TITLE: convergence and divergence of sequences
QUESTION [5 upvotes]: If we have a sequence $\{a_n\}$ in $R$,
$b_n = (a_1 + a_2 + \cdots + a_n)/n$
(a) an example of a sequence $a_n$ in which $a_n$ diverges while $b_n$ converges
(b) an example of a bounded sequence $a_n$ such that $b_n$ diverges
Could you give me some help?
I  thought about a sequence $\{1,2,3,4,\ldots, n\}$ but this obviously doesn't work.

REPLY [2 votes]: We give a detailed  answer to (b). Too detailed, perhaps, since it may make a simple idea look more complicated than it is.
Consider the sequence $a_1,a_2,a_3,\dots$ that goes as follows:
$$1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,\dots,1,0,0,0, \dots,$$
where we have $1$ one, then $9$ zeros, then $90$ ones, then $900$ zeros, then $9000$ ones, and so on. The sequence $(a_n)$ is bounded, since every entry is zero or one.
Note the following features of this sequence. The term $a_{10}$ is at the end of a string of $9$ consecutive zeros, the term $a_{100}$ is at the end of a string of $90$ consecutive ones, the term $a_{1000}$ is at the end of a string of $900$ consecutive zeros, and so on.
(i) We will estimate $b_{10}$, $b_{1000}$, $b_{100000}$, and so on.
(ii) We will separately estimate $b_{100}$, $b_{10000}$, $b_{1000000}$, and so on.
Estimates for (i): The number $b_{10}$ is $\frac{a_1+a_2+\cdots +a_{10}}{10}$, so it is the average of the first $10$ terms of the sequence $(a_n)$. Of these $10$ terms, $9$ are $0$'s, so $b_{10}\le \frac{1}{10}$ (of course we have equality).
Now look at $b_{1000}$. This is the average of the first $1000$ terms of the sequence $(a_n)$. Of these terms, at least $900$ are zeros. (Actually, $909$ are zeros.) So no more than a $100$ are ones, and therefore $b_{1000}\le \frac{1}{10}$.
Now look at $b_{100000}$. This is the average of the first $100000$ terms. Of these, the last $90000$ are zeros, so at most $10000$ are ones, and therefore $b_{100000}\le \frac{1}{10}$.
The same reasoning shows that if $n=10^k$, where $k$ is odd, then $b_n\le \frac{1}{10}$.  
Estimates for (ii): Look at $b_{100}$. The last $90$ of the terms we are averaging are $1$ (actually, exactly $91$ of the terms are $1$). So $b_n\ge \frac{9}{10}$.
Look at $b_{10000}$. The last $90000$ terms we are averaging are $1$, so $b_{10000}\ge \frac{9}{10}$.
Similarly, if $n=10^k$ where $k$ is even, then $b_n\ge \frac{9}{10}$.
Concluding the non-existence of the limit: We show that $\lim_{n\to \infty} b_n$ does not exist. One can do a formal $\epsilon$-$N$ argument or do it informally. For the informal version, there are arbitrarily large $n$ of Type (i), where $b_n\le \frac{1}{10}$. There are also arbitrarily large $n$ of Type (ii), where $b_n\ge \frac{9}{10}$.
So the limit of the $b_n$, it it existed, would have to be simultaneously $\le \frac{1}{10}$ and $\ge \frac{9}{10}$. That's impossible. 
If you need the $\epsilon$-$N$ version of non-existence, I can supply it. But as a hint we can take $\epsilon$ say $1/4$.