans to find subsequences of this adulterated sequence that converge. Mind it, there are not just four convergent subsequences. There are many many more, but essentially four; in the sense that the convergent ones have only four distinct limits and the two having the same limit differ only at finite number of terms.
Let us plot the adulterated sequence, and observe how there are basically four streams in the mixed sequence flowing four distinct ways: (Figure 6 and Figure 7).
 |
| Figure 6 |
 |
| Figure 7: The sequence at a later stage |
We can catch hold of the limit superior and the limit inferior of a sequence $\{a_n\}$ even without finding all the limits of the possible convergent subsequences. Let us now try to catch hold of the limit superior, that is to say the ultimate biggest of the sequence (approached by the sequence in the sense of the limit). First crude attempt to catch the ultimate biggest (supremum) can be to find the supremum of the entire sequence, that is $\sup \{a_n | n\geq1\}$. Having taken the first step, we get a direction for finding a better estimate for the ultimate supremum as $\sup \{a_n | n \geq 2\}$, which purifies the earlier effort somewhat, in case $\sup \{a_n | n \geq 1\}$ was influenced by $a_1$ being the biggest. (We must not forget that the ultimate limits are not affected by first, second, or any finite number of terms of the sequence). Proceeding further, better and better declining estimates for the limit superior can be arranged as
\begin{align*}
&\sup \{a_k | k \geq 1\}\\
&\geq \sup \{a_k | k \geq 2\}\\
&\geq \sup \{a_k | k \geq 3\} \\
&\hspace{.1cm}\vdots\\
&\geq \sup \{a_k | k \geq n\} \\
&\geq\ldots \\
&\hspace{.1cm}\vdots
\end{align*}
And to catch the best, the limit of the above or the smallest (infimum) of them (the ultimate among the suprema) is seen to be
\[
\limsup_{n \to \infty} \{a_k | k \geq n\}= \inf_n \sup \{a_k | k \geq n\}.
\]
We may well define (in notation)
\[
\overline{\lim_{n\to\infty}} a_n= \underset{n\to\infty}\limsup\ {a_n} =\inf_{k\geq n} {\sup a_k}.\]
In our example above:
\begin{align*} &\sup \{a_k | k \geq 1\} = 2,\\
&\sup \{a_k | k \geq 2\} = 3/2,\\
&\sup \{a_k | k \geq 3\} = 3/2,\\
&\sup \{a_k | k \geq 4\} = 3/2,\\
&\sup \{a_k | k \geq5\} = 3/2,\\
&\sup \{a_k | k \geq 6\} = 4/3.
\end{align*}
Observe how from the biggest in the entire sequence, step by step, we are approaching to catch the ULTIMATE BIGGEST, as the
\begin{align*}
\lim_{n\to\infty}\sup \{a_k| k \geq n\} &= \lim_{n\to\infty} \left\{\sup \{a_k| k \geq 1\}, \sup \{ a_k| k \geq 2\},\ldots, \sup\{ a_k| k \geq n\},\ldots\right\} \\
&= \lim \{ 2, 3/2, 3/2, 3/2, 3/2, 4/3,4/3 \ldots\}=1.
\end{align*}
We may as well take the earlier sequence with modification at three stages, so that the new sequence is (knowing well that the introduction of three members is not going to affect any of the limits):
$2, 1/3, 1/5, 50, 1/4, 3/2, 60, 1/ 3, 1/5,14, 1/5, 4/3,\ldots$
We find
\begin{align*}
& \sup \{a_k | k \geq1\}\\
&=\sup \{a_k| k\geq 2\} \\
&= \ldots\\
&=\sup \{a_k | k \geq 7\} = 60\\
\text{and}\\
&\sup \{a_k | k \geq 8\} \\
&= \sup \{a_k | k\geq 9\}\\
&=\sup \{a_k | k\geq10\} = 14\\
\text{and} \\
&\sup \{a_k | k \geq 11\} = 4/3\\
&= \sup \{a_k | k\geq 12\} = 4/3.
