Limit of a Sequence 07

Calculus Level 3

Let a n a_n be a real-valued sequence such that a n > 0 \displaystyle{a_n > 0} for all n ≥ 1 n \geq 1 , and lim ⁡ n → ∞ a n = L \displaystyle{\lim_{n \rightarrow \infty} a_n = L} , for some finite L > 0 L > 0 . Which of the following is the STRONGEST true statement that can be made about b n = a 1 + a 2 + . . . + a n n \displaystyle{b_n = \frac{a_1+a_2+...+a_n}{n}} ?

lim ⁡ n → ∞ b n \displaystyle \lim_{n \to \infty} b_n exists. None of these are necessarily true. lim ⁡ n → ∞ b n \displaystyle \lim_{n \to \infty} b_n exists and is equal to some finite value b > 0 b > 0 . lim ⁡ n → ∞ b n \displaystyle \lim_{n \to \infty} b_n exists and is equal to L L .

Daniel Juncos by Daniel Juncos , 45, USA

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1 solution

Daniel Juncos
Sep 19, 2017

Let ϵ > 0 \epsilon > 0 . Then there is some N 0 N_0 such that for all n ≥ N 0 n \geq N_0 we have ∣ a n − L ∣ < ϵ 2 \displaystyle{\left| a_n - L \right| < \frac{\epsilon}{2}} .

Meanwhile, ∣ b n − L ∣ = ∣ a 1 + a 2 + . . . + a n n − n L n ∣ \displaystyle{\left| b_n - L \right| = \left| \frac{a_1 + a_2 + ... + a_n}{n} - \frac{nL}{n}\right| }

= ∣ a 1 − L + a 2 − L + . . . + a N 0 − L n + a N 0 + 1 − L + . . . + a n − L n ∣ \displaystyle{= \left| \frac{a_1-L+ a_2-L + ...+ a_{N_0}-L}{n}+ \frac{a_{N_0+1}-L+...+a_n-L}{n}\right|}

≤ ∣ a 1 − L + a 2 − L + . . . + a N 0 − L ∣ n + ∣ a N 0 + 1 − L ∣ + . . . + ∣ a n − L ∣ n \displaystyle{\leq \frac{\left| a_1-L+ a_2-L + ...+ a_{N_0}-L \right|}{n}+\frac{\left|a_{N_0+1}-L \right|+ ... + \left| a_n - L \right|}{n}}

There is an N 1 > N 0 N_1 > N_0 such that for all n ≥ N 1 n \geq N_1 we have ∣ a 1 − L + a 2 − L + . . . + a N 0 − L ∣ n < ϵ 2 \displaystyle{\frac{\left| a_1-L+ a_2-L + ...+ a_{N_0}-L \right|}{n}< \frac{\epsilon}{2}} .

So for all n ≥ N 1 n \geq N_1 , we have

∣ b n − L ∣ < ϵ 2 + ∣ a N 0 + 1 − L ∣ + . . . + ∣ a n − L ∣ n < ϵ 2 + ϵ 2 + . . . + ϵ 2 n = ϵ 2 + ϵ 2 ( n − N 0 n ) < ϵ 2 + ϵ 2 ( n n ) = ϵ \displaystyle{\left| b_n - L \right| < \frac{\epsilon}{2} + \frac{\left|a_{N_0+1}-L \right|+ ... + \left| a_n - L \right|}{n} < \frac{\epsilon}{2}+ \frac{\frac{\epsilon}{2}+...+ \frac{\epsilon}{2}}{n}= \frac{\epsilon}{2}+\frac{\epsilon}{2}\left(\frac{n-N_0}{n}\right)<\frac{\epsilon}{2}+\frac{\epsilon}{2}\left(\frac{n}{n}\right) = \epsilon}

Thus b n → L b_n \rightarrow L as n → ∞ n \rightarrow \infty .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

This is also known as Cauchy's first limit theorem . Nice proof . It also has a corollary about the GM . That if bn was equal to the Geometric mean of a1 a2 ..... an . Then too it would converge to the limit L (L>0)

Arghyadeep Chatterjee - 2 years, 1 month ago

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