Asked by Ashley
Question :
It is given that lim (an) n->infinity=a.
Define bn=(a1+a2+…………+an)/n
What can you say about the convergence of {bn}?
Is (a1+a2+…………+an)= lim(an) as n-->infinity or? I'm confused!
It is given that lim (an) n->infinity=a.
Define bn=(a1+a2+…………+an)/n
What can you say about the convergence of {bn}?
Is (a1+a2+…………+an)= lim(an) as n-->infinity or? I'm confused!
Answers
Answered by
oobleck
My gut feeling is that
consider the sequence 1,1,1,... where the limit is 1
Dividing each term by n just gives terms of the Harmonic Series, which diverges.
On the other hand, if you start with something like
a_n = e^-n cos(nπ)
then b_n =
n
∑ e^-k cos(kπ)
k=1
and the terms converge to -1/(e+1)
consider the sequence 1,1,1,... where the limit is 1
Dividing each term by n just gives terms of the Harmonic Series, which diverges.
On the other hand, if you start with something like
a_n = e^-n cos(nπ)
then b_n =
n
∑ e^-k cos(kπ)
k=1
and the terms converge to -1/(e+1)
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