Sometimes we cannot use mathematical induction to prove a...

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise $74-76$ . Suppose that we want to prove that $$ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n}} $$ for all positive integers $n .$ a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. b) Show that mathematical induction can be used to prove the stronger inequality $$ \frac{1}{2} \cdot \frac{3}{4} \dots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n+1}} $$ for all integers $n$ greater than $1,$ which, together with a verification for the case where $n=1$ , establishes the weaker inequality we originally tried to prove using mathematical induction.

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise $74-76$ .
Suppose that we want to prove that
$$
\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n}}
$$
for all positive integers $n .$
a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails.
b) Show that mathematical induction can be used to prove the stronger inequality
$$
\frac{1}{2} \cdot \frac{3}{4} \dots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n+1}}
$$
for all integers $n$ greater than $1,$ which, together with a verification for the case where $n=1$ , establishes the weaker inequality we originally tried to prove using mathematical induction.

Key Concepts

Mathematical Induction

Mathematical induction is a fundamental method of proving statements that are formulated for all natural numbers. It consists of verifying the base case and then proving that if the statement holds for an arbitrary natural number n, it must also hold for n + 1, thereby establishing the truth of the statement for all natural numbers.

Basis Step

The basis step in an inductive proof involves verifying that the statement is true for the initial value, often n = 1. This step lays the foundation for the induction process by ensuring that the property holds at the starting point.

Inductive Step

The inductive step involves assuming that the statement is true for a particular natural number n (called the inductive hypothesis) and then using this assumption to prove that the statement must also be true for n + 1. This step is critical because it shows the propagation of truth from one case to the next.

Inductive Loading

Inductive loading is a technique used when a direct inductive proof of a given statement fails. In this strategy, one proves a stronger result than originally intended. The stronger inductive hypothesis provides additional leverage needed in the inductive step, and since the stronger statement implies the original weaker statement, it effectively proves the original claim.

Stronger Inductive Hypothesis

A stronger inductive hypothesis is an enhanced version of the original statement that is proved by induction when the original form is too weak to support the inductive step. By proving a more robust inequality or property, one can overcome obstacles in the proof process, and this stronger result ensures the validity of the original claim as a corollary.

Transcript

00:01 We are asked to explain why the first form of argument, if this is p of n, cannot be proved using induction, but the second form can.

00:16 And essentially, we change just 3n to 3n plus 1, right? a minor chain, but it affects how the proof works.

00:28 And we're going to show that the inductive step is fail kind of fail in the first case but it works in the second case so by that you will see in a minute first oh we we just skip basic step because they both five right first look at the inductive step usually when we have the information from p of n here so the first first term in the point product is from p of n's right and we multiply this last term to both side to make to make the left hand side what we want and then we what we want to do here what's left is to have this inequality right less than 1 over 3n plus 3 which is the argument which is the right -hand side of pn plus 1 argument if we have if we have this then everything will work and the inductive step will be able to do it i will show that this will lead to a problem so the last inequality give rise to something that is not true so sorry first we move move this there and move the second term to the other side so we arrive at this we square both side since they are positive there's no problem there so left -hand side is left with 1 plus 1 over n.

02:15 Right -hand side we can square this out and become this.

02:20 Now we queue 1, right, both sides.

02:23 And we move 2 over 2n plus 1 to the left -hand side and just add subtract the fraction.

02:34 You will find that we arrive at 1 over n times 1 over 1.

02:42 2n plus 1 is less than 1 over 2 n plus 1 square but this cannot happen why because it's implied that 1 over n is less than 1 over 2 n plus 1 which is clearly not the case because 2n plus 1 is greater than n so it should be the other way around and so we cannot we cannot have this this this part that we want and the proof kind of doesn't work.

03:22 So let's see what happened when we change the argument here.

03:30 We're still doing the same thing.

03:32 The first chunk is from p of n and we multiply the new term here on both sides.

03:39 We want this part right and this be 3n plus 4.

03:45 Now let's look at this inequality.

03:51 We did the same thing.

03:52 We moved through 3n plus 4 to the other side...