Question

Check whether the following sequences{sn} are Cauchy, where
(i) sn = 1+2+3+....+n
(ii) sn = [4n^(3)+3n]/[3n^(3)+n^(2)]

Answer 1

Answer on Question #69107 – Math – Real Analysis

Question

Check whether the following sequences{sn} are Cauchy, where

(i) $sn = 1 + 2 + 3 + \dots + n$ ;

(ii) $sn = [4n^{\wedge}(3) + 3n] / [3n^{\wedge}(3) + n^{\wedge}(2)]$ ;

Solution

Definition. Sequence $\{s_n\}$ is Cauchy if

$$\forall \varepsilon > 0 \ \exists N \ \forall n, m > N: |s_n - s_m| < \varepsilon$$

(i) First method.

For the sequence $s_n = 1 + 2 + 3 + \dots + n$, $n \geq 1$, the smallest difference between its two elements is $1$ ($s_{n+1} - s_n = 1$). It cannot be less than any $\varepsilon > 0$ by the definition.

So this sequence is not Cauchy.

Second method.

Every real Cauchy sequence is convergent, but the sequence $s_n = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$ does not converge $\left(\lim_{n \to \infty} \frac{n(n+1)}{2} \neq 0\right)$, so this sequence is not Cauchy.

(ii) We have the sequence $s_n = \frac{4n^3 + 3n}{3n^3 + n^2} = \frac{4n^2 + 3}{3n^2 + n}$.

Its limit is

$$\lim_{n \to \infty} \frac{4n^2 + 3}{3n^2 + n} = \lim_{n \to \infty} \frac{4 + \frac{3}{n^2}}{3 + \frac{1}{n}} = \frac{4}{3}.$$

By the definition of a convergent sequence, let $\varepsilon > 0$. Choose $N$ so that if $n > N$, then

$$\left| s_n - \frac{4}{3} \right| < \varepsilon$$

Now let's check if this sequence is Cauchy:

$$\forall \frac{\varepsilon}{2} > 0 \ \exists N \ \forall n, m > N: |s_n - s_m| < \frac{\varepsilon}{2};$$ $$|s_n - s_m| = \left| s_n - \frac{4}{3} - \left(s_m - \frac{4}{3}\right) \right| \leq \left| s_n - \frac{4}{3} \right| + \left| s_m - \frac{4}{3} \right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

Hence this sequence is Cauchy.

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