Question

Qno.1) a) check whether the following sequences are cauchy sequence or not ?
I) (n+1/n) ii) (1+1/2!+.....+1/n!)
iii) (-1)^n iv) (n+ (-1)^n/n)
b) let x1<x2 be arbitrary real number and Xn= 1/2(Xn-2+Xn-1) for n>2. Show that (Xn) is convergent ? What is it limit ?

Answer 1

Answer on Question # 82678 - Math - Real Analysis

Question 1. a) Is the sequence a Cauchy sequence or not?

b) $x_{n} = \frac{1}{2} (x_{n - 2} + x_{n - 1})$, $x_{1} < x_{2}$. Show that $\{x_{n}\}_{n\in \mathbb{N}}$ converges and find the limit.

Solution. a) We can use the statement: a sequence converges if and only if it is a Cauchy sequence. Proof of this statement: http://www.math.stonybrook.edu/~mde/319S_08/testsolII.pdf

(i) $\left\{n + \frac{1}{n} : n \in \mathbb{N}\right\}$ diverges because $n + \frac{1}{n} \to \infty, n \to \infty$. So it is not a Cauchy sequence.

(ii) $\left\{1 + \frac{1}{2!} + \dots + \frac{1}{n!} : n \in \mathbb{N}\right\}$ converges because we have the Taylor series $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ and for $x = 1$ we have $1 + \frac{1}{2!} + \dots + \frac{1}{n!} + \dots \to e - 1, n \to \infty$. So it is a Cauchy sequence.

(iii) $\{(-1)^n : n \in \mathbb{N}\}$ diverges because $(-1)^{2k} \to 1$ and $(-1)^{2k-1} \to -1, k \to \infty$. So it is not a Cauchy sequence.

(iv) $\left\{n + \frac{(-1)^n}{n} : n \in \mathbb{N}\right\}$ diverges because $n + \frac{(-1)^n}{n} \to \infty, n \to \infty$. So it is not a Cauchy sequence.

b) You can prove convergence using the Cauchy criterion - the distance between consecutive terms halves every term: $|x_{n} - x_{n - 1}| = \frac{1}{2} |x_{n - 1} - x_{n - 2}|$, so iterating this equality to obtain $|x_{n} - x_{n - 1}| = 2^{-n + 2}|x_{2} - x_{1}|$ we can use a geometric series bound (i.e. use $\sum_{n = 0}^{\infty}r^{n} = \frac{1}{1 - r}$ for $|r| < 1$) to show the sequence is Cauchy:

$$| x _ {n + m} - x _ {n} | \leqslant \sum_ {k = 1} ^ {m} | x _ {n + k} - x _ {n + k - 1} | \leqslant 2 ^ {- n} | x _ {2} - x _ {1} | \sum_ {k = 1} ^ {m} 2 ^ {- k} \leqslant 2 ^ {- n} | x _ {2} - x _ {1} | \to 0, n \to \infty .$$

To find the limit, write the sequence as a geometric series similar to the above:

$$\lim _ {n \to \infty} x _ {n} = x _ {1} + \sum_ {n = 2} ^ {\infty} (x _ {n} - x _ {n - 1}) = x _ {1} + (x _ {2} - x _ {1}) \sum_ {n = 2} ^ {\infty} (- 2) ^ {- n + 2} = x _ {1} + (x _ {2} - x _ {1}) \frac {1}{1 - (- 2)} = \frac {x _ {1}}{3} + \frac {2 x _ {2}}{3}.$$

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