Question

i must find lim of a_n for x_n = 1/n( 1/ln(2) + ... + 1/ln(n) ) using stoltz cesaro theorem

Answer 1

xn>int 1/xlnx]1 to inf=ln(lnx)]1 to inf=inf
so limxn=inf

Answer 2

(1/n)*( 1/ln(2) + ... + 1/ln(n) ) is the equation sorry

Answer 3

I must use stoltz cesaro \[\frac{a_{n+1}-a ^{n} }{ b_{n+1}-b ^{n} }\]
for \[x _{n}=\frac{ 1 }{ n}*(\frac{ 1 }{ \ln_{2} }+ ... +\frac{ 1 }{ \ln_{n} })\]
and i don t know how to find \[a_{n}\] and \[b _{n}\]

Answer 4

Please help

Answer 5

try b_n = n, and a_n = 1/ln2 + ... + 1/ln(n)

Answer 6

b_n is n or 1/n :-S

Answer 7

with n get lim = 0 which is good but can i take only n and not 1/n as b_n? :D

Answer 8

I compare with integral which it approach to infinity then I conclude the series is also infinity or it is diverged.

Answer 9

i have a hint and says that the result is 0 but if i consider b_n 1/n the result is infinity so i don't know if the hint is good or not or whick one is correct

Answer 10

ugh you want x_n in the form of a_n/b_n... so tell me, what would a_n be if b_n = 1/n?

Answer 11

you' re right now i saw that if u write them your way is the same thing, is correct, i understood and the result is 0 as it should be thank you very muc :D i apreciate it