# Calculate lim [arctg(1/(1+1+1^2) +arctg(1/(1+2+2^2)+...+arctg(1/(1+n+n^2) ] !

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Student Comments

neela | Student

Call arctg(1/(1+r+r^2)= Tr.

Summation arctg(1/(1+r+r^2)=Summation Tr; r=1,2..n.

To find limt{T1+T2+T3+.....+Tn} as n-> inf.

Tr = arctg{1/1+(r(r+1)}=arctg{((r+1)-r)/[1+(r+1)r]}

=arctg(r+1)-arctg r.= tr+1 - tr say, where tr =arctg r

Therefore, Putting r=1,2,3 ... succesively in Tr =tr+1 - tr:

T1= t2-t1

T2= t3-t2

T3=t4-t3

T4=T5-T4

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Tn= (tn+1)-(tn)

Adding , we see on the right omle -t1 and tn remains , the others getting cancelled, altertvely.

T1+T2+T3...+Tn = -t1+tn+1 = -arctg 1+arctg(n+1)

= arctg{ [-1+(n+1)]/(1-(-1)(n)}=arctg{n/((1+n)}

In limit as n->inf, ( n/n+1) = 1/(1+(1/n)) =1, which is the argument of arctg function.

Therefore, Limit(T1+T2+T3+...+Tn )= a**rctan(1)= Pi/4** is the required limit.