Prove that, if $\lim_{j\rightarrow\infty}a_j=l$ then $\lim_{j\rightarrow\infty}m_j=l$.

Question

So I was wondering if anybody could help me tackle this problem that I've been assigned in my real analysis course (book used is Foundations of Analysis by Steven G. Krantz). From my understanding, this text was just recently written.

First, I could use some help understanding exactly what this problem is asking. The way it's written (perhaps the notation) is confusing to me. We are in the chapter involving series, and this particular exercise is found in the section on elementary convergence tests. I found a similar question but none of the answers there made sense to me, so I figured I'd see if somebody could do it more justice and explain what is being asked and help me start it.

Second if one could maybe give me a hint on how to proceed, as I am somewhat at a loss as to how to prove this particular statement, I would be most grateful.

The problem is as follows.

"Let $a_j$ be a sequence of real numbers. Define

$m_j=\frac{a_1+a_2+\cdots a_j}{j}$.

Prove that, if $\lim_{j\rightarrow\infty}a_j=l$ then $\lim_{j\rightarrow\infty}m_j=l$. Give an example to show that the converse is not true."

I tried playing around with a few of the different elementary convergence tests (such as the root test and comparison test) and seeing how they could apply here, but in all honesty I don't understand how to apply them to this question. I just feel really off base with everything I've tried, so I figured I'd see if someone here could set me on the right path.

Thank you all for your time!

Answer 1

Let $\varepsilon>0$. Write first: $$\left|\dfrac{1}{j}\sum_{k=1}^j a_k-l\right|\leq\dfrac{1}{j}\sum_{k=1}^j |a_k-l|=\dfrac{1}{j}\sum_{k=1}^{N}|a_k-l|+\dfrac{1}{j}\sum_{k=N+1}^{j}|a_k-l|\,(\ast)$$ where $1\leq N\leq j$. Since $a_j\rightarrow l$ we can choose $N$ such that $$|a_k-l|<\dfrac{\varepsilon}{2}\text{ if }k>N$$ Next choose $j$ large enough so that $$\dfrac{1}{j}\sum_{k=1}^{N}|a_k-l|<\dfrac{\varepsilon}{2}$$ This is possible since $N$ has been fixed, so $\sum_{k=1}^{N}|a_k-l|$ is a fixed constant. For this choice of $N$ and if $j$ is large enough we obtain from $(\ast)$: $$\left|\dfrac{1}{j}\sum_{k=1}^j a_k-l\right|<\dfrac{\varepsilon}{2}+\dfrac{1}{j}\sum_{k=N+1}^j\dfrac{\varepsilon}{2}<\varepsilon$$ since there are fewer than $j$ terms in that latter sum. This shows that $m_j\rightarrow l$.

Answer 2

First of all, the result should be intuitive: if $a_j \to a$, then $a_j$ becomes closer and closer to $a$, so as $j \to \infty$, the averages $\frac{a_1+a_2+\dots+a_n}{n}$ should tends toward $a$. We can make this intuition rigorous and thus prove the result by using epsilons and deltas; we don't need any convergence tests.

First take $\epsilon > 0$ and $M$ such that $|a_m-a| < \epsilon$ for all $m \ge M$. Take $N > M$. We want to eventually choose $N$ large enough so that enough of the terms in the numerator of our average are $\epsilon$ close to $a$. So,

$|\frac{a_1+\dots+a_N}{N}-a| = |\frac{a_1+\dots+a_M}{N}+\frac{a_{M+1}+\dots+a_N}{N}-a| \le |\frac{(a_1-a)+\dots+(a_M-a)}{N}|+|\frac{(a_{M+1}-a)+\dots+(a_N-a)}{N}| \le \frac{C_M}{N}+\frac{(N-M)\epsilon}{N}$

where $C_M$ is just a constant depending on $M$. We thus see that $\limsup_{N \to \infty} |\frac{a_1+\dots+a_N}{N}-a| \le \epsilon$. Since, $\epsilon$ is arbitrary, we get that

$0 \le \liminf_{N \to \infty} |\frac{a_1+\dots+a_N}{N}-a| \le \limsup_{N \to \infty} |\frac{a_1+\dots+a_N}{N}-a| = 0$

implying the limit exists and is $0$.

Answer 3

A nice (and immediate) proof can be given using Stolz-Cesaro Theorem. This result is quite useful and helps solve rapidly many such problems. It is almost like l'Hopital's rule, which allows to "differentiate" the numerator and denominator to (hopefully) reduce the problem to a simpler question.

Let $A_n = (a_1+...+a_n)$ and $B_n=n$. $B_n$ is strictly increasing and $B_n \to \infty$.

Compute $\frac{A_{n+1}-A_n}{B_{n+1}-B_n} = a_n \to a$. The referenced theorem states that $\frac{A_n}{B_n}$ is also convergent and has the same limit.