Analyze the convergence or divergence of $\{1/n^2\}$

$$\left|\frac{1}{m^2} - \frac{1}{n^2}\right| < \left|\frac{1}{m^2}\right| < \frac{1}{N^2} < \varepsilon$$ whenever $N > \dfrac{1}{\sqrt{\varepsilon}}$ (first step because $n^2 > 0$ and hence $\dfrac{1}{n^2} > 0$ and $\dfrac{1}{m^2} - \dfrac{1}{n^2} < \dfrac{1}{m^2}$)

So the sequence is Cauchy and it certainly converges to zero.

Is my approach correct?

  • $\begingroup$ 1) Please learn how to type in LaTeX in this site: it's pretty simple! 2) What's the question? $\endgroup$ – DonAntonio May 24 '12 at 2:47
  • $\begingroup$ Is it right to my procedure? $\endgroup$ – Daniela del Carmen May 24 '12 at 2:51
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    $\begingroup$ Please: learn some simple LaTeX instead of relying on others to make your posts intelligible, and make sure to include in the body both your question, and what it is you are doing. The post does not begin in the title and continue in the body, it begins at the first character of the body. $\endgroup$ – Arturo Magidin May 24 '12 at 2:58
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    $\begingroup$ Are you sure it is the convergence of the sequence that is asked for? The notation $\{1/n^2\}$ is unusual for sequences. In the version of the definition of Cauhy sequence that is in your book, probably $m<n$, but you should not take it for granted that it is. Also, the definition probably says that there is an $N$ such that if $n\gt m \gt N$, then $\dots$. But again you should be explcit, as if the person reading your proof did not have your book in front of them. Apart from these flaws, your solution is fine. $\endgroup$ – André Nicolas May 24 '12 at 3:12
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    $\begingroup$ @AndréNicolas: Some authors use this notation instead of $(\cdot)$ for a sequence, and it is not even that unusual. See for example Falconer's book, the geometry of fractal sets, which is a classic piece. He uses $\{\cdot\}$ notion for sequences, and there are plenty of others. Hence I would not conclude that this question is not about sequences based on that notation. $\endgroup$ – T. Eskin May 24 '12 at 4:04

The reasoning is not quite correct, but it is in fact a Cauchy sequence that converges to $0$.

$\frac{1}{n^2}>0$ is true when $n$ is a positive integer, and this does imply that $\frac{1}{m^2}-\frac{1}{n^2}<\frac{1}{m^2}$. However, that does not mean that $\left|\frac{1}{m^2}-\frac{1}{n^2}\right|<\frac{1}{m^2}$. For example, consider $n=1$, $m=4$.

Instead, you could note that $\left|\frac{1}{m^2}-\frac{1}{n^2}\right|$ is equal to either $\frac{1}{m^2}-\frac{1}{n^2}$ or $\frac{1}{n^2}-\frac{1}{m^2}$, and find a number that is an upper bound for both of these.

The reasoning could also be made more complete by specifying how $m$ and $n$ are related to $N$.



To answer straight from the definition, take $N$ such that $1/N < \sqrt{\epsilon/2}$

Then using the triangle inequality for $m,n\geq N$

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

Your approach is flawed because your first inequality is incorrect. The easiest way to show this sequence is Cauchy is to show it converges to 0. (A convergence sequence is Cauchy)

Fix $\epsilon>0$. There exists an $N$ (by the Archimedean property) such that for all $n\geq N$, $1/n \leq 1/N < \sqrt{\epsilon}$

Then $$|1/n^{2} - 0| = 1/n^{2} <\epsilon$$

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    $\begingroup$ If the problem is to specifically learn how to use the Cauchy definition with the $\varepsilon$'s-$N$'s the whole point is to stick with it, as it is usually when you want to prove trivialities such as $\frac 1{n^2} \to 0$... I think OP knew that but wanted help on how to show it, which is a different game. $\endgroup$ – Patrick Da Silva May 24 '12 at 3:36
  • $\begingroup$ Thanks @PatrickDaSilva, I updated my post. $\endgroup$ – toypajme May 24 '12 at 12:45
  • $\begingroup$ I think it is worth saying that depending on the wording of your Archimedean property, using it is essentially equivalent to using convergence of the sequence to 0. In which case, my "direct" strategy is just using the strategy behind proving an arbitrary Cauchy sequence converges in $\mathbb{R}$. $\endgroup$ – toypajme May 24 '12 at 14:03
  • $\begingroup$ Indeed, that answers OP's question better. +1 $\endgroup$ – Patrick Da Silva May 24 '12 at 16:20

I assume that you want to prove that the sequence $\{a_n \}$ given by $a_n = \frac{1}{n^2}$ is convergent. Indeed as you see from the other answer, you can first prove that the sequence is Cauchy, and then conclude that it is convergent. So that approach is correct. I just wanted to point out that you actually can prove "directly" that the sequence is convergent.

You all ready have guessed that the limit of the sequence is zero. So let $\epsilon > 0$ be given. We want to find $N>0$ such that if $n \geq N$, then $\lvert a_n \lvert < \epsilon$. So we need $$\frac{1}{n^2} < \epsilon. $$ Note that this equivalent to saying that $$ \frac{1}{\sqrt{\epsilon}} < n. $$ Hence we can pick any $N$ such that $\frac{1}{\sqrt{\epsilon}} < N$. Then indeed if $n \geq N$ then you have $$\begin{align} \frac{1}{\sqrt{\epsilon}} < N \leq n &&\Rightarrow \\ \lvert a_n \lvert = \frac{1}{n^2} < \epsilon. \end{align} $$ And we are done.


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