Crack the Inverse!

$\large{\arctan \frac{(n+1)-n}{1+n(n+1)}=\arctan (n+1)-\arctan (n)}$

Also $\arctan x=\text{arccot}(\frac{1}{x})$

$\large{\arctan (x)+\text{arccot} (x)=\frac{\pi}{2}}$

Now we have

$\large{\color{#D61F06}{\arctan 2}-\arctan 1+\color{#3D99F6}{\arctan 3}-\color{#D61F06}{\arctan 2}+.......+\color{#20A900}{\arctan n}-\color{#E81990}{\arctan (n-1)}+\arctan(n+1)-\color{#20A900}{\arctan n}+\text{arccot} (n+1)}$

This reduces to

$\large{\arctan(n+1)+\text{arccot} (n+1)-\arctan 1}$

$\large{\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}=\boxed{0.785}}$