Suppose $\lim_{n\to \infty} \frac{a_n}{n}=0 $

Question

Suppose $$\lim_{n \to \infty} \frac{a_n}{n}=0. $$ Prove $$\lim_{n\to \infty} \frac{\max\{a_1,a_2,\ldots, a_n \}}{n}=0. $$

Below is what I tried , but I am not sure about my proof.

Denotes that $$a_{n_1}=\max\{a_1\}$$ $$a_{n_2}=\max\{a_1,a_2\}$$ $$\vdots$$ $$a_{n_k}=\max\{a_1,a_2,\ldots, a_k \}$$ Then what we want proof can write as $$\lim_{k\to \infty} \frac{a_{n_k}}{k}=0 $$ It's easy to see $n_k\ge n_{k-1} $, hence $\{a_{n_k}\}$ is a subsequence of $\{a_n\}$. Also $\{\frac{a_{n_k}}{k}\} $ is a subsequence of $\{\frac{a_n}{n} \} $. Then we have $$\lim_{k\to \infty} \frac{a_{n_k}}{n_k}=0 $$ Since $k\ge n_k $, then $$\frac{|a_{n_k}|}{k}\le \frac{|a_{n_k}|}{n_k} $$ Here we can have $$\lim_{k\to \infty} \frac{a_{n_k}}{k}=0 $$ So any problem with my proof, since I am not so satisfied with this kind of notation, any more convenient ways to do it ?

Answer 1

$\lim_\limits{n\to \infty} \frac{a_n}n = 0$

For any $\epsilon >0$, there is an $N$ such that when $n> N, |\frac {a_n}{n}|< \epsilon.$

Let $a_m = \max \{a_1,\cdots, a_N\}$, and $M = |\frac {a_m}{\epsilon}|$,

$n> \max (N,M) \implies |\frac {\max \{a_1,\cdots, a_n\}}{n}|<\epsilon$

Answer 2

For each $n\in\Bbb N,$ let $k(n)$ be the smallest index $k\le n$ such that $$\max(a_1,\dots,a_n)=a_k.$$ Then, $$\left|\frac{\max(a_1,\dots,a_n)}n\right|=\left|\frac{a_{k(n)}}n\right|.$$ Since $(k(n))$ is a non-decreasing sequence, as $n\to\infty$ we have

• either $k(n)\to+\infty$ and $$\left|\frac{\max(a_1,\dots,a_n)}n\right|\le\left|\frac{a_{k(n)}}{k(n)}\right|\to0$$
• or $k(n)$ is eventually constant and for $n\ge n_0,$ $$\left|\frac{\max(a_1,\dots,a_n)}n\right|=\left|\frac{a_{k(n_0)}}n\right|\to0.$$

Answer 3

Let $\epsilon >0$ be fixed and arbitrary. Find $N_1$ so that $n>N_1$ implies $\Big|\frac{a_n}{n}\Big|<\epsilon$. Next find $N_2$ so that $n>N_2$ implies $\frac{\max\{|a_1| , \dots , |a_{N_1}|\}}{n}<\frac{\epsilon}{2}$. Put $N=\max\{N_1,N_2\}$. So we get for $n=N+1,N+2,\ldots$ that \begin{eqnarray*} \Bigg|\frac{\max\{a_1, \dots , a_n\}}{n}\Bigg| & \leq & \frac{\max\{|a_1|, \dots , |a_n|\}}{n} \\ & = & \max\Big\{\Big|\frac{a_1}{n}\Big|, \dots ,\Big|\frac{a_n}{n}\Big|\Big\} \\ & \leq & \max\bigg\{\frac{\max\{|a_1| , \dots , |a_{N_1}|\}}{n},\epsilon\bigg\} \\ & \leq & \epsilon \end{eqnarray*}

Answer 4

If your max operation changes the limit at all, this happens only because $a_n$ might be replaced but another term.

When you increment $n$ to $n+1$, a new $$\max\{a_1,a_2,a_3...a_{n+1}\}$$ could be the new term $a_{n+1}$, but notice that if that is the case, $a_{n+1}$ never appears before $n+1$th position.

If it happens that you have one fixed position $k < \infty$ that remains maximum $a_k=\max\{a_1,a_2,a_3...a_{n}\}$ after $k$ for all $n>k$, the series will definitely converge to $0$ since $a_k$ is then just a constant appearing after $k$th position, so it is simply

$$\lim_{n \to \infty}\frac{a_k}{n+k}=0$$

If there is an infinite number of changes of $\max\{a_1,a_2,a_3...a_{k}\}$ as $k$ increases, none of them is appearing before the intended position $k$. If $a_k$ appears from $k$ to $k+m_k$ as the maximum, the maximal term is still where it was $\frac{a_k}{k}$ since:

$$\frac{a_k}{k}\geq\frac{a_k}{k+1}\geq...\geq\frac{a_k}{k+m_k}$$

and since $a_k$ is the max for all elements from $k$ to $k+m_k$ no term in between $\frac{a_k}{k}$ and $\frac{a_{k+m_k}}{k+m_k}$ is greater than $\frac{a_k}{k}$.

Extract all these terms that are the maximum in the segments from $k_i$ to $k_{i+1}-1$. However, this is just an infinite subseries of the existing series:

$$a_{k_i} = \max\{a_{k_i},a_{k_i+1},a_{k_i+2},....,a_{k_{i+1}-1}\}$$

$$\frac{a_{k_1}}{k_1},\frac{a_{k_2}}{k_2},\frac{a_{k_3}}{k_3},...$$

so it has to have the same limit.