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If the ssequence of $a_n$ goes to infinity as $n$ goes to infinity, then does $$\lim_{n\to\infty} \frac{a_1 + \dots + a_n}{n} = \infty?$$

I know this sequence converges to a finite value if the sequence $a_n$ converges to a finite value, but I don't know if that helps. I've tried using the definition a sequence converging to infinity, I've also tried using the convergence of $\frac{1}{a_n}$ to $0$ to show $\frac{n}{a_1 + \dots + a_n}$ converges to $0$, but no luck. Do I utilize the arithmetic mean inequality? Any hints are more than welcome (only hints please!).

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    $\begingroup$ 1) If ${b_n}$ converges to a finite limit , what will $\frac{b_1+\cdots+b_n}{n}$ converge to ? 2) If you take $b_n = \frac{1}{a_n}$ what can you say , if you apply AM-HM inequality ? $\endgroup$ – John Sep 17 at 2:09
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    $\begingroup$ For 1), I believe I loosely stated that it would converge to the finite limit of $b_n$. Now if $b_n = \frac{1}{a_n}$, I can't apply the arithmetic mean inequality because I don't know if the $a_n$'s are non-negative $\endgroup$ – user439126 Sep 17 at 2:22
  • $\begingroup$ Ow , sorry , I misread the conditions :( $\endgroup$ – John Sep 17 at 2:23
  • $\begingroup$ Terminology comment: the limit doesn't go to infinity, it's the sequence $a_n$ itself that does. The limit is what the sequence tends to, so in this case the (improper) limit is infinity. $\endgroup$ – Hans Lundmark Sep 17 at 8:57
  • $\begingroup$ Thanks, that's absolutely right $\endgroup$ – user439126 Sep 17 at 21:36
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If the limit goes to infinity, then for any constant $C$. All terms bigger than some constant $N_C$ are bigger than $C$. Taking the arithmetic mean of the first $10N_C$ terms give you an arithmetic mean as large as $.9C$. Since I can make $C$ as large as I like, I can make $.9C$ as large as I like, so the arithmetic mean must go to infinity.

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  • $\begingroup$ What do you mean by the first $10N_C$ terms? $\endgroup$ – user439126 Sep 17 at 3:29
  • $\begingroup$ So, like, say the sequence is $a_n=n^2$. Since $\lim_{n \to \infty} a_n = \infty$ for any $C$, say $C=100$, every term bigger than. Some constant term index (in this case $N_C = 10$ )is bigger than 100. I can always find such a number, since thats a definition of saying the limit is infinite. The idea then, is that I can look at the first 100 total terms. We've just said the last 90 of them are at least 100. So the AM is at least 90. Making $C$ bigger gives the desired result $\endgroup$ – Cade Reinberger Sep 17 at 3:35
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Indeed,

THEOREM If the limit of $a_n$ goes to infinity as $n$ goes to infinity, then $$\lim_{n\to\infty} \frac{a_1 + \dots + a_n}{n} = \infty$$

PROOF Let $\ C>0.\ $ There exists natural $\ N_C\ $ such that $\ a_k>2\cdot C\ $ for every $\ k>N_C.\ $ Let

$$ B_C\ :=\ \sum_{n=1}^{N_C}\ a_n $$ and let natural $\ m_C\ $ satisfy

$$ m_C\ >\ N_C-\frac {B_C}C $$

so that $$ B_C+2\cdot C\cdot m_C\ >\ (N_C+m_C)\cdot C $$

Now, let $\ n>N_C+m_C.\ $ Then

$$ \frac{a_1 + \dots + a_n}n\,\ > \,\ \frac{B_C\ +\ 2\cdot C\cdot m_C\ +\ \sum_{k=N_C+m_C+1}^n a_k}n $$

$$ >\,\ \frac{(N_C+m_C)\cdot C\,\ +\,\ (n-(N_C+m_C))\cdot2\cdot C}n \,\ >\,\ C $$

Since $\ C>0\ $ is arbitrary, the theorem holds.   Great!

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  • $\begingroup$ How did you get down to the first inequality? How do you toss out $a_{N_C + 1} + \dots + a_{N_C +m_C}$ and replace it with $2Cm_C$? $\endgroup$ – user439126 Sep 17 at 4:33
  • $\begingroup$ @user439126, at the very start of the proof we had: $\ a_k>2\cdot C\ $ for every $\ k>N_C.\ $ (Is there also the first part to your question/comment? If yes then please state it explicitly). $\endgroup$ – Wlod AA Sep 17 at 4:39
  • $\begingroup$ It does, thank you. $\endgroup$ – user439126 Sep 17 at 12:27
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This might be a bit overkill but I think it is worth noting it:

Your question can be dealt with using the general form of the Cesaro-Stolz theorem:

$$+\infty=\lim_{n\to\infty}a_n=\liminf_{n\to \infty}\frac{a_n}{1} \leq \liminf_{n\to \infty}\frac{a_1 + \cdots + a_n}{n}$$

So, it follows immediately that $\boxed{\frac{a_1 + \cdots + a_n}{n}\stackrel{n \to \infty}{\longrightarrow}+\infty}$.

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Let $S(n)=\frac {1}{n}(a_1+...a_n).$ Given $R>0,$ we want some $n_1$ such that $n\ge n_1\implies S(n)>R.$

First, take some $n_0\in \Bbb N$ such that $n> n_0\implies a_n>2R.$ Now if $n>n_0$ then $$S(n)=\frac {1}{n}(a_1+..+a_{n_0})+ \frac {1}{n}(a_{n_0+1}+...+a_n)>\frac {1}{n}(a_1+...+a_{n_0})+\frac {(n-n_0)}{n}\cdot 2R=T(n).$$ With $n_0$ fixed, we have $\lim_{n\to \infty}\frac {1}{n}(a_1+...+a_{n_0})=0$ and $\lim_{n\to \infty}\frac {(n-n_0)}{n}\cdot 2R=\lim_{n\to \infty}(1-\frac {n_0}{n})\cdot 2R=2R.$ Therefore $\lim_{n\to \infty}T(n)=2R.$

So there exists $n_1>n_0$ such that $$n\ge n_1\implies |T(n)-2R|<R/2\implies R<3R/2<T(n)<S(n)\implies R<S(n).$$

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Intuitively, as $a_n$ goes to infinity, $a_n>2M$ finishes to hold for any $M$, how ever large. This means that the average $\overline{a_n}>M$ finishes to hold as well, because it is larger than the average of a finite sum and as many $2M$ terms as you want.

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