Help with a Mathematical Induction Problem

In summary, the conversation discusses the use of mathematical induction to prove an inequality involving fractions. The basis step is shown to work, but the inductive step fails. It is then suggested to use mathematical induction to prove a stronger inequality, where the inductive step is successful. However, further steps are needed to prove this second inequality.
  • #1
Walshy1
3
0
Suppose we want to prove that: 1/2 * 3/4 ... 2n-1/2n < 1/sqrt(3n)

for all positive integers.
(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: 1/2 * 3/4 ... 2n-1/2n < 1/sqrt(3n+1)

So far I have proven the basis step works in part a by plugging in 1, however, I do not know how to say the inductive step fails. I have no clue on part b. Thanks.
 
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  • #2
The inductive step would require you to show that

$$\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \cdot \frac{2n+1}{2(n+1)} < \frac{1}{\sqrt{3(n+1)}}.$$

Following the inductive step, we know that

$$\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n-1}{2n} < \frac{1}{\sqrt{3n}},$$

therefore we can conclude that

$$\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \cdot \frac{2n+1}{2(n+1)} < \frac{1}{\sqrt{3n}} \cdot \frac{2n+1}{2(n+1)}.$$

Now do a simple test: take $n=1$. Our expression yields $\approx 0.43$, whereas the proposed one generates $\approx 0.40$. Our inductive step fails! :p

You should proceed the same way for the next item, bearing in mind that now the inequality will be true. ;)
 

FAQ: Help with a Mathematical Induction Problem

1. What is mathematical induction?

Mathematical induction is a method of proving a statement or formula for all natural numbers by showing that it holds for the smallest number (usually 0 or 1) and then proving that if it holds for any number, it also holds for the next number.

2. How do I know when to use mathematical induction?

Mathematical induction is typically used to prove statements or formulas that involve natural numbers, such as those in algebra or number theory. It is also useful for proving properties of sequences and series.

3. What are the steps for solving a mathematical induction problem?

The steps for solving a mathematical induction problem are: 1) Prove that the statement holds for the base case (usually n = 0 or 1), 2) Assume that the statement holds for some arbitrary number k, 3) Use this assumption to prove that the statement also holds for the next number, k+1, and 4) Conclude that the statement holds for all natural numbers using the principle of mathematical induction.

4. Can mathematical induction be used to prove all statements?

No, mathematical induction can only be used to prove statements that involve natural numbers. It cannot be used to prove statements about real numbers or other mathematical objects.

5. What are some common mistakes to avoid when using mathematical induction?

Some common mistakes to avoid when using mathematical induction include: 1) Assuming that the statement holds for all natural numbers without proving it, 2) Using the wrong base case, 3) Making incorrect assumptions about k+1 based on the assumption for k, and 4) Using circular reasoning.

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