Example of $(b_n)$ such that $\lim_{n\to\infty} {\frac1n}\sum_{i=1}^{n-1}b_i$ does not exists and $0\le b_n\le 1$ Find a ${{b_n}}$ $n\in\Bbb N$ and $0\le b_n\le 1$ such as  the limit
$$\lim_{n\to\infty} {\frac1n}\sum_{i=1}^{n-1}b_i$$
does not exist.
I don't know how to deal with this problem, it seems to me that this limit can not be undetermined because the terms of the sum are all positive, so how is it possbile to obtain an "oscillation" of the limit?
Maybe it can be  undetermined because of $1/n$ but  I can't find any ${b_n}$, I tried whit absolute value of sine and cosine but I think that I should consider a  ${b_n}$ defined for recurrence or defined by intervals
Thanks for your help
 A: You can do it by only setting $b_n$ to either $0$ or $1$.

First of all, let's call $$B_n = \frac1n\sum_{i=1}^{n-1} b_i$$
The important thing is, that let's say we already determined $b_1,\dots, b_k$. Then, $B_k$ is a fixed number.
Now let's say we add $N$ ones to the end of the sequence, so $b_{k+1}, b_{k+2}\dots b_{k+N} = 1$.
Then, 
$$\frac1{k+N}\sum_{i=1}^{k+N-1} b_i =\frac{1}{k+N}\left(\sum_{i=1}^{k-1} b_i + \sum_{i=k}^{k+N-1} b_i\right)  \\ 
=\frac{1}{k+N}\left(k\cdot B_k + \sum_{i=k}^{k+N-1} b_i\right)  \\
=\frac{k}{k+N} B_k + \frac{N-1}{k+N}$$
which, when $N$ becomes large, approaches $1$.

So, no matter what the current value of $B_k$ is, if we add enough ones to the end of the sequence, we can get close to $1$. So, say, we can reach $\frac34$. Then, you can add enough zeroes to reach $\frac14$ and continue.
A: Given an integer $m\ge2$, let
$$
b_k=\frac12\left(1+(-1)^{\left\lfloor\log_m(k)\right\rfloor}\right)
$$
Then we have that $b_k$ is $m-1$ ones followed by $m^2-m$ zeroes followed by $m^3-m^2$ ones followed by $m^4-m^3$ zeroes, etc.
Thus, we get
$$
\sum_{k=1}^{m^n-1}b_k=\left\{\begin{array}{}
\frac{m^{n+1}-1}{m+1}&\text{if $n$ is odd}\\
\frac{m^n-1}{m+1}&\text{if $n$ is even}
\end{array}\right.
$$
Therefore,
$$
\frac1{m^n-1}\sum_{k=1}^{m^n-1}b_k=\left\{\begin{array}{}
\frac{m^{n+1}-1}{(m+1)(m^n-1)}&\text{if $n$ is odd}\\
\frac1{m+1}&\text{if $n$ is even}
\end{array}\right.
$$
Thus,
$$
\limsup_{n\to\infty}\frac1n\sum_{k=1}^nb_k\ge\frac{m}{m+1}
$$
and
$$
\liminf_{n\to\infty}\frac1n\sum_{k=1}^nb_k\le\frac1{m+1}
$$
Therefore, the limit doesn't exist. Choosing $m$ appropriately, the liminf and limsup can be made as close to $0$ and $1$ as desired.

