If you want a headache

$\displaystyle \lim_{n \to \infty} n \prod_{m=1}^n \Bigg( 1-{1\over m} + {5\over 4{m^2}}\Bigg)$ $\displaystyle \lim_{n\to \infty} {1\over {n{((n-1)!^2)}}} \prod_{k=1}^n \Bigg[ 1+ \Big( k - {1\over 2} \Bigg)\Bigg]$ $\displaystyle \lim_{n\to \infty} {1\over {n{((n-1)!^2)}}} \prod_{k=1}^n \Bigg( k-{1\over 2} +i \Bigg) \prod_{k=1}^n \Bigg( k - {1\over2} - i \Bigg)$ $\displaystyle \lim_{n\to \infty} {1\over {n{((n-1)!^2)}}} \prod_{k=0}^{n-1} \Bigg( k+{1\over 2} +i \Bigg) \prod_{k=0}^{n-1} \Bigg( k + {1\over2} - i \Bigg)$ $z_1 ={1\over 2} + i$ $z_2= {1\over 2} -i$ $\displaystyle \lim_{n\to \infty} {1\over {n{((n-1)!^2)}}} \prod_{k=0}^{n-1} \Bigg( k+ z_1 \Bigg) \prod_{k=0}^{n-1} \Bigg( k + z_2 \Bigg)$ $\displaystyle \frac {\lim_{n\to \infty} \Bigg( { n^{-1}} {(n-1)^{z_1}} {(n-1)^{z_2}} \Bigg)}{\lim_{n\to \infty}{{ (n-1)! {(n-1)^{z_1}}}\over \prod_{k=0}^{n-1} ({z_1}+ k )} \frac{(n-1)! {(n-1)^{z_2}}}{ \prod_{k=0}^{n-1} ({z_2}+ k )}}$ $1\over \Gamma {2i+ 1\over 2} \Gamma { 2i+ 1\over 2}$ $\cos(i\pi)\over \pi$ $cosh(-\pi)\over \pi$ = 3.689 ..... $i = \sqrt{-1} \text{ and } \Gamma \text{ is the Gamma function}$

And the problem is solved :)