tan−113+tan−117+tan−1118+…+tan−11n2+n+1+ ..to ∞ is equal to
Let tn=tan−11n2+n+1
=tan−1(n+1)−nn2+n+1=tan−1(n+1)−n1+(n+1)n=tan−1(n+1)−tan−1(n),n=1,2,3,…n
⇒t1=tan−12−tan−11
t2=tan−13−t2
⋮ ⋮
tn=tan−1(n+1)−tan−1n.
On adding, we get
t1+t2+…tn=tan−1(n+1)−tan−11
=tan−1(n+1)−11+n+1=tan−1n2+n
As n→∞, it becomes tan−1(1)=π4,
Hence,
tan−113+tan−117+…+tan−11n2+n+1+… upto ∞=π4.