Using Induction to prove a product series equality?

Question

I'm getting confused by these types of problems where "n" appears in the general term:

Use Induction to prove: $$\prod_{i=1}^n \frac{n+i}{2i-3}= 2^n(1-2n)$$

Would my P(n+1) be this?:

$\mathbf P(n+1):$ $$\prod_{i=1}^{n+1} \frac{n+1+i}{2i-3}= 2^{n+1}(1-2(n+1))$$

If this is the case, I dont know how to use the Inductive hypothesis to prove it since the product sequence is now a "different" sequence.

How would I go about solving this?

Many thanks in advance.

Edit: the index was indeed wrong, thanks T Ford, changed it to i=1

Answer 1

Here is a go: $$\prod_{i=0}^n\frac{n+i}{2i-3}=\frac{\prod_{i=0}^nn+i}{\prod_{i=0}^n2i-3}$$ Firstly: $$\prod_{i=0}^nn+i=n(n+1)(n+2)(n+3)...(2n)=\frac{(2n)!}{(n-1)!}$$ and: $$\prod_{i=0}^n2i-3=(-3)(-1)(1)(3)=3*n!!$$ so: $$\prod_{i=0}^n\frac{n+i}{2i-3}=\frac{(2n)!}{3(n-1)!*n!!}$$ since in this case our double factorial is odd, $$n!!=\frac{(n+1)!}{2^{(n+1)/2}((n+1)/2)!}$$ giving: $$\prod_{i=0}^n\frac{n+i}{2i-3}=\frac{(2n)!}{(n-1)!}*\frac{3*2^{(n+1)/2}((n+1)/2)!}{(n+1)!}$$ Not sure if this is on the right path but its my attempt

Answer 2

Your stated equality is false. In order to fix this the index $i$ has to start with $1$ instead of $0$ otherwise the RHS will have a slightly different shape. However, first of all I have to say that this problem is quite unfitting to be attack via induction, at least in the way you tried to do. More convenient is the way J.R. proposed here.

Using the idea of Henry Lee we may rewrite the product as the following

$$\prod_{i=0}^n\frac{n+i}{2i-3}=\frac{\prod\limits_{i=0}^nn+i}{\prod\limits_{i=0}^n2i-3}$$

The upper product can be intepreted as Pochhammer Symbol whereas the denominator can be rewritten as double factorial. Using these functions we further get

$$\frac{\prod\limits_{i=0}^nn+i}{\prod\limits_{i=0}^n2i-3}=\frac{(n)_{n+1}}{(-3)(-1)(2n-3)!!}=\frac13\frac{(2n)!}{(n-1)!}\frac{2^{n-1}(n-1)!}{(2n-2)!}=\frac{2^{n-1}}{3}(2n)(2n-1)$$

$$\therefore~\prod_{i=0}^n\frac{n+i}{2i-3}~=~\frac{2^n}{3}n(2n-1)$$

The same technique can be applied for $i=1$ as start index. From hereon we get

$$\prod_{i=1}^n\frac{n+i}{2i-3}=\frac{\prod\limits_{i=1}^nn+i}{\prod\limits_{i=1}^n2i-3}=\frac{(n+1)_{n+1}}{(-1)(2n-3)!!}=-\frac{(2n+1)!}{n!}\frac{2^{n-1}(n-1)!}{(2n-2)!}=-2^{n-1}(2)(2n-1)$$

$$\therefore~\prod_{i=0}^n\frac{n+i}{2i-3}~=~2^n(1-2n)$$