How should you prove product rules by induction?

Question

For example: $$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$ For every $n$ greater than or equal to $2$

my approach for this was that I need to prove that: $$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$

is this the right approach? Because when i try and work out the algebra i keep on hitting a wall.

\begin{align} \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)&=1-\frac 1{(1-n)^2}-\frac 1{n^2}-\frac 1{n^2(n+1)} \\ &=\frac{n^2}{(n+1)^2}-\frac 1{(n+1)^2} \\ &=\frac{n^2-1}{(n+1)^2}-\frac 1{n^2}+\frac 1{n^2(n+1)^2} \\ &=\frac{n^2(n^2-1)}{n^2(n+1)^2}-\frac{(n+1)^2}{n^2(n+1)^2} \\ &=\frac{n^2(n^2-1)-(n+1)^2}{n^2(n+1)^2}+\frac 1{n^2(n+1)^2} \\ &=\frac{n^2(n^2-1)-(n+1)^2+1}{n^2(n+1)^2} \end{align}

Answer 1

Check that the equation holds for $n=2$. Assuming that it holds for some $n$:

$$\begin{align}\prod_{i=2}^n\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\\ \end{align}$$

And so

$$\begin{align} \prod_{i=2}^{n+1}\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\left( 1-\frac1{(n+1)^2}\right)\\ &=\frac{n+1}{2n}-\frac{n+1}{2n(n+1)^2}\\ &=\frac{(n+1)^2}{2n(n+1)}-\frac{1}{2n(n+1)}\\ &=\frac{(n+1)^2-1}{2n(n+1)}\\ &=\frac{n^2+2n}{2n(n+1)}\\ &=\frac{n+2}{2(n+1)}\\ \end{align}$$

If the equation holds for some $n$, it also holds for $n+1$. Since it holds for $2$, it also holds for $3$, and since it holds for $3$, it also holds for $4$, and so on. The statement is true for all natural numbers greater than or equal to $2$.

Answer 2

Hint:

$$P_{n-1}\left(1-\frac1{n^2}\right)=\frac{(n-1)+1}{2(n-1)}\left(1-\frac1{n^2}\right)=\frac{n+1}{2n}=P_n.$$

Answer 3

$\begin{aligned}& \prod_{2 \le k \le n} \bigg(1-\frac{1}{k^2} \bigg) = \exp\bigg[\ln{\prod_{2\le k\le n}\bigg(1-\frac{1}{k^{2}}\bigg)}\bigg] = \exp\bigg[\sum_{2 \le k \le n} \ln{\bigg(\frac{k^2-1}{k^2} \bigg)}\bigg] = \exp(S_{n}). \\& \begin{aligned} & \begin{aligned} S_{n} & = \sum_{2 \le k \le n}\ln(k-1)+ \sum_{2 \le k \le n}\ln(k+1)-2 \sum_{2 \le k \le n}\ln(k) =\sum_{2 \le k+1 \le n}\ln(k)+ \sum_{2 \le k-1 \le n}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) \\& = \sum_{1 \le k \le n-1}\ln(k)+ \sum_{3 \le k \le n+1}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) = \ln(1)-\ln(n)-\ln(2)+\ln(n+1)\\& +\sum_{2 \le k \le n}\ln(k)+ \sum_{2 \le k \le n}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) = \ln\bigg(\frac{1}{2}\bigg)+\ln\bigg(1+\frac{1}{n}\bigg) .\end{aligned} \\& \begin{aligned} \therefore \prod_{ 2 \le k \le n} \bigg(1-\frac{1}{k^2} \bigg) = \exp\bigg[\ln\bigg(\frac{1}{2}\bigg)+\ln\bigg(1+\frac{1}{n}\bigg)\bigg] = \frac{n+1}{2n}. \end{aligned} \end{aligned} \end{aligned} $

Answer 4

Hint $\ $ No ingenuity is required: by telescopy the proof reduces to this one-line calculation

$\qquad\ \ $ if $\rm\, \ f(k) = \dfrac{k\!+\!1}{2k\ }\ $ then $\rm\,\ \dfrac{f(k)}{f(k\!-\!1)} =\, \dfrac{k\!+\!1}{2k\ }\dfrac{2(k\!-\!1)}{k\, }\,=\,\dfrac{k^2\!-\!1}{k^2}\, =\, 1-\dfrac{1}{k^2}\,\ $ thus

Multiplicative Telescopy $\ \ \rm\displaystyle f(a\!-\!1) \prod_{\large k\,=\,a}^{\large n} \dfrac{f(k)}{f(k-1)}\, =\ f(n) $

Proof $ $ Induct on $\rm\,n.\,$ Base is $\rm\, f(a\!-\!1)\frac{f(a)}{f(a-1)}=\,f(a)\,$ at $\rm\,n\!=\!a.\,$ Induction $\rm\,\color{#0a0}{P(n)}\Rightarrow\, P(n\!+\!1)\,$ is

$\quad\ \displaystyle\rm f(a\!-\!1)\prod_{\large k\,=\,a}^{\large n+1}\dfrac{f(k)}{f(k\!-\!1)}\, =\, \left[f(a\!-\!1)\prod_{\large k\,=\,a}^{\large n}\dfrac{f(k)}{f(k\!-\!1)}\right]\dfrac{f(n\!+\!1)}{f(n)}\,\overset{\rm\color{#0a0}{ P(n)}} =\, \color{brown}{f(n)}\dfrac{f(n\!+\!1)}{\color{brown}{f(n)}} \, =\, f(n\!+\!1) $

Remark $\ $ Unwinding the induction yields a vivid depiction of the telescopic cancellation

$\quad \rm\displaystyle f(a\!-\!1)\prod_{\large k\,=\,a}^{n} \frac{f(k)}{f(k\!-\!1)}\, = \ \frac{\color{#c00}{\rlap{---}f(a\!-\!1)}}{1}\frac{\color{green}{\rlap{--}f(a)}}{\color{#C00}{\rlap{---}f(a\!-\!1)}}\frac{\color{royalblue}{\rlap{---}f(a\!+\!1)}}{\color{green}{\rlap{--}f(a)}}\frac{\phantom{\rlap{--}f(3)}}{\color{royalblue}{\rlap{---}f(a\!+\!1)}}\, \cdots\, \frac{\color{brown}{\rlap{---}f(n\!-\!1)}}{\phantom{\rlap{--}f(n\!-\!1)}}\frac{f(n)}{\color{brown}{\rlap{---}f(n\!-\!1)}}\, =\ \frac{f(n)}{1} $

You can find many further examples of multiplicative telescopy in other posts here.