lim
x-infinity 1/n(tanpi/4n +tan2pi/4n+..........tan npi/4n) equals...
is it '2pilog2'?
1is it n→∞ ? then perhaps,we cn write the rth 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
1\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
1-log | cos ax | = log ( 1/|cos ax| ) = log | sec ax |
and we both are getting same answer.. i just did it in latex to clarify :)
1Its 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 :(
