For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$

Question

For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$

We need to use induction.

The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:

1. $P(n_o)$
2. $( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$

Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.

Suppose, $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.

$\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$

The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.

$(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$

$\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$

$\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$

$\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$

$\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$

$\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$

We need to demonstrate that

$(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$

holds true for $n \geq 2$

The question is do I just substitute a number that is greater than two after I did the induction?

Answer 1

Without Induction: $$ \begin{align} \prod_{k=2}^n\left(1-\frac1{k^2}\right) &=\prod_{k=2}^n\left[\left(1-\frac1k\right)\left(1+\frac1k\right)\right]\\ &=\prod_{k=2}^n\left(\frac{k-1}{k}\right)\prod_{k=2}^n\left(\frac{k+1}{k}\right)\\ &=\prod_{k=1}^{n-1}\left(\frac{k}{k+1}\right)\prod_{k=2}^n\left(\frac{k+1}{k}\right)\\ &=\frac12\color{#C00000}{\prod_{k=2}^{n-1}\left(\frac{k}{k+1}\right)}\cdot\frac{n+1}{n}\color{#C00000}{\prod_{k=2}^{n-1}\left(\frac{k+1}{k}\right)}\\ &=\frac{n+1}{2n} \end{align} $$

---

Comment on the Induction: The formula for the product of $n-1$ terms is is $\frac{n+1}{2n}$.

For $n=1$, the product of $0$ terms is $1$ and that is what the formula gives.

Term $n$ is $1-\frac1{(n+1)^2}$ so the product of $n$ terms is $$ \begin{align} \frac{n+1}{2n}\left(1-\frac1{(n+1)^2}\right) &=\frac{n+1}{2n}\cdot\frac{n(n+2)}{(n+1)^2}\\ &=\frac{n+2}{2(n+1)} \end{align} $$ which is the formula for $n$ terms.

Answer 2

The principle of Induction must be used in two steps, called basis and induction step.

For basis, you must check that the theorem or identity you want to prove is true for the "starting point" $n$; in your case, for $n = 2$.

So, you must check that :

$(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$

is true (and this is simple: put $n = 2$ inside the formula and see that LHS and RHS both give you $\frac{3}{4}$).

For the induction step, you must assume (i.e.suppose) that identity holds for $n$ (we call it induction hypotheses) and prove that (under this assumption) it holds also for $n+1$.

This is made by writing the LHS up to the $(n+1)$-th term as made of the sum or product or ... (it depends on your problem; it's harder to write in words than to calculate it) of the first $n$ terms and the $(n+1)$-th.

Then you exploit the induction hypotheses and substitute into the LHS the RHS of the induction hypo, add or sum to it the $(n+1)$-th term ... and see what happens.

So, in the end, your method is the correct one; provide that you correct the typo in the third line of the induction step, you wull get the correct result, as soon as (I prefer to work with $n$ in place of $k$) you realize that :

$\frac{n+2}{2(n+1)} = \frac{(n+1)+1}{2(n+1)}$

is exactly yhe RHS of your formula for the case $n+1$.

Answer 3

My way doesn't use induction. It uses formulas $a^2-b^2=(a-b)(a+b)$ $$(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) =\\=((1- \frac{1}{2})(1- \frac{1}{3})(1- \frac{1}{4})...(1- \frac{1}{n}))((1+ \frac{1}{2})(1+ \frac{1}{3})(1+\frac{1}{4})...(1+ \frac{1}{n})) = \\=(\frac12\cdot\frac23\cdot\frac34 \ldots \cdot\frac{n-1}{n}) (\frac32\cdot\frac43\cdot\frac54 \ldots \cdot\frac{n+1}{n})= \frac{1}{n}(\frac{1}{2}(n+1))=\frac{n+1}{2n}$$

Answer 4

Here is an answer that uses logarithmic functions.

Let $$ P_n=\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right) $$ and $$ S_n=\log P_n. $$ Then \begin{eqnarray} S_n&=&\log\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)\\ &=&\sum_{k=2}^{n}\log\left(1-\frac{1}{k^2}\right)\\ &=&\sum_{k=2}^{n}\log\left(\frac{k^2-1}{k^2}\right)\\ &=&\sum_{k=2}^{n}\log\left[\left(\frac{k-1}{k}\right)\left(\frac{k+1}{k}\right)\right]\\ &=&\sum_{k=2}^{n}\left[\log\left(\frac{k-1}{k}\right)+\log\left(\frac{k+1}{k}\right)\right]\\ &=&\sum_{k=2}^{n}\left[\log(k-1)-\log(k)\right]+\sum_{k=2}^n\left[\log(k+1)-\log(k)\right]\\ &=&\sum_{k=1}^{n-1}\log(k)-\sum_{k=2}^{n}\log(k)+\sum_{k=3}^{n+1}\log(k)-\sum_{k=2}^{n}\log(k)\\ &=&\log(1)+\sum_{k=2}^{n-1}[\log(k)-\log(k)]-\log(n)+\log(n+1)+\sum_{k=3}^{n}[\log(k)-\log(k)]-\log(2)\\ &=&0+0-\log(n)+\log(n+1)+0-\log(2)\\ &=&\log\left(\frac{n+1}{2n}\right). \end{eqnarray}

Since $$ S_n=\log P_n=\log\left(\frac{n+1}{2n}\right), $$ it follows that $$ P_n=\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right) =\frac{n+1}{2n} $$