Product

We need to find: $\prod_{k=1}^{2015}\Bigg(1-\frac{1}{(k+1)^2}\Bigg)$ We rewrite as follows: $\prod_{k=1}^{2015}\Bigg(\frac{(k+1)^2-1}{(k+1)^2}\Bigg)$ Notice that the numerator can be factored as a difference of squares: $\prod_{k=1}^{2015}\Bigg(\frac{((k+1)+1)((k+1)-1)}{(k+1)^2}\Bigg)=\prod_{k=1}^{2015}\Bigg(\frac{k(k+2)}{(k+1)^2}\Bigg)$ Now we write out some terms: $\prod_{k=1}^{2015}\Bigg(\frac{k(k+2)}{(k+1)^2}\Bigg)=\Bigg(\frac{1\cdot 3}{2^2}\Bigg)\cdot \Bigg(\frac{2\cdot 4}{3^2}\Bigg)\cdot...\cdot \Bigg(\frac{2015\cdot 2017}{2016^2}\Bigg)$ It follows that the product is: $\Bigg(\frac{1\cdot 3}{2^2}\Bigg)\cdot \Bigg(\frac{2\cdot 4}{3^2}\Bigg)\cdot...\cdot \Bigg(\frac{2015\cdot 2017}{2016^2}\Bigg)=\frac{(1\cdot 2\cdot...\cdot 2015)(3\cdot 4\cdot...\cdot2017)}{(2\cdot 3\cdot...\cdot2016)^2}=\frac{(2015!)(\frac{2017!}{2})}{(2016!)^2}$ We finally reduce this to: $\frac{(2015!)(\frac{2017!}{2})}{(2016!)^2}=\frac{2017}{4032}$ So we conclude: $\prod_{k=1}^{2015}\Bigg(1-\frac{1}{(k+1)^2}\Bigg)=\boxed{\frac{2017}{4032}}$

Notice that every fraction in the form $(1-\frac{1}{k^2})$ can be written as $(1-\frac{1}{k})(1+\frac{1}{k})$ . Expanding for the first few yields $\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{2016}\right)\left(1+\frac{1}{2016}\right)$ $=\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{2}{3}\right)\left(\frac{4}{3}\right)...\left(\frac{2015}{2016}\right)\left(\frac{2017}{2016}\right)$ Note that all consecutive terms starting from the second cancels out, so we are left with $\left(\frac{1}{2}\right)\left(\frac{2017}{2016}\right)$ = $\boxed{\frac{2017}{4032}}$