limits....

is it n→∞ ? then perhaps,we cn write the r^th term as 1/n tan rpi/4n.

so its summation from r=1 to r=n cn be converted into definite integral,by replacing 1/n with dx and r/n with x.

Lt. = ∫ dx .tan xpi/4 = -log |cos xpi/4|*4/pi,

note that its a definte integral with limits 0 to 1.

now,put the limits to get the answer: 2/pi * log2

\sum_{r=1}^{\infty}{\frac{1}{n}tan\frac{\pi r}{4n}}=\int_{0}^{1}{tan\frac{\pi x}{4}}dx

=\frac{4}{\pi }log\left( sec\frac{\pi x}{4}\right)_{0} ^{1}

=\frac{2}{\pi }log2

Excuse me,sir.But ∫ tan ax dx = - log |cos ax| / a

oh,how silly ? sorry,sir for inconvinience

Its alright yaar... i also have a habit to commit silly mistakes...

in JEE 2005 i calculated 14 x 4 = 42 and lost away 2 marks... this pushed my rank downward by 200 :(