epsilon-N proof of divergent sequence with another divergence assumed

Question

I'd like to prove the following proposition: $$ \lim_{n\rightarrow\infty} a_n = \infty \Rightarrow \lim_{n\rightarrow\infty} b_n = \infty \\ \text{where}\;b_n = \dfrac{1}{n}\sum_{k=1}^{n}a_k, $$

but my proof was stuck. Would you tell me how should I finish the proof?

Suppose $\lim_{n\rightarrow\infty} a_n = \infty$ ($\forall\epsilon\in\mathbb{R},\,\exists N\in\mathbb{N},\,\forall n\in \mathbb{N},\,n\geq{}N\Rightarrow a_n > \epsilon$), then I'll prove $\lim_{n\rightarrow\infty} b_n = \infty$.

Let $\epsilon\in \mathbb{R}$. For $n\geq N$, $b_n=\dfrac{1}{n}\sum_{k=1}^{N}a_k+\dfrac{1}{n}\sum_{k=N+1}^{n}a_k>\dfrac{1}{n}\sum_{k=1}^{N}a_k+\dfrac{n-N}{n}\epsilon=\dfrac{1}{n}\sum_{k=1}^{N}(a_k-\epsilon) + \epsilon,$ from the supposition.

In order to prove $b_n>\epsilon$, I only prove $\dfrac{1}{n}\sum_{k=1}^{N}(a_k-\epsilon)\geq0$ ($N_0$ exists and for all $n\geq N_0$), but $a_n-\epsilon$ for $n < N$ can't be proved. What's wrong? And how do I let $N_0$?

Answer 1

Note that $x_n > y_n \implies \liminf x_n \geqslant \liminf y_n$ and for all $\epsilon > 0$

$$\limsup_{n \to \infty} b_n \geqslant \liminf_{n \to \infty} b_n \geqslant \lim_{n \to \infty} \frac1{n}\sum_{k=1}^Na_k + \lim_{n \to \infty}(1 - N/n)\epsilon = \epsilon.$$

Hence, $\lim b_n = \limsup b_n = \liminf b_n = + \infty.$

Alternatively, with $N$ fixed,

$$\lim_{n \to \infty} \frac1{n}\sum_{k=1}^Na_k = 0 , \\ \lim_{n \to \infty} (1 - N/n) = 1,$$

and, for $n$ sufficiently large

$$\frac1{n}\sum_{k=1}^Na_k > -\epsilon/4, \\ (1 - N/n) > 1/2,$$ and

$$b_n > \frac1{n}\sum_{k=1}^Na_k + (1 - N/n)\epsilon > -\epsilon/4 + \epsilon/2 = \epsilon/4$$

Answer 2

You just need to go than further $N$.

Let $A > 0$ be arbitrary ($\epsilon$ usually denotes a small quantity, I prefer to use $A$ instead).

Since $a_n\to \infty$, there is some $N$ such that $n\geq N\implies a_n\geq 2A$.

Even if it's out of the blue at the moment, let $\displaystyle N'=\max(1+\lfloor \frac{\sum_{k=1}^N (2A-a_k)}{A}\rfloor,0).$

Let $N''=\max (N,N')$ and consider some $n\geq N''$.

Note that$$\begin{align}\sum_{k=1}^n a_k&\geq \sum_{k=1}^{N^{''}} a_k + (n-N'')2A \;\; \text{because}\; n\geq N \\ &= An + \left(An - \sum_{k=1}^{N^{''}}(2A-a_k) \right) \\ &\geq An \;\; \text{because}\; n\geq N'\end{align}$$

Hence $\frac {1}n \sum_{k=1}^n a_k \geq A$.