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I am asked to verify that the sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$:

$$\lim \frac{1}{6n^2+1}=0.$$

Here is my work:

$$\left|\frac{1}{6n^2+1}-0\right|<\epsilon$$

$\frac{1}{6n^2+1}<\epsilon$, since $\frac{1}{6n^2+1}$ is positive

$$\frac{1}{\epsilon}<6n^2+1$$

$$\frac{1}{\epsilon}-1<6n^2$$

$$\frac{1}{6\epsilon}-\frac{1}{6}<n^2$$

At this point, I am stuck. I'm not sure if I take the square root of both sides if I then have to deal with $\pm\sqrt{\frac{1}{6\epsilon}-\frac{1}{6}}$. That doesn't seem right.

The book provides the answer:

$$\sqrt{\frac{1}{6\epsilon}}<n$$

But I don't understand (1) what happened to the $\frac{1}{6}$, and (2) why there's not a +/- in front of the square root.

Any help is greatly appreciated.

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  • $\begingroup$ You need to BOUND $|s_n-L|$ by something. You are not solving for n. You can use crude estimates. $\endgroup$ – user223391 May 30 '15 at 21:32
  • $\begingroup$ Important thing to remember about choice of such values: There is not the answer. There can be a - in some sort - optimal answer (wich is not $n > \frac1{\sqrt{6\epsilon}}$ here) and some constants for wich it is easy to prove that they work ($\frac1{\sqrt{6\epsilon}}$ falls into this category). You generally want to go for the latter in these limit proofs, since the former are usually much harder to find. $\endgroup$ – AlexR May 30 '15 at 21:32
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    $\begingroup$ Be lazy. ${1\over 6n^2+1}<{1\over n}$. $\endgroup$ – David Mitra May 30 '15 at 21:43
  • $\begingroup$ If you want to show $a_n < \epsilon$, it's often easier to find a $b_n$ (Which is simpler than $a_n$) such that $a_n<b_n< \epsilon$. This is a common trick with proving limits from first principles, and means the expressions are simplified like in Simon's proof. $\endgroup$ – Omar Haque May 30 '15 at 23:39
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Note that $\left| {1 \over 6n^2 + 1} - 0 \right| < \left| 1 \over 6n^2 \right|$. Hence if we can bound the larger term by $\epsilon$ you're done. I.e., given an arbitrary $\epsilon > 0$, we can choose a lower bound $N$ by

$$N = \sqrt{\frac{1}{6\epsilon}}$$

Then

$$n > N \Rightarrow \left| {1 \over 6n^2 + 1} - 0 \right| < \left| 1 \over 6n^2 \right| < {1 \over 6N^2} = \epsilon$$

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Let $\epsilon > 0$. Let's sketch it first. $$\frac{1}{6n^2+1} < \frac{1}{6n^2} < \epsilon.$$ Look: $$\frac{1}{6n^2} < \epsilon \iff 1 < 6n^2\epsilon \iff \frac{1}{6\epsilon} < n^2 \iff n > \frac{1}{\sqrt{6\epsilon}}.$$


Now we begin. Let $\epsilon > 0$. Then exists $n_0 \in \Bbb N$ such that $n_0 > 1/\sqrt{6\epsilon}$. If $n \geq n_0$, then: $$\left|\frac{1}{6n^2+1}\right| = \frac{1}{6n^2+1} < \frac{1}{6n^2} \leq \frac{1}{6n_0^2} < \epsilon.$$

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Hy!

I will take a similar approach to that of Simon's and that's to find a "similar" N for which numbers greater than him are smaller than an upper bound defined in N(that's our epsilon).

It's actually a convergence test done with comparison.

$$\text{Given we have two sequences, } a_n \text{ and } b_n \text{ if} \\ \sum_0^na_n \to convergent \quad \text{than} \quad \sum_0^nb_n \to convergent \quad if\\$$ $$\lim_{x\to\infty} \dfrac{a_n}{b_n} = c, \quad where \quad c\in(0, \infty)$$ That means for your problem that one similar sequence might be: $$\dfrac{1}{6n^2} \quad or \quad \dfrac{1}{n^2}$$ $\dfrac{1}{n}$ is not good, or better said not helpful. These methods have their root in your epsilon, and their true just because of what Simon said.

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