Is the sequence $0,1,0,0,1,0,0,0,1,0,0,0,0,1,\ldots$ convergent? The sequence $0,1,0,0,1,0,0,0,1,0,0,0,0,1,\ldots$ is divergent according to a video I watched on WebAssign but the comment was very informal. Could someone provide a hint about the formal proof that the sequence is divergent?
 A: In fact, $t=0,100100010000\ldots$ is irrational, and your sequence is s.t. $x_n$ gives the $n$-th digit of $t$ and thus can't converge. But anyway, it's not the exercise to show that it's irrational. First, it's clear that $0\leq x_n\leq 1$ and that $x_{n_k}\in \{x_1,x_4,x_8,x_{13},\ldots\}$ is a subsequence that is s.t. $x_{n_k}=1$ for all $k$. And $x_{m_k}\in\{x_0,x_2,x_5,x_9,x_{14},\ldots\}$ is a subsequence s.t. $x_{m_k}=0$ for all $k$. Therefore $$\limsup_{n\to\infty }x_n=1$$
and $$\liminf_{n\to\infty }x_n=0$$
and thus, the sequence doesn't converge.
A: Every convergent sequence must be Cauchy, right? 
A: Let's assume that $\lim_n x_n=L$, this means that
$$\forall \epsilon>0\exists N\in \mathbb{N}\ : \ n\geqslant N \Rightarrow |x_n-L|<\epsilon$$
Let's fix a random $N_0\in \mathbb{N}$. We notice that if $n\geqslant N_0$ then $|x_n-L|<\epsilon$ can't hold for any $\epsilon <\frac12$. Therefore the sequence doesn't converge.
A: Observe that the $n$-th term of the given sequence $(x_n)$ is 1 if $n+1$ is a triangular number, and is 0 otherwise.
Suppose that $(x_n)$ is convergent, then $\exists a \in \Bbb R$ such that $x_n \to a$.
i.e. $\forall \epsilon > 0, \exists N \in \Bbb N$ such that $\forall n \ge N, |x_n - a| < \epsilon$.
Case 1: $a = 0$
Take $\epsilon_0 = \dfrac12$.  Then it's easy to see that if $n+1$ is a triangular number and $n \ge N$, then $|x_n - a| = 1 - 0 > \dfrac12 = \epsilon_0$.
Case 2: $a = 1$
Take $\epsilon_0 = \dfrac12$.  Then it's easy to see that if $n+1$ is not a triangular number and $n \ge N$, then $|x_n - a| = |0 - 1| > \dfrac12 = \epsilon_0$.
Case 3: $a \ne 0$ and $a \ne 1$
Take $\epsilon_0 = \min\{|a|,|a - 1|\} > 0$.  Then $|x_n - a| = |1 - a|$ or $|a - 0| \ge \epsilon_0$.
In either case, the $\epsilon$-$N$ definition of the convergence of sequence is contradicted since we can find an $\epsilon_0$ which doesn't fit into the definition.
A: If the series converges to some limit $L$, then all but finitely many of its terms are within a distance of $0.1$ from $L$, i.e. between $L-0.1$ and $L+0.1$.  But $0$ and $1$ cannot both be between $L-0.1$ and $L+0.1$.  The triangle inequality shows that.
