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How do I show if $x_n = \frac{n^2}{n^2 -1/2}$ is a Cauchy sequence? (using the definition of Cauchy sequence)

My attempt: A sequence is Cauchy if $ \forall \epsilon>0$ $ \exists N \in \mathbb N$ $\forall m,n \geq N$ :|$x_n -x_m$|$\leq \epsilon$

|$x_n -x_m|=|\frac{n^2}{n^2 -1/2} -\frac{m^2}{m^2 -1/2}$| $\leq|\frac{n^2}{n^2 -1/2}-1|+|1-\frac{m^2}{m^2 -1/2}|= 1/2|\frac{1}{n^2 -1/2}|+1/2|\frac{1}{m^2 -1/2}|$

$1/2|\frac{1}{n^2 -1/2}| \leq \epsilon$ and also $1/2|\frac{1}{m^2 -1/2}| \leq \epsilon$

So $1/2|\frac{1}{n^2 -1/2}|+1/2|\frac{1}{m^2 -1/2}| \leq 2 \epsilon$ if we choose $N \in \mathbb N$ such that $N >\sqrt{\frac{\epsilon}{2}+\frac{1}{2}}$

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4 Answers 4

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Easier:

$$x_n = \frac{1}{1-\frac{1}{2n^2}} \stackrel{n \to \infty}\longrightarrow 1$$ and convergent sequences are Cauchy sequences.

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Your penultimate line is unnecessary. You actually want $N>\sqrt{\frac{1}{2\epsilon}+\frac12}$ to ensure $|x_n-x_m|<2\epsilon$. Usually, we'd instead prove $|x_n-x_m|<\epsilon$ is achievable, with $\min\{m,\,n\}>\sqrt{\frac{1}{\epsilon}+\frac12}$.

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The definition you posted is correct. Note though that $$ x_n = \frac{n^2}{n^2-1/2} = 1 + \frac{1/2}{n^2-1/2}, $$ and it's clear this is a sequence which decreases to $1$, so $x_{n+1} < x_n$ for all $n$. (You could at this point claim that since the sequence is convergent, it is also Cauchy, but if you need a proof from the fundamentals, read on).

Assuming $N<m<n$, $$ \begin{split} \left|x_m - x_n\right| &= x_m - x_n \\ &= \left(1 + \frac{1/2}{m^2-1/2} - 1 - \frac{1/2}{n^2-1/2}\right) \\ &= \frac12 \left(\frac{1}{m^2-1/2} - \frac{1}{n^2-1/2}\right) \\ &= \frac12 \left(\frac{n^2 - m^2} {\left(m^2-1/2\right)\left(n^2-1/2\right)}\right) \\ &\le \frac12 \left(\frac{n^2} {\left(m^2-1/2\right)\left(n^2-1/2\right)}\right) \\ &\le \frac{1/2}{m^2-1/2} \left(\frac{n^2}{n^2-1/2}\right)\\ &\le \frac{1/2}{m^2-1/2} \left(1 + \frac{1/2}{n^2-1/2}\right)\\ &\le \frac{1}{m^2-1/2}\\ &\le \frac{1}{N^2}. \end{split} $$ Can you find what $N$ you need to pick in terms of $\epsilon$ to have that expression $< \epsilon$ in the end?

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  • $\begingroup$ so we need $N> \sqrt{1/ \epsilon}$ right? $\endgroup$
    – Bob
    Jul 1, 2020 at 20:39
  • $\begingroup$ @Bob correct :) $\endgroup$
    – gt6989b
    Jul 2, 2020 at 14:09
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In this case, you can easily conclude that it is Cauchy because it converges for $n \rightarrow \infty$

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