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Problem: Given $a_n \to c \in \overline{\mathbb{R}}$, prove that $\displaystyle \lim_{n \to \infty} \frac{a_1 + a_2 + \cdots + a_n}{n} = c$.

I have read many of the related posts on this topic:

These all attempt to split the summation in the numerator like so: $\sum_{i=1}^N + \sum_{i=N}^n$ and then bring in the c, so that eacn term is $\frac{(a_1 - c) + (a_2 - c)}{n}$.

However, all of these previous proofs assume that the sequence is convergent, which means that $\frac{(a_1 - c) + (a_2 - c)}{n} = 0$. In my case, the sequence converges to a number in the extended reals and can thus diverge to $-\infty$ or $\infty$, meaning $c$ could be $-\infty$ or $\infty$, and we can't assume $\lim_{n \to \infty}\frac{(a_1 - c) + (a_2 - c)}{n} = 0$.

Are the methods in the above links still applicable? If not, how should I approach this proof?

Michael Hardy
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Sarah
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  • Converging to a finite number and diverging to (positive or negative) infinity are two completely different things as far as calculus, real analysis, and $\epsilon$-$N$ proofs go. So it shouldn't be surprising that the two cases need separate proofs. Also, $\pm \infty$ is not a number, so doing arithmetic with it is crazy (in other words, $\frac{(a_1 - c) +\cdots + (a_n - c)}{n}$ makes no sense if $c$ is infinite). – Arthur Sep 14 '17 at 06:57
  • @Arthur I'm not sure it makes no sense, actually. "Given a convergent sequence of points on the unit sphere, show that the partial means converge to the same point." – Patrick Stevens Sep 14 '17 at 07:20
  • @PatrickStevens We are talking about the (extended) number line here, not the unit sphere. Of course, one can apply all manner of transformations, and change to equivalent metrics to make converging to a finite limit and diverging to infinity have the same definition. But that wouldn't be the extended number line any more, would it? It would just become some line segment or something. – Arthur Sep 14 '17 at 07:27
  • Oh, sorry; I consistently misread that question with $\bar{\mathbb{C}}$ rather than $\bar{\mathbb{R}}$. I think it was a mental misstep based on the fact that the letter used for the limit was $c$. – Patrick Stevens Sep 14 '17 at 07:27
  • The proposed duplicate addresses the cases where $c=+\infty$. When that is applied to the sequence $-a_n$, the cases where $c=-\infty$ are solved. – hardmath Sep 27 '17 at 06:20

1 Answers1

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We can do one-sided limits in a uniform way:

Lemma. Let $a_1,a_2,\dots\in\mathbb R$ and define $m_n=(a_1+\dots+a_n)/n$. Then $$\limsup_{n\to\infty} m_n \leq \limsup_{n\to\infty} a_n\tag{1}$$ where we allow limits in $\overline{\mathbb R}$.

Proof. For every finite $U>\limsup_{n\to\infty} a_n$ there exists $N$ such that for all $n> N$ we have $a_n\leq U$, so $$m_n = \frac{Nm_N + a_{N+1}+\dots+a_n}{n} \leq \frac {Nm_N} n + \frac{n-N}{n} U \to U\text{ as $n\to\infty$}.$$ taking $\limsup_{n\to\infty}$ of both sides we get $\limsup_{n\to\infty} m_n\leq U$ for all $U>\limsup_{n\to\infty} a_n$, which gives (1).

Note that if the right-hand-side of (1) is $+\infty$, this argument is vacuous - there is no such $L$ - but it's still a valid argument.

We can apply this lemma to $a_n$ and the negated sequence $-a_n$ to get:

$$\liminf_{n\to\infty} a_n \leq \liminf_{n\to\infty} m_n\leq \limsup_{n\to\infty} m_n \leq \limsup_{n\to\infty} a_n$$

In particular if $\lim_{n\to\infty} a_n=c\in\overline{\mathbb R}$ then $\lim_{n\to\infty} m_n$ exists and equals $c$.

Dap
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