Prove that $\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$

Question

Prove $$\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$$ for all n greater or equal to 2.

First of all, I'm well aware this exact question exists here. Unfortunately, this post didn't help me, and is much too old for me to resurrect. For reference, I'm a undergraduate college student, in an introductory proof writing course.

Currently, I've made the following progress:

We wish to show that $\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$ for every integer n greater than or equal to 2. We will proceed via induction.

For the base step, let n be equal to 2. Then, $1 – (1/2^2) = ¾$, and $(2+1)/2(2) = ¾$. Therefore, $\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$ at n = 2.

Next, assume that $\prod_{i=2}^k (1-1/i^2) = {k+1\over 2k}$ up until some integer k. We wish to show that this is true at k+1 as well.

$\prod_{i=2}^k (1-1/i^2) = {k+1\over 2k}$ can be written as ${(k+1)-1\over k} * {(k+1)+1\over k}$, or ${k\over k} * {k+2\over k}$.

k/k is 1, of course. but I'm not sure where to go from k+2/k. Or maybe I made some other mistake? I can't decipher user17762's comment under his answer to the linked question, nor the OP's edit after the "so". Any guidance is appreciated!

Answer 1

You have an algebra mistake: $\frac{k+1}{2k}$ cannot be written as $\frac kk \frac{k+2}k$.

You've established the base step for $n=2$. To finish the proof by induction, first you assume the claim holds for $n=k$, so substitute $k$ in place of $n$ in the assertion: $$\prod_{i=2}^{k}\left(1-\frac1{i^2}\right) = \frac{k+1}{2k}\tag1$$ You must now show that the claim holds for $n=k+1$: $$\prod_{i=2}^{k+1}\left(1-\frac1{i^2}\right) = \frac{(k+1)+1}{2(k+1)}\tag2$$ The key is to observe that the LHS of (1) lives inside of the LHS of (2): $$\prod_{i=2}^{k+1}\left(1-\frac1{i^2}\right)=\left[\prod_{i=2}^{k}\left(1-\frac1{i^2}\right)\right] \cdot \left(1-\frac1{(k+1)^2}\right)\tag3$$ In (3), the quantity in square brackets equals $\frac{k+1}{2k}$, by assumption (1). Given this, can you obtain the RHS of (2)?

Answer 2

For induction you need to multiply $\frac{n+1}{2n}$ by $1-\frac{1}{(n+1)^2}=\frac{n(n+2)}{(n+1)^2}$. What do you mean by $k$?

Without induction: By using $1-1/i^2=\frac{(i-1)}{i}\times\frac{(i+1)}{i}$,

$$\prod_{i=2}^n(1-1/i^2)=\frac{1}{2}\times\underline{\left[\frac{3}{2}\frac{2}{3}\times\frac{4}{3}\frac{3}{4}\times\frac{5}{4}...\frac{(n-3)}{(n-2)}\times\frac{(n-1)}{(n-2)}\frac{(n-2)}{(n-1)}\times\frac{(n)}{(n-1)}\frac{(n-1)}{n}\right]}\times\frac{(n+1)}{n}$$ Hence, we have $$\prod_{i=2}^n(1-1/i^2)=\frac{n+1}{2n}$$

Answer 3

for reference, I'm an economist so don't mind if my notation is not a precise one. However, I'm usually trying to keep up with the mathematicians. Let us begin with your proof. We'll do the proof by induction.

1. The case where $k=2$, we have that $1-\frac{1}{2^2}=\frac{3}{4}=\frac{1+2}{2(2)}$
2. Let $n\rightarrow k$, with this we make the following inductive assumption:

\begin{equation} \prod_{i=1}^{k}\left(1-\frac{1}{i^2}\right)=\frac{k+1}{2k} \end{equation} 3. Let us now check whether $P(k)\Longrightarrow P(k+1)$, which we'll do in the following way:

\begin{equation} \prod_{i=1}^{k+1}\left(1-\frac{1}{i^2}\right)=\underbrace{\frac{k+1}{2k}}_{\prod_{i=1}^{k}\left(1-\frac{1}{i^2}\right)}\left(1-\frac{1}{(k+1)^2}\right)=\\ \frac{k+1}{2k}\left(\frac{(k+1)^2-1}{(k+1)^2}\right)=\\ \frac{1}{2k}\left(\frac{(k+1)^2-1}{(k+1)}\right)=\\ \frac{1}{2k}\left(\frac{(k+1-1)(k+1+1)}{(k+1)}\right)=\\ \underbrace{\frac{k+2}{2(k+1)}}_{P(k+1)}\\ \therefore P(k)\Longrightarrow P(k+1) \end{equation}