My book says that this sequence diverges because it "takes on only two values, 0 and 1, and never stays arbitrarily close to either one for n sufficiently large." However, I don't understand why this cannot converge. My intuition says that for an infinite n, we will reach a point where there are infinitely many 0's before hitting a 1, because each repetition, we get more and more zeroes. Eventually, wouldn't we have infinite zeros, and the pattern would converge to 0?
[Math] Why does {0,1,0,0,1,0,0,0,1,0,0,0,0,1…} diverge
sequences-and-series
Related Solutions
In textbooks, the convergence of a sequences is often discussed before series.
If a sequence is monotonous, say $a_{n+1} \geq a_n$, and bounded from above, yes, it will converge. This is proven by showing it has an upper bound (reformulation of "bounded from above"), and using the fact that every bounded set has a smallest upper bound (called supremum). This smallest upper bound of the sequence is its limit.
Next, by considering the partial sums of a series, you will notice that the partial sums form a sequence. If this sequence of partial sums converges, the series is said to converge.
Your textbook seems to be confusing for you for not repeating the partial sum notation in the lines you quote. Just consider that writing the partial sum with a sum-sign is the shorthand for the many terms of the sum added together. The $s_n$ form the sequence of partial sums of the $a_n$.
If a sequence of partial sums $s_n$ converges, it follows that the sequence of its terms $a_n$ converge to 0. Proof. For all n>N, $|s_n - s_{n+1}|$ will be arbitrarily small, hence $|a_{n+1}|$ will be arbitrarily small, i.e., the sequence of $a_n$ converges to 0.
Caution. The reverse is not true. If the sequence $(a_n)_{n \geq 1}$ converges to 0, then it does NOT follow that the sequence of its partial sums $(s_n)_{n \geq 1}$ converges.
Example. The sequence $(a_n)_n = (1/n)_n$ converges to 0, however the sequence of the partial sums $s_N = \sum_{n=1}^N a_n$ is unbounded (hence, the series diverges). Note that $\sum_{n=2^{k}+1}^{2^{k+1}} 1/n > 1/2$ for all k.
You have a sequence of partial sums $\{S_n\}_{n=1}^{\infty}$.
By adding an arbitrary but finite number of zeros after each term, you have new partial sums $\{\tilde{S}_n\}_{n=1}^{\infty}$. The new partial sum sequence is the same as the old except each term $S_n$ is repeated however many (finite) times you like. For example $$\{\tilde{S}_n\} = \{S_1, S_2, S_2, S_3, S_3, S_3, S_3, S_4, ...\}$$
Fix $c \in \mathbb{R}$. If $S_n\rightarrow c$ then $\tilde{S}_n\rightarrow c$.
On the other hand, suppose we only know: $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n S_i = c$$ Then $\frac{1}{n}\sum_{i=1}^n S_i$ may not converge. Try $$a_i = (-1)^{i+1} \quad \forall i \in \{1, 2, 3, ...\}$$ Then $S_i=1$ if $i$ is odd and $S_i=0$ if $i$ is even, and $$ \lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n S_i = \frac{1}{2}$$ However, we can just repeat terms to make $\frac{1}{n}\sum_{i=1}^n \tilde{S}_i$ converge to something other than $\frac{1}{2}$, or not converge at all. For example \begin{align} &\{S_n\} = \{1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...\}\\ &\{\tilde{S}_n\} = \{1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...\} \end{align} We can make $\{\tilde{S}_n\}$ any arbitrary binary-valued sequence that starts with 1 and contains an infinite number of 0s and an infinite number of 1s.
Best Answer
No, you will not. The number of zeros between each pair of ones gets bigger and bigger, but it is always finite. Consider that each term in the sequence only has a finite number of terms that come before it. So if after some instance of 1, you have infinitely many zeros, how many zeros precede this 1?
Moreover, we have a definition for a real sequence converging:
So if $\epsilon=1/2$, how big must $N$ be?