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EDIT: #8 th now :)

ok ok I'll try :P

is ans 1/2 ?

No, but for \(n=10000\) the answer is \(\frac 12 \)

hmm,,i used hit and trial i.e. checked each term..sum approaches 0.5 only when it gets bigger..

Nopes :) You have to derive a term of \(S_n \) summation upto \(n\) terms

damn how do you easily see that?

My head hurt but isn't it somewhere near 1/2 and 1/3

@zzr0ck3r
i somehow saw that it is (n^2 +1)^2 - (n^2)
dont know how,,even that amazes me!! :D

nice work

but @FoolForMath says its not 1/2..i must be wrong somewhere..hmm

and how do we know its 1/2 only upto n=10000?? sir?? you there?

i think no he is not here

code it and check

im guessing he would not have said it if it was not true:)

hmmm.. :)

\[S_{n} = \frac{n(n+1)}{2(n^{2}+n+1)}\]

ohh..i see,,i calcuated for infinite terms i.e n tends to infinity..hmm..not the general formula

how do you find summation (n/n^4 + n^2 +1) ??

ohh lol..yes,,damn i missed that..nice :)