i am getting π
\hspace{-16}\bf{\lim_{n\rightarrow \infty}\frac{1}{n}\tan\left\{\sum_{k=1}^{4n+1}\tan^{-1}\left(1+\frac{2}{k.(k+1)}\right)\right\}=}
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\hspace{-16}\bf{\lim_{n\rightarrow \infty}\frac{1}{n}\tan\left\{\sum_{k=1}^{4n+1}\tan^{-1}\left(1+\frac{2}{k.(k+1)}\right)\right\}=}