3

Let $(a_n)$ be a sequence of real numbers. Prove that $$\limsup_{n\to \infty}\frac{a_n}{n}\leq \limsup_{n\to\infty}\, (a_{n}-a_{n-1}). $$

If the sequence $(a_n)$ converges, then this is immediate: both sides are $0$. If it is not convergent, how should I approach this problem?

  • 2
    It might be easier to think in terms of the sequence $b_n$ with $b_n=a_{n+1}-a_n$; then the LHS is the lim sup of $\frac1n\sum_{k\leq n}b_k$ and the RHS is just the lim sup of $b_n$. – Steven Stadnicki Jul 05 '17 at 15:56
  • 2
    I would not say this is a duplicate because the point is to identify that this problem can be reduced to the other one! –  Jul 06 '17 at 13:00

2 Answers2

3

First suppose $\limsup\limits_{n\rightarrow\infty} (a_n - a_{n-1}) < \infty$ (otherwise the statement is certainly true), and say it is equal to $M$. Then, intuitively speaking, as we look further along the sequence, each successive term is at most approximately $M$ more than the previous one. Looking further and further along, the worst case scenario is that approximately $M$ is continued to be added to each term. As more terms are added, any aberrant earlier behaviour will be nullified by the increasing $n$ in the denominator, which will make the inequality true.

Let's prove it. First, fix $\varepsilon > 0$. It is my aim to prove that, $$\limsup_{n\rightarrow \infty} \frac{a_n}{n} \le M + \varepsilon.$$ By the definition of $M$, it follows that we may find some $N \in \mathbb{N}$ such that, $$n \ge N \implies a_{n+1} \le a_n + M + \varepsilon.$$ It therefore follows by induction that, \begin{align*} & a_{N+k} \le a_N + k(M + \varepsilon) \\ \implies& \frac{a_{N+k}}{N+k} \le \frac{a_N}{N + k} + \frac{k}{N + k}(M + \varepsilon). \end{align*} Take the limit superior of both sides, as $k \rightarrow \infty$. The left side will simplify to $\limsup\limits_{n\rightarrow\infty} \frac{a_n}{n}$, as we are considering a tail of the full sequence. The right side is convergent as $k \rightarrow \infty$, and converges to $M + \varepsilon$. Thus, $$\limsup_{n\rightarrow\infty} \frac{a_n}{n} \le M + \varepsilon,$$ as required.

Theo Bendit
  • 50,900
  • Take the limit superior of both sides, as $k \rightarrow \infty$. The left side will simplify to $\limsup\limits_{n\rightarrow\infty} \frac{a_n}{n}$, as we are considering a tail of the full sequence. How the left side simplified like this please explain? –  Aug 12 '17 at 09:32
  • 1
    @N.Maneesh The limsup of a sequence is the limit of the suprema of tails of the sequence, that is,$$\limsup\lim_{n\rightarrow\infty} a_n=\lim_{n\rightarrow\infty}\sup_{k\ge n} a_k.$$The terms of the sequence $\left(\frac{a_{N+k}}{N+k}\right){k=1}^\infty$ are simply the terms of $(a_n / n){n=1}^\infty$, but with the first $N$ terms removed. The suprema of tails of the sequence will eventually coincide for the two sequences, and hence the limits of the suprema are the same. – Theo Bendit Aug 12 '17 at 12:49
0

Thanks to Theo's answer and Steven's comment, I have a slightly different interpretation of the proof. Let $b_1=a_1$ and $b_n=a_n-a_{n-1}$ for $n>1$. Then it suffices to prove that $$ \limsup_{n\to\infty}\frac{1}{n}\sum_{k=1}^nb_k\leq \limsup_{n\to\infty}b_n=:M<\infty $$ Given $\epsilon>0$, there exists $N>0$ such that for all $n\geq N$, $$ b_n< M+\epsilon $$ and thus for all $n\geq N$, $$ \frac1n\sum_{k=1}^n b_k\leq\frac{1}{n}(b_1+\cdots+b_N)+\frac1n\big[M(n-N)+\epsilon(n-N)\big] $$ It follows that $$ \limsup_{n\to\infty}\frac1n\sum_{k=1}^n b_k\leq M+\epsilon. $$ Since $\epsilon$ is arbitrary, we are done.