limn→∞ (n!/(nk)n)1/n
please do solve this even if it seems simple
i don't think answer should be 1/(2k) but if you get this somehow it'll be very interesting
21@subhash.... Dude pl. chk out LIMIT of sum type definite integrtn....
if u know then c : log(n!) = log1 + log2 + log3 +... logn; {so these are the n terms }
and we hav -nlog(nk) = - {log(nk) + log(nk) + log(nk) +log(nk) + .... n terms }
so u get lny = 1/n(σ [log(r/nk)] )
wer r varies frm 1 to n.......
13don't get u Prajith .....how can u integrete0∫1ln x dx??whatz the result??
1ok but im getting a different answer tapu
how come you getting difft when same till this point
21ln(x/k) = lnx - lnk;
therfor ∫lnx - ∫lnk
= -1 - lnk * (1-0)
= -lnk - 1
=-ln(ke)
= ln(1/ke)
21Thnx Cele!!!!
well abt the source of questn...
philip will be able to answer better
1i got it jus written somewhere u understand ??
jus like this for riemanns sum probs
any thing special about the source celes ??
1hmm...
so abhishek in your method we can safely substitute 2 for e when we see that ur method will be fundamentally wrong ...
[6]
jus joking yaar...
but i sincerely hope it was this easy [4]
but definitely this isn't so ... [2]
9if the source is a test series q then we can be assured that the sol theyve given is rong thats why ;)
1no not a test series question and "they"(i don't know them) didn't give any solution
1dunno tapans method is rite but i don't know about the final answer
33don't ask how... its very controversial and wrong method for this..
1hey abhi I didn't expect this answer from u
anyways i'm sure that 1/2k is wrong only bcoz of that method of urs
any other method... someone
ok try it with Riemann's sum ... i couldn't get it with that
33[3]
lol
i thought u were asking for a method for getting that wrong answer...
:D
and i have that..
1i guess the ans should be
1/k i.e a constant
tell me if its rite...???
1hey abhishek your method isn't wrong only it won't work over here for a valid reason
I dont know the correct answer so everyone please post your solution also
btw abhishek i posted this to get the right answer and see if the right answer magically matches with the "wrong answer"
then it will be really interesting...
21i had thot of doing it by using LIMIT of a sum n solving....
applying log on both sides : => lny = 1/n ( log(n!) - nlog(n) - nlog(k) )
log(n!) is same as log1 + log2 + .... log(n);
so => lny = 1/n{ln(1/kn) + ln(2/kn) + ln(3/kn) + ...... ln(n/kn)}
therefor applyin liimit of sum and treating 1/n as ∂x
lny = 0∫1 ln(x/k) ∂x
lny = - ln(ke)
therfor y = 1/(ke)
