How you calculate this?#sum_(n=1)^oo arctan((n+1)/(n^3+1))#

1 Answer
Ratnaker Mehta
Feb 25, 2018

# pi/2#.

Explanation:

Prerequisite :

#(0) : arc tanx-arc tany=arc tan{(x-y)/(1+xy)}, (x>0,y>0)#.

We have, #(n+1)/(n^3+1)=(n+1)/{(n+1)(n^2-n+1)}#

#=1/(n^2-n+1)............[because, n ne -1]#,

#=1/{1+n(n-1)}#

Hence, if #S=sum_(n=1)^oo arc tan{(n+1)/(n^3+1)}#, then,

#S=sum_(n=1)^oo arc tan{1/{1+n(n-1)}}#,

#=sumarc tan[{n-(n-1)}/{1+n(n-1)}]#,

#=lim_(m to oo)sum_(n=1)^m{arc tann-arctan(n-1)}...[because, (0)] #,

#=lim_(m to oo)[{cancel(arc tan1)-arc tan0}+{cancel(arc tan2)-cancel(arc tan1)}+{arc tan3-cancel(arc tan2)}...+{cancel(arc tan(m-1))-cancel(arc tan(m-2))}+{arc tanm-cancel(arc tan(m-1))}]#,

#=lim_(m to oo){arc tanm-0}#.

#rArr S=sum_(n=1)^oo arc tan{(n+1)/(n^3+1)}=pi/2#.

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