b) Converges to an irrational number.
$x_n < 1,\;\forall n \geq 1$. So, the sequence is bounded.
$x_{n+1} > x_n,\;\forall n \geq 1$. So, the sequence is monotone.
Since any bounded monotone sequence converges (Monotone Convergence Theorem), the given sequence converges.
Any rational number produces a repeating decimal sequence. Since the decimal sequence of $x_{\infty}$ never repeats (the number of 0's before the next 1 increases), $x_{\infty}$ is an irrational number.
Another way to prove that the sequence converges is as follows:
$$x_n = \sum_{i=1}^{n}10^{-i(i+1)/2}$$
Theorem: $x_{\infty} = \sum_{i=1}^{\infty}\underbrace{10^{-i(i+1)/2}}_{a_i}$ converges.
Proof by Ratio Test:
$$\begin{align}&\lim_{i \to \infty}\frac{a_i}{a_{i-1}} \\[1em]= &\lim_{i\to\infty}\frac{1/10^{i(i+1)/2}}{1/10^{(i-1)i/2}} \\[2em]= &\large \lim_{i\to\infty} 10^{\left (\frac{(i-1)i}{2}-\frac{i(i+1)}{2} \right)} \\[2em] = &\lim_{i\to\infty}10^{-i} = 0\end{align}$$
Since this limit it $0$, by the Ratio Test of Series Convergence, the series $x_{\infty} = \sum_{i=1}^{\infty}10^{-i(i+1)/2}$ converges.
Since $x_{\infty}$ converges, the sequence $\left \{x_n \right\}$ converges.