Notice: In the case of $m=2$, this sequence is the same sequence that appears in Andreas' earlier answer.
Notice: In the case of $m=2$, the upper and lower limits, although a bit different, are also given in Andreas' earlier answer.
A: EDIT #2 related references
I noticed meanwhile that the sequence $b(n)$ I proposed here to illustrate the validity of the statement in the OP is contained in another context in OEIS (https://oeis.org/A086694,  A run of 2^n 1's followed by a run of 2^n 0's, for n=0, 1, 2, ... ).
There also an explicit formula is given
$$b(n)=\left\lfloor \log _2(n+1)\right\rfloor -\left\lfloor \log _2\left(\frac{4 (n+1)}{3}\right)\right\rfloor +1$$
and an interesting recursion relation
$$b(1) = 1, b(2) = 0, b(2n+1) = b(n), b(2n) = b(n-1)$$
Many other sequences of this type, so called "divide-and-conquer" sequences, studied by R. Stephan can be found in this reference.
EDIT
Notice that, in the following example, it is the pattern that matters, not the value. Indeed, the b's could change between $0$ and any value $c$ with $0<c<1$. The limits mentioned below then have to be multiplied by $c$.
Original post
Let me illustrate the solution idea of 5xum of using a sequence of $0$s and $1$s with an explicit example in some more detail than is possible in a comment.
We use a pattern with powers of 2: 
$$b(0) = 1, b(1) = 0, $$
$$b(2)=b(3) = 1, b(4)=b(5) = 0, $$
$$b(6)=b(7)=b(8)=b(9) = 1, b(10)=b(11)=b(12)=b(13) = 0 $$
and so on. 
The generating function of this sequence is
$$g(z)=\frac{1}{1-z}\sum _{n=0}^{\infty } z^{2 \left(2^n-1\right)} \left(1-z^{2^n}\right)$$
The sequence of 
$$B(n)=\frac{1}{n}\sum _{k=0}^n b(k)$$
has no limit because oscillates for large $n$ between 1/2 and 2/3. 
A proof of the latter statement can be done as follows
Let us consider groups of elements and refer a group by an index $m$ stating at $m=0$.
group $m = 0$ starts at $k1 = 0$ with a $1$,  followed at $k0=1$ by a $0$
group $m = 1$ starts at $k1 = 2$ with two $1$s, followed at $k0 = 4$ by two $0$s
In general,
group $m$ starts at $k1= 2(2^m-1)$ with $2^m$ $1$s, followed at $k0= 2(2^m- 1) + 2^m  = 3(2^m-1) + 1$ by $2^m$ $0$s 
The sum of all elements of a particular group $m$ is obviously $2^m$.
Hence the sum of all groups up to group $M$ inclusively, is $2^{M+1} - 1$.
This corresponds to the index start of next group $M+1$ minus $1$:
$$n =  2(2^{M+1}-1)-1$$
Hence $B(n)$, the sum divided by $n$, goes very quickly to 1/2. This is a minimum of $B(n)$ with respect to $n$ because the next goup comes in with $1$s.
The maximum value of $B(n)$ is reached at the end of the $1$s of group M. This is at 
$$n = k1 + 2^M =  2(2^M-1)+2^M = 3*2^M - 2$$
Hence the maximum of $B(n)$ is
$$B(n) =(2^{M+1} - 1)/ (3*2^M - 2)$$
the limit of which is 2/3.
Q.E.D.
A: Here is a rigorous answer, building upon the idea of @Dr. Wolfgang Hintze.
Let us consider summing $b_i=1$'s only for indices $i$ with $2^{2k} \leq i <  2^{2k+1}$, $k \in \cal N_0$, otherwise we sum $b_i=0$. The length of the $k$th summation interval is  $2^{2k+1} - 2^{2k}=   2^{2k} = 4^k$. Adding all intervals from $k=0$ to $k=L$ gives 
$$
\sum_{k=0}^{L} 4^k = \frac{4^{L+1}-1}{4-1} =
\frac{4}{3} \frac{4^{L}-1/4}{1}
$$
This corresponds to having added all numbers $b_i$ from $i=1$ up to  $i=2^{2L+1} -1 = 2 \cdot 4^L - 1$. Since we are continuing adding zeroes until $i=2^{2(L+1)} -1 = 4 \cdot 4^L - 1$, the value of the sum won't change. Since the OP's question asks for the mean, we have that the mean oscillates between 
$$
\frac{4}{3} \frac{4^{L}-1/4}{2 \cdot 4^L - 1}
$$
and 
$$
\frac{4}{3} \frac{4^{L}-1/4}{4 \cdot 4^L - 1}
$$
Large $n$ is the same as large $L$, so taking $L\to \infty$, the mean oscillates between $2/3$ and  $1/3$, hence the limit does not exist. 