\end{align*}
Thereafter the earlier pattern repeats, leading finally to
$\limsup a_n =\inf \sup \{a_k | k \geq n \} = 1$.
We may remember that the limit superior (say $M$) being the infimum (the greatest lower bound) of $\{\sup \{a_k | k \geq n\}, n\in \mathbb{N}\}$, satisfies the following:
$\bullet$ For $\epsilon> 0, M + \epsilon$ is not a lower bound for
$\big[\sup \{a_k | k \geq n\}\big]_n$; so $\sup \{a_k | k \geq n \} < M + \epsilon$ for some $n = n_0$, so, $a_n < M + \epsilon \quad $for $n \geq n_0$, that is, some $n_0$-th stage onwards every element of the sequence is less than $ M +\epsilon$. In our example, for $\epsilon = 1/10$ we find that $a_n < 1+ 1/10$ for every $n > 37$. Looking graphically, after the 37-th stage, the sequence is below the horizontal line at $1 + 1/10$ (see Figure 8).
 |
| Figure 8 |
$\bullet$ $M$ is a lower bound for $\big[\sup \{a_k | k \geq n\}\big]_n$, so $M < \sup \{a_k | k \geq n\}$ for all $n$. For $\epsilon > 0, \, M - \epsilon $ is not an upper bound for $\{a_k | k \geq n\}$ for any $n$. So given integer $n_0$, we can find $n >n_0$ such that $a_n > M - \epsilon$. That is to say, we obtain $a_n > M - \epsilon$ for infinitely many arbitrarily large values of $n$. Taking $\epsilon =1/10$ in our example we find that $a_n > 1- 1/10$ for $n = 1, 5, 9,\ldots.$
The missing members of our sequence $\{a_2, a_3, a_4, a_6, a_7,\ldots\}$ are essentially not part of the sequence converging to 1 (the limit superior), and so need not fall in the strip $(1- \epsilon, 1+\epsilon)$.
We may analogously catch hold of limit inferior and define it as $\underline{\lim} \{ a_n\} =\underset{n}\sup \inf \{a_k | k \geq n\}$. The reader may try the above definition in finding the limit superior and limit inferior of the function $f(x)$, as $x$ is tending to zero, where $f$ is defined on $[0,1]$ by:
\begin{equation}
f(x)= \begin{cases}
1, &\text{for $x$ a rational number in [0,1] except 1/3,\,1/5,\,1/7}\\
30, &\text{for $x= 1/3$}\\
-50, &\text{for $x= 1/5$}\\
-70, &\text{for $x= 1/7$}\\
2, &\text{for $x=\sqrt{2}/n$ for $n\in \mathbb{N},\, n\geq 2$}\\
3, &\text{for $x=\sqrt{3}/n$ for $n\in \mathbb{N},\, n\geq 2$}\\
5, &\text{for $x=\sqrt{5}/n$ for $n\in \mathbb{N},\, n\geq 3$}\\
3/2, &\text{elsewhere in [0,1].}
\end{cases}
\end{equation}
Let us close the discussion with the following two remarks:
We may observe that the factor $(-1)^n$ makes sequences $\{(-1)^nn\}$, $\{(-1)^nn^2\}$ oscillate wildly (unboundedly). A similar role is played by factors of the form $\sin n$ or $\cos n$ to make $\{ n \sin n\}$ or $\{n^2 \cos n \}$ oscillate.
It is interesting to observe that the series $1+1/3+1/5+1/7+\ldots $ and $1/2+1/4+1/6+\ldots$ are both divergent, and yet the series $1-1/2+1/3-1/4+1/5- \ldots$ converges.
It is like the case of two drunkards, not able to walk on their own yet managing to walk along balancing each other's step. Rearrangement on the steps can bring consequences unimagined. Both the drunkards leaning on the same side, making all terms of the series positive, is one such disaster.
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