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Define $\left \{ x_{n} \right \}$ as $x_{1}=0.1,x_{2}=0.101,x_{3}=0.101001,\dots$ Then the sequence $\left \{ x_{n} \right \}$.

  1. Converges to a rational number.
  2. Converges to a irrational number.
  3. Does not coverage.
  4. Oscillates.
asked in Calculus by Veteran (48k points) | 210 views

1 Answer

+3 votes
Best answer

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.


answered by Veteran (20.8k points)
Is sequence part of GATE syllabus ?

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