# Help with a Mathematical Induction Problem

#### Walshy

##### New member
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.

#### Fantini

##### "Read Euler, read Euler." - Laplace
MHB Math Helper
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!

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