1how come denominator is 3+2 ???
given
S = tan-1(1/3)+tan-1(1/5)+tan-1(1/7)+..........+tan-1(1/(2n-1))
if S = Î /4
find n
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30 Answers
62Only debotosh can answer his logic to #2.. but I think what he has done is exactly what I was doing (earlier)
he must have ended up with \lim_{n->\infty}tan^{-1}(n)-tan^{-1}1 = \pi/2-\pi/4
21teh question is rong i guess or atleast it needs some improvement
a simple usage of calculator wud tell u.
1debotosh's answer is definitely wrong as
we can clearly see fron nishant bhai post #22 that the seriess
diverges
tan-11-tan-12/3 =finite
tan-11-tan-13/4 =finite
.
.
.
so sum of series goes to ∞ as n→∞
i think we have to prove
that we can make n finetly large to get that value
i think we have to nest general term b/w two inequalities whioch in turn gives telescoping summation
just as we do in case of Σ1√k
23sir i still feel that there is something more here ..... or is it dat i shud take rest [2]
62\\tan^{-1}\frac{1-1/2}{1+1.1/2}+tan^{-1}\frac{1-2/3}{1+2/3.1}+tan^{-1}\frac{1-3/4}{1+3/4.1}+.... \\=(tan^{-1}1-tan^{-1}1/2)+(tan^{-1}1-tan^{-1}2/3)+(tan^{-1}1-tan^{-1}3/4)+...
Now I hope sanity has returned :P
13OMG...It seems tht this thread was completely wiped-out frm my memory after 14-Dec...never bothered 2 look in.
Gud work tracking it dwn RPF, but i had never thot/done anything beyond #8... :P
62dude.. this person went mad on "#9 Posted 00:03am 15-12-09"
and was mad until he saw "#16 Posted 6:08pm 03-05-10"
This person was never seen sane in between.... After that he realized his mistake and apologizes [3]
62Where's the issue?
\\tan^{-1}\frac{2-1}{2+1}+tan^{-1}\frac{3-2}{3+2}+tan^{-1}\frac{4-3}{4+3}+.... \\=(tan^{-1}2-tan^{-1}1)+(tan^{-1}3-tan^{-1}2)+(tan^{-1}4-tan^{-1}3)+...
1that is ok
but clearly here it is not
of that form
how can we convert avik's step to that form ?
62why dude?
tan^{-1}a-tan^{-1}b=tan^-1\frac{a-b}{1+ab}
put a=1 and see what happens..
13This helping in any way.....??
tan^{-1}\frac{2-1}{2+1} + tan^{-1}\frac{3-2}{3+2} + tan^{-1}\frac{4-3}{4+3}+....
or...maybe i am 1/2 asleep... [13]
1rhs is constantly decreasing,so it ought to be infinity
1hmmmm.......
your answer is correct, but can anyone (other than debotosh)
tell the steps of the solution..