Suppose we have a problem we would like to approach by Induction, but are not given a hypothesis that works directly. What approach should we take? In this post, learn about the technique of Stronger Induction.

You may first choose to read about Induction and Strong Induction, if you haven't already done so.

How would you use Stronger Induction to solve the following problem?

Comment below to discuss your approach.Prove that for all positive integers \(n\),

\[\frac{1}{2} \times \frac{3}{4} \times \cdots \times \frac{2n-1} {2n} \leq \frac {1} {\sqrt{3n}} \]

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## Comments

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TopNewest\[\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \ldots \times \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n}}\]

Consider: \(A_{n}=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \ldots \times \frac{2n-1}{2n}\)

Because we want to prove for all \(positive\) integer \(n\), we can say that

\[A_{n} \leq \frac{1}{\sqrt{3n+1}} \ldots (1)\]

For \(n = 1,\)

\[A_{1} = \frac{1}{2} = \frac{1}{\sqrt{3 \times 1 + 1}}\]

For \(n = 2,\)

\[A_{2} = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8} < \frac{1}{\sqrt{3 \times 2 + 1}} = \frac{1}{\sqrt{7}}\]

\(\vdots\)

For \(n = k, (1)\) holds

\[A_{k} < \frac{1}{\sqrt{3k+1}}\]

Now, prove that for \(n = k+1, (1)\) also holds

\[A_{k+1} < \frac{1}{\sqrt{3(k+1)+1}} = \frac{1}{\sqrt{3k+4}} \ldots (2)\]

Since \(A_{k+1} = A_{k} \times \frac{2k+1}{2k+2},\) then

\[A_{k+1} < \frac{2k+1}{2k+2} \times \frac{1}{\sqrt{3k+1}} \ldots (3)\]

By squares RHS in \((3)\)

\[(\frac{2k+1}{2k+2} \times \frac{1}{\sqrt{3k+1}})^2 \]

\[= \frac{(2k+1)^2}{(2k+2)^2(3k+1)} = \frac{(2k+1)^2}{(2k+1)^2(3k+4)+k} < \frac{(2k+1)^2}{(2k+1)^2(3k+4)} = \frac{1}{3k+4}\]

Which proved RHS in \((2).\) Hence,

\[A_{n} \leq \frac{1}{\sqrt{3n+1}} \leq \frac{1}{\sqrt{3n}} \]

Edit:

I just realized in this post it is mention there are \(3\) types of induction. Then what is the induction I use here? It is confusing :/

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Read the posts and you will know what type of induction this is called. Note that there are actually much more than these 3 types of induction.

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How can come up with:

\[A_{n} \leq \frac{1}{\sqrt{3n+1}} \leq \frac{1}{\sqrt{3n}}\]

So cool!

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The basic idea is that we want a function \( f(n) \geq \sqrt{3n} \) such that \[ \frac{2n+1}{2n+2} \leq \frac{ f(n) }{f(n+1)} \]

To make our life simple, we can try \( f(n) = \sqrt{3n+a} \) where \(a \) is a positive number. We can then solve this as a usual inequality, and find that \( a > \frac{6}{7} \) works if we want \( n \geq 1 \). Of course, choosing \(a= 1 \) simplifies the algebra, while still making the base case true.

As you can see, the simplification of \( f(n) = \sqrt{3n+a} \) could also be modified. Can we use \( f(n) = \sqrt{ \pi n + a } \) instead?

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Wow, that took me a very long time to solve, but it was worth it! For that exact reason I won't post my solution here, as the temptation to read it without trying long enough for yourself might be too big.

Here's my (cryptic) hint: proving something harder might be easier!

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Indeed, and as I stated:

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Oh, I'm sorry, I wasn't aware of the difference between "Stronger Induction" and "Strong Induction", I thought they linked to te same article. That does make the problem a little easier.

It's quite surprising that trying to prove the 'easy' statement is a lot harder than proving the 'hard' one.

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Well one of the best ways to figure out the induction hypothesis is to try the sum or product to a certain extent and try to see a pattern. Sometimes one can even find the pattern through finite differences.

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For example, trying out the sum of the first \(k\) odd integers seems to always give \(k^2\). This is simple to prove by induction.

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You are referring to the basic idea of induction. In this post, we are exploring scenarios where Induction doesn't immediately work, and it's uncertain if / how it could work.

In the example question that I gave, since \( \frac{1}{ \sqrt{3n} } \times \frac{ 2n+1}{2n+2} > \frac {1} { \sqrt{3n+3} } \), you would be unable to prove the induction step. What else can you do?

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Another hint would be to try to replace the RHS with a constant and try to prove the inequality.

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That wouldn't make much sense, as the smallest constant that would work is \(\frac{1}{2}\), and that doesn't really help.

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