Proof by Induction $\frac{n+1}{2n} = \left(1-\frac{1}{2^2}\right)\dotsc \left(1-\frac{1}{n^2}\right)$

Question

proof by induction. I'm pretty close to done with this proof but got stuck at the end

$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{n^2}\right) = \frac {n+1}{2n}$$

for all integers $n \geq 2$

1. basis step $$1-\frac{1}{2^2} = .75 \qquad \frac{2+1}{2\times2} = 3/4= .75$$

2. set $n = k$ assumed to be true $$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$

3. set $n = k+1$ $$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1}$$

from step 2 we know that

$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$

so we are going to plug that in

$$\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1} $$

now to get both sides equal we multiply by $2k$ to get common denominators

$$\frac{k+1}{2k} \times {\left(1-{1\over(k+1)^2}\right) \times 2k \over 2k}$$ This is the left hand side

now to combine it

$${\left((k+1) \left(1- {1\over(k+1)^2}\right) \times 2k \right)\over 2k}$$ - this is where I got stuck...

I wanted to cancel the $2k$'s on the top and bottom but then there would be no way to get it to equal the $(k+1)+1\over2(k+1)$ because the denominator required a $2k$. Any help is appreciated thank you.

Answer 1

You are close.

You want to show that $\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+2} $. Note that the denominator on the right is $2(k+1) = 2k+2$, not the $2k+1$ you have. That is why you were unable to complete the proof - you were trying to prove something that was false.

$\begin{array}\\ \frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) &=\frac {k+1}{2k} \times \left(\frac{(k+1)^2-1}{(k+1)^2}\right)\\ &=\frac {1}{2k} \times \left(\frac{k^2+2k+1-1}{k+1}\right)\\ &=\frac {1}{2k} \times \left(\frac{k^2+2k}{k+1}\right)\\ &=\frac {1}{2k} \times \left(\frac{k(k+2)}{k+1}\right)\\ &=\frac{k+2}{2(k+1)}\\ \end{array} $

which is what you want.

Answer 2

Why not multiply the whole thing by $2k\cdot (k+1)^2\cdot 2(k+1)$ and see if you get an identity (after correcting the denominator of the RHS) ...