m = 1 ∑ ∞ [ tan − 1 ( m 6 − 3 m 5 + 3 m 4 − m 3 + 1 3 m 2 − 3 m + 1 ) ] = ?
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As you have said "other approaches are highly welcomed!!(sic)" I think if there are any other approach.
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You are never going to know which approach can blow a seemingly 'one way solvable' question. A need for alternative can sometimes be a storehouse of greater creative insights.
Write the general term of the sum as m = 1 ∑ ∞ tan − 1 ( 1 + ( m ) 3 ( m − 1 ) 3 m 3 − ( m − 1 ) 3 )
so, now it is clear that it is a telescoping sum.
m = 1 ∑ ∞ t a n − 1 ( m ) 3 − t a n − 1 ( m − 1 ) 3
Hence,
tan − 1 ( ∞ ) 3 − t a n − 1 ( 1 − 1 ) 3 = π / 2
This is an IIT previous year question.
No its not. They guys can't copy it from me and the chances of them making exactly the same question are low.
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Actually the problem was not this
The IIT Problem was
n = 0 ∑ ∞ tan − 1 ( n 2 + n + 1 1 ) or something like this
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We have ( m 6 − 3 m 5 + 3 m 4 − m 3 + 1 3 m 2 − 3 m + 1 ) = ( m 6 − m 3 − 3 m 3 ( m 2 − m ) + 1 m 3 − m 3 + 3 m 2 − 3 m + 1 ) = ( 1 + ( m 2 − m ) 3 m 3 − [ m 3 − 1 − 3 m ( m − 1 ) ] ) = ( 1 + ( m ) 3 ( m − 1 ) 3 m 3 − ( m − 1 ) 3 ) . So that m = 1 ∑ ∞ tan − 1 ( m 6 − 3 m 5 + 3 m 4 − m 3 + 1 3 m 2 − 3 m + 1 ) = m = 1 ∑ ∞ tan − 1 ( 1 + m 3 ( m − 1 ) 3 m 3 − ( m − 1 ) 3 ) = m = 1 ∑ ∞ ( t a n − 1 m 3 − t a n − 1 ( m − 1 ) 3 ) Telescoping the sum we finally get, = tan − 1 ( ∞ ) − tan − 1 0 = 2 π
Other approaches are highly welcomed!!