If the average sequence converges to zero, does a subsequence converge to zero?

Question

Let $(a_n)_{n\in\mathbb{N}}$ be a positive sequence $0\leq a_n$, and suppose that the sequence of averages converges to zero: $$0\leq \left(\frac{1}{t}\sum_{i=0}^{t-1}a_n\right)_{t\geq 1}\to 0.$$ I want to prove that there exists at least a subsequence $(a_{n_k})_{k\in\mathbb{N}}$ such that $a_{n_k}\to 0$. Below is my proof attempt, but I'm not entirely convinced it's correct. Could someone please check it?

Suppose, for the sake of contradiction, that, for every subsequence $(a_{n_k})_{k\in\mathbb{N}}$, there exists some $\epsilon>0$ and $K\in\mathbb{N}$ such that we are bounded away from zero beyond $K$: $$K\leq k\quad\Rightarrow\quad 0<\varepsilon\leq a_{n_k}.$$ Let $\varepsilon$ and $K$ satisfy the above for the subsequence $(a_{n_k})_{k\in\mathbb{N}}=(a_n)_{n\in\mathbb{N}}$. Then, for every $t\in\{K, K+1,...\}\subset\mathbb{N}$,

$$0 < \frac{t-1-K}{t}\varepsilon=\frac{1}{t}\sum_{n=K}^{t-1}\varepsilon\leq\frac{1}{t}\sum_{n=K}^{t-1}a_{n}.$$ Since $a_n$ is positive, we can add the rest of the sequence to the righthand side $$0 < \frac{t-1-K}{t}\varepsilon\leq\frac{1}{t}\sum_{n=0}^{t-1}a_{n}.$$ Taking the limit as $t$ tends to $\infty$ implies that $0 < \varepsilon \leq 0$, a contradiction. [end of proof]

I am unsure about maintaining the strictness of the inequality in the the final step where we apply the limit -- usually, when applying a limit to an inequality, any strict inequality $<$ turns into a wide one $\leq$. However, since the coefficient of $\varepsilon$ goes to $1$, I think we're okay here? Does this need any further justification?

Otherwise, if I am wrong here, is there a nice counterexample?

Answer 1

Your argument is essentially right but you are lacking some clarity on the issue of when to pass to a subsequence.

Firstly, the direct negation of "there is a convergent subsequence" is only "every subsequence does not converge to $0$", which means for every subsequence, there is a further subsequence that is bounded away from $0$ by some $\epsilon$. A subsequence not converging does not by itself mean it is bounded away from $0$.

Now, you can argue pretty quickly that for every subsequence to have this property, the original sequence must be bounded away from $0$, and at some levels of math you might even state that without proof. When you are just starting off with proofs, however, I would recommend writing the argument out.

But in any case, it is key for your argument that you don't need to pass to a subsequence, since once you pass to a subsequence you no longer control the average. What you do is state without proof a fact (every subsequence has an $\epsilon>0$ and a $K>0$) that is stronger than you need - you just need it for the original sequence.

With all of that said, here is how you could tighten it up a bit.

Proof

Suppose by contradiction that the sequence of averages converges to $0$, but no subsequence converges.

We first claim this implies that there is some $\epsilon>0$ and $N\in\mathbb N$ for which $n>N$ implies $a_n>\epsilon$. For otherwise, for each $k\in \mathbb N$ we could choose $n_k>k$ satisfying $a_{n_k}<\frac{1}{k}$, whereby the sequence $\{a_{n_k}\}$ would converge, a contradiction.

Now we estimate \begin{align*} \frac{\sum_{k=1}^{n} a_k}{n} & \geq \frac{\sum_{k=N+1}^n a_k}{n} \geq \frac{ (n-N)\epsilon}{n}\to \epsilon>0\text{,} \end{align*} contradicting the assumption that the sequence of averages converges to $0$.

Answer 2

A better proof can be stated by rephrasing your attempt and a bit more rigor.

Obviously, $\liminf_{n\to\infty}(a_n)=0$. To prove it, suppose not. Since $a_n\ge 0$ for all $n$, we have $\liminf_{n\in \Bbb N}(a_n)\ge0$ and consequently $\liminf_{n\to\infty}(a_n)=l>0$. This implies that there exists some $N$ such that $a_n>\frac{l}{2}$ for $n>N$. This is in contradiction to $\frac{1}{t}\sum_{i=0}^{t}a_n\to 0$ as $t\to\infty$. Therefore, $\liminf_{n\to \infty}(a_n)=0$ $\blacksquare$

Now, the proof of your question can be completed by noting that since $\liminf_{n\to\infty}(a_n)=0$, for each $\epsilon>0$, there exists some $N_\epsilon$ such that $a_{N_\epsilon}<\epsilon.$ Now, gradually reduce $\epsilon$ to $0$ and construct such a subsequence $a_{N_\epsilon}$ that tends to $0$.