# Arithmetic mean sequence

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!).

• 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 ? – John Sep 17 '19 at 2:09
• 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 – user439126 Sep 17 '19 at 2:22
• Ow , sorry , I misread the conditions :( – John Sep 17 '19 at 2:23
• 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. – Hans Lundmark Sep 17 '19 at 8:57
• Thanks, that's absolutely right – user439126 Sep 17 '19 at 21:36

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!

• 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 $2Cm_C$? – user439126 Sep 17 '19 at 4:33
• @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). – Wlod AA Sep 17 '19 at 4:39
• It does, thank you. – user439126 Sep 17 '19 at 12:27

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.

• What do you mean by the first $10N_C$ terms? – user439126 Sep 17 '19 at 3:29
• 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 – Cade Reinberger Sep 17 '19 at 3:35

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 $$\boxed{\frac{a_1 + \cdots + a_n}{n}\stackrel{n \to \infty}{\longrightarrow}+\infty}$$.

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|

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.