tan-1(4/7) + tan-1(4/19) + tan-1(4/39) + tan-1(4/67) +...to infinite terms is? Share with your friends Share 2 Manosh T M answered this Dear student Here we we have of sum of infinte seriestan-147+tan-1419+tan-1439+...This sum can be written as ∑n=1∞tan-144n2+3Now when n reaches a finite value m that is ∑n=1mtan-144n2+3 we have ∑n=1mtan-144n2+3=tan-12m-34(m+1)+tan-134In the limit as m→∞We get limm→∞tan-12m-34(m+1)=limm→∞tan-12m-2+2-34(m+1)limm→∞tan-12m-34(m+1)=limm→∞tan-12(m+1)-2-34(m+1)⇒limm→∞tan-12m-34(m+1)=limm→∞tan-12(m+1)4(m+1)+-2-34(m+1)⇒limm→∞tan-12(m+1)4(m+1)+-2-34(m+1)=limm→∞tan-124+-2-34(m+1)Now give the limit limm→∞tan-12m-34(m+1)=tan-112Thus we have ∑n=1∞tan-144n2+3=limm→∞∑n=1mtan-144n2+3⇒∑n=1∞tan-144n2+3=limm→∞tan-12m-34(m+1)+tan-134Thus ∑n=1∞tan-144n2+3=tan-112+tan-134So the sum istan-147+tan-1419+tan-1439+...=tan-112+tan-134and the value is converging to tan-147+tan-1419+tan-1439+...≈1.1071 Regards -9 View Full Answer
