Question : if $ \lim_{n\to \infty} a_n = \infty $ then $ \lim_{n\to \infty} \frac 1n\sum_{k=1}^n a_k = \infty$

My attempt for solving : $$ \lim_{n\to \infty} a_n = \infty \Rightarrow \forall M \exists N\in \Bbb N : \forall n > N \Rightarrow a_n >M $$

To prove by definition we need to show that $$ \forall M \exists n_0\in \Bbb N : \forall n > n_0 \Rightarrow \frac 1n\sum_{k=1}^n a_k >M$$

Now I am trying to find $n_0$ to satisfy the conditions and write in the formal proof :

Let us define $ a_m = min \{ a_1 , a_2 , \ldots ,a_N\}$

$$ \frac 1n\sum_{k=1}^n a_k = \frac 1n\sum_{k=1}^N a_k + \frac 1n\sum_{k=N+1}^n a_k >\frac 1nNa_m + \frac 1n\left( n - N \right)M$$

From here on I got stuck and couldn't figure how to find my $n_0$ .

Another thought : I tried using a previous proof saying : if $ \lim_{n\to \infty} a_n = \infty $ and for each $n\in \Bbb N : a_n \le b_n$ Then it follows that $ \lim_{n\to \infty} b_n = \infty $ .

I defined the AM sequence as $b_n$ but I failed to find a sequence $a_n$ that satisfies these conditions .


First, you can take $n = n(M)$ large enough, so that

$\frac{1}{n}\sum_{k=1}^na_k > \frac{n-N}{n}M = \big(1-\frac{N}{n}\big)M > \frac{M}{2}$

and you can take $M$ as large as you want

  • $\begingroup$ Thank you for the answer, can you maybe elaborate more on the reasoning for the first > sign? I failed to follow $\endgroup$ – Shahar Palensya Jul 27 at 9:19
  • $\begingroup$ Correct me if i'm wrong, but if we assume that the first N-1 elements are negative then it's incorrect to say that the AM_n > (n-N)M/n . $\endgroup$ – Shahar Palensya Jul 27 at 9:33
  • $\begingroup$ I assumed that $a_k > 0$; otherwise the statement is wrong. The counterexample: a sequence: $$1, 1, -2, 2, 2, -4, \ldots, 2^k, 2^k, -2^{k+1}, \ldots$$ Every sum of first $3k$ terms is 0. The first $>$ sign follows from your inequality $\frac{1}{n}\sum_{k=1}^na_k > \frac{1}{n}Na_m + \frac{1}{n}(n-N)M$ $\endgroup$ – diplodoc Jul 27 at 9:49
  • $\begingroup$ I see ! You are right , it makes sense now . Thank you ! $\endgroup$ – Shahar Palensya Jul 27 at 9:52

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