Limit of the mean value of partial sums?

Question

I'm supposed to construct a sequence of real numbers ${a_{n}}$ such that $\limsup_{n\to\infty} a_n=\infty$ but $\lim_{n\to\infty} b_n= \frac {a_{1}+a_{2}+...+a_{n}}{n} =0$.

Answer 1

Hint: Use the sequence $$\sqrt{1},-\sqrt{1}, \sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3},\sqrt{4},-\sqrt{4},\sqrt{5},\dots.$$

Answer 2

What you need is that $a_n$ is big only from time to time.

One example I can think of is to take $$ a_n=\begin{cases}n^{1/2},&\ n=2^{k/2}\ \text{ for some }k\in\mathbb Z\\0,&\ \text{ otherwise }\end{cases} $$ Then $\limsup a_n=\infty$, but if $2^{k}\leq n\leq 2^{k+1}$ then $$ \frac{a_1+\ldots+a_n}n\leq\frac{a_1+\ldots+a_{2^{k+1}}}{2^k} =\frac1{2^k}\,\sum_{j=0}^k2^{j/2}=\frac{2^{(k+1)/2}-1}{2^k(\sqrt2-1)} \leq\frac{2^{(k+1)/2}}{2^k}\leq\frac1{2^k}\leq\frac2n $$

Answer 3

Let $a_n = (-1)^{n-1}\ln(\lfloor \frac{n+1}{2} \rfloor)$.

$\forall k , \,\, a_{2k} = -\ln(k) = - a_{2k-1}$. Thus

$$\forall k , \,\, m_{2k} = \frac{1}{2k}\sum_{i=1}^{2k} a_{i} = 0$$

and

$$\forall k , \,\, m_{2k+1} = \frac{1}{2k+1}\sum_{i=1}^{2k+1} a_{i} = \frac{a_{2k+1}}{2k+1}$$

Clearly $a_{2k+1} \to +\infty$ and $\frac{a_{2k+1}}{2k+1} \to 0$ so the sequence $(a_n)$ is what you are looking for.

P.S. The general idea here is to pick an increasing sequence $(s_n)$ that tends to infinity at a rate slower than $n$. For example you can take $s_n = \ln(n)$ or $s_n = n^\gamma$ with $0 < \gamma < 1$. Just make sure that $s_n \to +\infty$and that $\frac{s_n}{n} \to 0$. Then the sequence defined as $s_1, -s_1, s_2, -s_2, \cdots$ will be a good one.

Answer 4

Intuitively, hardly any $a(n)$ are large so that $b(n)$ can go to $ 0$, and when $a_n$ is large, it is not "large compared to $n$". For example if $a(n)=0$ when $n$ is not a power of $2$ and $a(2^n)=n$. Note that the sequence $(a_n)_n$ cannot be monotonic. Remark : In general if a sequence $(x_n)_n$ converges to a finite limit $L$ then the sequence $y_n=(1/n)\sum_{j=1}^{j=n}x_n$ also converges to $L$, but the converse statement is false.

Answer 5

Let $a_{2^n} = n$ and $a_k = 0$ for all $k$ that are not a power of 2. The partial averages behave like $O(n^2/2^n)$ but $\limsup_n a_n = 0$.