Can you see the attachment?

Question

ganeshie8 · Answer

nope, 
iphone again ?

dls · Answer

Parth is posting the screenshot of the problem :p

parthkohli · Answer

Hi

dls · Answer

I am getting after simplification, [ integral of tan x from 0 to 1 ] which is 0. I doubt my limits are wrong.How do I check that? Answer isn't 0. :[

xapproachesinfinity · Answer

The picture isn't clear by the way,  the question is it about the limit of that expression of sums?

dls · Answer

Yeah limit of sum.

ikram002p · Answer

limit of sum is integral right ?

ikram002p · Answer

let 2/n inside the sumation mmmm

ikram002p · Answer

what about cos (2r/n ) the pracits are gretest integer function ?
so its descrete ??

ikram002p · Answer

then cos (2r/n ) is always 1

ikram002p · Answer

so u would have 
$lim  2/n \sum_1^n  (  1/2 ) $ as n goes to$ \infty$

ikram002p · Answer

seems solved nw ^^ nothing more

ikram002p · Answer

@ganeshie8 size of series is it n or n/2  ?

ikram002p · Answer

ok i think its n :P
so u would have 
lim  2/n * n/2 =1

kainui · Answer

This looks good but how did you know that cos(2r/n) is always 1? I can see that as n goes to infinity we have cos(0)=1 I guess? 

The only problem I have is that when I think of the sum backwards I see that r=n as the last term and have cos(2n/n)=cos(2) which is not dependent on n anymore. Although I suppose that in an infinite sum this could go away as being unimportant? I'm not sure I am looking at it quite right when I do this though.

ℐℵk℣ⱱøƴȡ◆ · Answer

I can't see the attachment..

anonymous · Answer

the given limit lies between |dw:1406462883994:dw|