Evaluating $\tan\left(\sum_{r=1}^{\infty} \arctan\left(\frac{4}{4r^2 +3}\right)\right)$ [duplicate]

$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{4}{4r^2 +3}\right)\right)= ?$$

I wrote it in the form:

$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{\dfrac43}{\dfrac{4r^2}{3} +1}\right)\right)$$ and tried to use: $$\arctan x- \arctan y = \arctan\left(\dfrac{x-y}{1+xy}\right)$$ but that trick doesn't help here. How to go about solving this problem then?

marked as duplicate by lab bhattacharjee calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 21 '18 at 18:18

You are on the right track. Just note that $$\frac{4}{4r^2+3} =\frac{1}{r^2+3/4}=\frac{1}{1+r^2-(1/2)^2}=\frac{(r+1/2)-(r-1/2)}{1+(r+1/2)(r-1/2)}$$ and therefore $$\arctan\left(\frac{4}{4r^2+3}\right)=\arctan\left(r+\frac{1}{2}\right)- \arctan\left(r-\frac{1}{2}\right)\\=\arctan\left((r+1)-\frac{1}{2}\right)- \arctan\left(r-\frac{1}{2}\right).$$ Can you take it from here?