Inverse Squares Product

$the\: given\: equation\: is\: f(n)=\prod_{i=1}^{n}\left ( 1-\frac{1}{(i+1)^{2}} \right )$

$Now,$ $f(1999)=\prod_{i=1}^{1999}\left ( 1-\frac{1}{(i+1)^{2}} \right )$ $=\prod_{i=1}^{1999}\left ( \frac{(i+1)^{2}-1}{(i+1)^{2}} \right )$ $=\prod_{i=1}^{1999}\left ( \frac{i^{2}+2i}{(i+1)^{2}} \right )$ $=\prod_{i=1}^{1999}\left ( \frac{i(i+2)}{(i+1)^{2}} \right )$ $=\prod_{i=1}^{1999}\left ( \frac{i}{i+1}\times\frac{i+2}{i+1} \right )$ $=\prod_{i=1}^{1999}\left ( \frac{i}{i+1} \right ) \times \prod_{i=1}^{1999}\left ( \frac{i+2}{i+1} \right )$

$now,\: the\: first\: term,$ $\prod_{i=1}^{1999}\left ( \frac{i}{i+1} \right ) = \frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times ... \times \frac{1998}{1999} \times \frac{1999}{2000}$ $=\frac{1}{2000} \: [Becuase\: each\: term\: telescopes]$

$again,\: the\: second\: term,$ $\prod_{i=1}^{1999}\left ( \frac{i+2}{i+1} \right ) = \frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times ... \times \frac{2000}{1999} \times \frac{2001}{2000}$ $=\frac{2001}{2} \: [Becuase\: each\: term\: telescopes]$

$Thus\: we\: get,$ $\prod_{i=1}^{1999}\left ( \frac{i}{i+1} \right ) \times \prod_{i=1}^{1999}\left ( \frac{i+2}{i+1} \right )=\frac{1}{2000}\times \frac{2001}{2}$ $=\frac{2001}{4000}$ $=\boxed{0.50025}$

Try the product for the first few terms. You will notice that the result is $f(n) = \frac {n+2}{2(n+1)}.$ Hence exact answer is 2001/4000.

We'll show that f(n) = (n+2)/(2(n+1)) and then let n = 1999 to get our answer of 0.50025.

First let's look at a single factor of our product series: (1 - 1/(i+1)^2) = ((i+1)^2 - 1)/(i+1)^2 = (i(i+2))/(i+1)^2 .

Now let's have a look at what our product series looks like with this substitution : f(n) = ((1(3))/(2^2)) ((2(4))/3^2) ... (((n-1)(n+1))/n^2) ((n(n+2))/(n+1)^2) = (n! * ((n+2)!)/2)/(n+1)!^2 =1/2 * n!/(n+1)! * (n+2)!/(n+1)! = 1/2* 1/(n+1) * (n+2) = (n+2)/(2(n+1)).

Plugging in 1999 into f(n), we have : f(1999) = (2001)/(2*(2000)) = 0.50025.

It's very interesting that f(n) has the form (1/2) GAMMA(n+1.) GAMMA(n+3.)/GAMMA(n+2.)^2

Let there be k terms .The answer is [{(k/2)+1}/k+1]

After rewriting the difference of squares as $(\frac{n + 2}{n + 1})(\frac{n}{n + 1})$ we evaluate the first terms getting this:

$(\frac{3}{2})(\frac{1}{2})(\frac{4}{3})(\frac{2}{3})(\frac{5}{4})(\frac{3}{4}) \ldots$ .

The idea is to note that for an specific $n$ , the only terms that will not cancel are $\frac{1}{2}$ and the penultimate term (excepting $n = 1$ , case in which both terms don't cancels). So for the case $n = 1999$ we're left with the terms $(\frac {1}{2})(\frac{2001}{2000})$ . Finally,

$f(1999) = \frac{2001}{4000} = \boxed{0.50025}$ .