Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$

Question

Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$$

How to start the proof by induction? Is there any way to show this?

Answer 1

Base case:

For $n=2$, $$LHS=(1-\frac{1}{4})=\frac{3}{4}$$ $$RHS=\frac{(2+1)}{2\cdot 2}=\frac{3}{4}$$

Hence for $n=2$, the equality holds.

Induction hypothesis:

Let it hold for some $n=k$. Then, $$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right) = \frac{k + 1}{2k}$$

For $n=k+1$, $$LHS=\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right)\left(1 - \frac{1}{(k+1)^2}\right)$$

Inductive step:

Using the induction hypothesis, for $n=k+1$, $$LHS=\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right)\left(1 - \frac{1}{(k+1)^2}\right)=\frac{k + 1}{2k}\left(1 - \frac{1}{(k+1)^2}\right)$$ $$=\frac{(k+1)^2-1}{2k(k+1)}=\frac{(k^2+2k)}{2k(k+1)}=\frac{k+2}{2(k+1)}=\frac{(k+1)+1}{2(k+1)}=RHS$$

Thus, whenever the equality holds for $n=k\ge2$, it also hold for $n=k+1$.

Using the principle of mathematical induction, it holds for all $n\ge2$.

Answer 2

Let $$P = \prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) =\prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \left(1+\frac{1}{r}\right) $$

So $$\prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) = \prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \prod^{n}_{r=2}\left(1+\frac{1}{r}\right)$$

Now Open These two products.

Answer 3

How to start a proof by induction?

What a straight line!

You start at the beginning.

...

Then you show that the middle flows, and conclude that the end is inevitable.

But seriously... You start with $n = 2$.

Prove that $(1 - 1/4) = (2+1)/2*2$.

That's the initial or base step.

Then you do the induction step. You prove that if $(1-1/4)....(1-\frac{1}{k^2}) = \frac{k+1}{2*k}$ for some $k$, then it follows that $(1-1/4) .....(1-\frac{1}{(k+1)^2}) = \frac{k + 2}{2(k+1)}$

And then you are done. It follows that if it is true for $n =2$, and you've proved that if it is true for $n = k$ then it must be true for $n = k + 1$, you have proven it is true for all $n \ge 2$.

The heart is showing that if $(1-1/4)....(1-1/k^2 = (k+1)/2*k$ for some $k$, then it follows that $(1-1/4) .....(1-\frac{1}{(k+1)^2}) = \frac{k + 2}{2(k+1)}$

For that note $(1-1/4) .....(1 - \frac{1}{k^2})(1-\frac{1}{(k+1)^2}) = [![![![![(1-1/4) .....(1 - \frac{1}{k^2})]!]!]!]!](1-\frac{1}{(k+1)^2}) = [\frac{k+1}{2*k}][1-\frac{1}{(k+1)^2})]$

Can you prove that that equals $\frac{k + 2}{2(k+1)}$?

Answer 4

Here is an inductive proof:

• For $n=2$ we get $$1-\frac{1}{2^{2}}=\frac{2+1}{2*2}=\frac{3}{4}$$

• Let $n=k>2$ and $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})=\frac{k+1}{2*k}$$ and now the induction step:

• Let $n=k+1$, then $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})(1-\frac{1}{(k+1)^{2}})=(\frac{k+1}{2*k})(1-\frac{1}{(k+1)^{2}})= \\ =(\frac{k+1}{2*k})\frac{(k+1)^{2}-1}{(k+1)^{2}}=(\frac{1}{2*k})\frac{k(k+2)}{(k+1)}=\frac{1}{2}\frac{(k+2)}{(k+1)}=\frac{(k+1)+1}{2*(k+1)}$$ thus the proof is complete.