Akis Androulakis

In the first article we introduced the concept of a limit and saw some examples of how it can be used in calculations. Before delving into further applications, we’ll encounter some counter-intuitive results and cases where limits don’t behave the way we might initially expect…

Consider the following questions:

1. What is the value of:  1-1+1-1+1-1+1-1+… ?

2. Can a sequence of rational numbers (those written as fractions) have an irrational limit (a number which cannot be written as a fraction)?

3. Let (aₙ) and (bₙ) be sequences with bₙ ≠ 0 for all n. We can then create two new sequences (aₙ / bₙ) and (aₙ – bₙ) [see below for an explicit example of how to calculate these two sequences]. If aₙ ≈ bₙ (approximately equal) as n gets large, we may express this as: (aₙ-bₙ) converges to 0, or (aₙ/bₙ) converges to 1. Are the two expressions equivalent?

Eg. if (aₙ)= 1, 2, 3, 4, …  and (bₙ)= 1, 1/2, 1/3, 1/4, … then (aₙ / bₙ)= 1, 4, 9, 16, …  and (aₙ – bₙ) = 0, 3/2, 8/3, 15/4, … 

We’ll address each question separately, but I first encourage you to think about how you would answer them before reading on.

For the first question, one possible answer could be 0 by grouping the terms into pairs and claiming that (1-1) + (1-1) + (1-1) + … = 0 + 0 + 0 + … = 0 which at first glance seems perfectly reasonable.

However, one can similarly claim: 1 + (−1+1) + (−1+1) + (−1+1) + … = 1 + 0 + 0 + 0 + … = 1. So, which is the correct limit and can we have even more plausible answers if we rearrange the terms? 

As intuition fails to answer these questions, we need to rely on the maths we learned in the first article. Specifically, the correct answer to the initial question can only be obtained by applying a method similar to the one we used to show that 0.999… = 1.

We need to consider the sequence of partial sums, a sequence whose nth term is the sum of first the n terms of our original sequence. For example, given the sequence (aₙ) = 1/n  for all n:  1, ½, ⅓, ¼, … , then the sequence of partial sums (Sₙ) is:

S₁ = 1
S₂ = 1 + 1/2 = 3/2
S₃ = 1 + 1/2 + 1/3 = 11/6
S₄ = 1 + 1/2 + 1/3 + 1/4 = 25/12
…

and so on. In the example we want to consider here, the (Sₙ) are:  

S₁ = 1
S₂ = 1-1 = 0
S₃ = 1-1+1 = 1
S₄ = 1-1+1-1 = 0
…

So we have 1’s and 0’s alternating which means that this sequence doesn’t converge! Whichever real number you try (maybe 0, ½ and 1 intuitively make sense), the terms of the sequence will never get “arbitrarily close” to your chosen limit in the way we defined in the first article. Infinite sums are in fact the limit of finite sums when the number of added terms tends to infinity. So in this case the infinite sum is undefined as the limit of (Sₙ) doesn’t exist. The reason we treat series this way, is to avoid any paradoxes such as the sum having two or more possible values.

The second question can be answered more quickly as we can just consider the sequence (a_n) where the nth term is the rational number with n digits, identical to the first n digits of pi.

a₁ = 3
a₂ = 3+1/10
a₃ = 3 + 14/100
a₄ = 3 + 141/1000
…

We can see that (a_n) is a sequence consisting only of rational numbers (every term can be written as a sum of whole numbers and fractions) but converges to pi which is irrational! Anyone interested in strictly proving that by the definition, can do it almost the same way we showed the sequence: 0.9, 0.99, 0.999, … converges to 1 in the first article. It is also interesting to note that in tis counterexample we could have chosen any irrational number x instead of pi and similarly constructed a rational sequence which converges to x.

The third question is a little trickier to handle because if we had real numbers a and b (with b non-zero), then:

 ab = 1   ⇔   a = b   ⇔   a – b = 0

However, convergence of a sequence doesn’t imply that the terms of the sequence ever become exactly the same, and in fact here is a counterexample to the claim in question 3:

Let (x_n) = n+1 and (y_n) = n  for all n.

Then (x_n – y_n) = 1 and (x_n / y_n) = (n+1)/n  for all n.

Note that (x_n – y_n) is not a number but a sequence whose terms are all equal to 1 and clearly doesn’t converge to 0, even though (x_n / y_n) = (n+1) / n = 1 + 1/n converges to 1 (as we saw in the first article). Thus, the two expressions are not equivalent, leaving us with the question of how to interpret aₙ ≈ bₙ (approximately equal) as n gets large.

(a_n – b_n) measures the exact difference between the terms of (a_n) and (b_n) so if it converges to 0, it means that the terms (a_n) and (b_n) eventually become arbitrarily close to one another. However, (a_n / b_n) converging to 1 is a tricky case. Consider the sequence (c_n) = (a_n / b_n) – 1 = (a_n – b_n)/b_n for all n, which converges to 0. Essentially, for (c_n) to converge to 0, (a_n – b_n) doesn’t necessarily have to converge to 0, but rather has to get “infinitely smaller” (in absolute value) than the denominator b_n. Take for example the case (a_n) = n+1 and (b_n) = n we used earlier, then (a_n – b_n) = 1 and c_n = 1/n for all n. Indeed, there is no constant limit to how many times the denominator n will get larger than the numerator 1.

Therefore, (a_n / b_n) converging to 1 means that a_n and b_n eventually become approximately equal in the sense that the difference between them gets “infinitely smaller” and thus insignificant compared to their value.

This whole idea was analysed not only to emphasise the counter-intuitive properties of limits, but also because it will be useful for the next article (COMING SOON). There we will see applications of limits in probability with real-life examples. 

As probability will be featured in the next article (COMING SOON), we’ll briefly present some of its basic principles through a fascinating paradox:

Suppose that a natural number (positive integer) is chosen randomly. What is the probability that 1 was chosen?

To begin with, our answer (as with any other probability question) must be a number between 0 and 1. A probability of 0 roughly means impossibility whereas 1 roughly denotes certainty (you might have seen this as 100%). This is why when there are multiple possible outcomes in an experiment, the total probability must add up to 1. For instance, when we toss a fair coin there is a ½ probability it lands head and a ½ probability it lands tails. 

Now, back to the question at hand. It is stated that the natural number is chosen randomly, which means that all numbers have the same probability of being chosen – let’s call this probability p. Now if p is positive, then the total probability adds up to infinity instead of 1. Specifically, the probability of a number between 1 and n being chosen is n × p (p added to itself n times) which will eventually exceed 1 as we increase n. On the other hand, if p = 0, then all the probabilities sum up to 0 and so we reach a contradiction again!

So, did we just find a fatal flaw in probability? Fortunately not, but we did in fact show that it is impossible (even in a hypothetical experiment) to pick a number from the naturals completely at random.

In conclusion, by improving our understanding of limits and familiarising ourselves with the basic principles of probability, we have now established the necessary background required, in order to delve deeper into applications and real-life examples. Come back soon for the third and final article (COMING SOON) where we’ll be doing just that!