# Describe series behaviour:`sum_(n=1)^oo (tg[(pi/4)(1 +2n)])/n`

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### 2 Answers

Given series is `sum_(n=1)^oo(tg((pi/4)(1+2n)))/n` .

which can be re-written as `sum_(n=1)^oo(tg((pi/4)+(pi/4)(2n)))/n`

or, `sum_(n=1)^oo(tg((pi/4)+(npi/2)))/n`

On expanding this series we get

`sum_(n=1)^oo(tg((pi/4)+(npi/2)))/n=(tg((pi/4)+(pi/2)))/1+(tg((pi/4)+(2pi/2)))/2+(tg((pi/4)+(3pi/2)))/3+(tg((pi/4)+(4pi/2)))/4+............`

`=-1+(1/2)-(1/3)+(1/4)-................`

Which is an alternating series whose terms are monotonically decreasing. So given series is convergent.

sum_(n=1)^oo (tg[(pi/4)(1 +2n)])/n

`sum_(n=1)^ooa_n` where

`a_n=(tan(((2n+1)pi)/4))/n=(tan((npi)/2+pi/4))/n`

`a_1=-1`

`a_2=1/2`

`a_3=-1/3`

`a_4=1/4`

`` Thus we have a series

`sum_(n=1)^ooa_n=a_1+a_2+a_3+a_4+............. to oo`

`=-1+1/2-1/3+1/4-1/5+........`

which is an alternating ,monotonically decreasing series.So by alternating series test ,it will converge.