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I am trying to to understand the limit definition. For the sequence $\dfrac{3n+1}{n}$ I can divide by the highest polynomial in denominator and get that 3 may be the limit.

Using the limit definition $$\left|\dfrac{3n+1}{n}-3\right|<\epsilon \implies \left|\dfrac{3n+1-3n}{n}\right|<\epsilon \implies \dfrac{1}{n}< \epsilon \rightarrow \dfrac{1}{\epsilon}<n $$ which means that whatever $\epsilon$ smallest as I want the will be $N<n$ that from then on all the element of the sequence will be close to $3$.

Now if I take a wrong limit like $2$ I get $$\left|\frac{3n+1}{n}-2\right|<\epsilon \implies \left|\frac{3n+1-2n}{n}\right|<\epsilon \implies\frac{n+1}{n}<\epsilon$$

Now because I can not isolate $n$ that mean that there is no $N<n$ that from than on all the element of the sequence will be close to $2$?

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    $\begingroup$ In the title, did you mean to have $n \to \infty$ instead of $x$? $\endgroup$ – JimmyK4542 Dec 24 '14 at 23:41
  • $\begingroup$ @JimmyK4542 Yes sorry fixed it $\endgroup$ – gbox Dec 24 '14 at 23:41
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    $\begingroup$ It's not because you can't "isolate" $n$. It's because $\frac{n+1}n$ does not tend to $0$ so you can't make it be arbitrarily small. $\endgroup$ – Mathmo123 Dec 24 '14 at 23:42
  • $\begingroup$ I think I have it now, if I manage isolate $n$ so for any $\epsilon$ I will take I will get $N<n$ that from than on all the elements of the sequence will fall in the epsilon area $\endgroup$ – gbox Dec 25 '14 at 0:13
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In general, $x_n \to \ell$ as $n \to \infty$ if and only if $x_n - \ell \to 0$ as $n \to \infty$.

So the way to find out if you have the correct limit is:

Do we have $x_n - \ell \to 0$?


In your example, $x_n = \frac{3n+1}n$ and $\ell = 3$. Notice that $x_n - \ell = \frac 1n \to 0$ as $n \to \infty$. This means that for any $\epsilon > 0$, we can find $N$ such that for every $n \ge N$, $\frac 1n \lt \epsilon$.

But $x_n - 2 = \frac{n+1}n \to 1 \ne 0$

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  • $\begingroup$ Thanks! do you know where I can read more about the theorem $x_n \to \ell$ as $n \to \infty$ if and only if $x_n - \ell \to 0$ as $n \to \infty$.? $\endgroup$ – gbox Dec 25 '14 at 0:02
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    $\begingroup$ It more or less follows directly from the definition. In showing that $\forall \epsilon \gt 0$, $\exists N$ such that for all $n \gt N$, $|x_n - \ell| < \epsilon$, you are showing exactly that $x_n - \ell \to 0$. (Write out the definition of the latter fact and you'll see it's the same) $\endgroup$ – Mathmo123 Dec 25 '14 at 0:04
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Note that $\frac{n+1}{n}=1+\frac 1n$. You'll have some trouble getting this to be less than $\epsilon$, since it's always greater than $1$ for positive integers $n$!

Let's say you want to find $\lim_{n \to \infty} \frac{p(n)}{q(n)}$, where $p$ and $q$ are polynomials of the same degree. Let's write $p(x) = a_0x^k + \dots a_k$ and $q(x) = b_0 x^k + \dots + b_k$. Then I claim the limit of $\frac{p(n)}{q(n)}$ is $a_0/b_0$. (I can write a proof if you want, but you can probably find it in your textbook.) In particular, when you're trying to use the $\varepsilon$-definition to see that $\lim \frac{p(n)}{q(n)} \to L$, you have an intermediate step where you write $\|\frac{p(n)}{q(n)} - L\| = \|\frac{p_2(n)}{q_2(n)}\|$ for some other polynomials $p_2$ and $q_2$. (In your example above, when you picked $L = 2$, you got $p_2(n) = n+1$ and $q_2(n) = n$.) So if you want $\frac{p(n)}{q(n)}$ to have any hope of converging to $L$, this new thing had better converge to zero - so $p_2$ had better have degree less than $q_2$! And that's the problem when you picked $L=2$ - the two polynomials had the same degree.

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    $\begingroup$ I am trying to generalize it, if I have a constant with the n-expression I know it is not the limit? $\endgroup$ – gbox Dec 24 '14 at 23:46
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Logic:

$\frac{3n+1}{n}$ well, $3n+1$ is only a little bit bigger than $3n$, take n being 100, then $3n+1$ is 301, which is basically 300, take 1,000,000, then 3,000,001 is basically 3,000,000

So $\frac{3n+1}{n}\approx\frac{3n}{n}=3$

That's the logic.

Formal:

Don't bother doing $\forall\epsilon>0\exists N\in\mathbb{N}:n>N\implies|a_n-a|<\epsilon$ where $a_n$ is the $n^\text{th}$ term and $a$ is the limit.

You should have done convergence of the sum of two convergent sequences, just be like:

BUT WAIT: $\frac{3n+1}{n}=\frac{3n}{n}+\frac{1}{n}=3+\frac{1}{n}$ and note that $\frac{1}{n}\rightarrow 0$ which is an easy one to prove (reply in a comment if you want the proof of this)

Job done!

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  • $\begingroup$ thanks Alec. What I am trying to understand is where does the formal definition of the limit "fail" when I enter a wrong limit $\endgroup$ – gbox Dec 24 '14 at 23:52
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    $\begingroup$ @gbox it did fail! Take your $\frac{n+1}{n}=\frac{n}{n}+\frac{1}{n}=1+\frac{1}{n}$ The definition must work FOR ALL $\epsilon$ so pick $\epsilon=2$ say, BAM, failure. Look at the definition as I wrote it - for all $\epsilon$ there must exist an $N$ such that for any number bigger than $N$ we have the sequence values at $n$ and the limit less than epsilon apart. $\endgroup$ – Alec Teal Dec 24 '14 at 23:56
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    $\begingroup$ @gbox sorry, take $\epsilon=\frac{1}{2}$ then $1+\frac{1}{n}<\epsilon=\frac{1}{2}$ is a fail. $\endgroup$ – Alec Teal Dec 24 '14 at 23:59
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    $\begingroup$ @gbox AHHH! I see your confusion, let me explain, you are not trying to show for really tiny epsilon that the definition works, you NEED TO SHOW IT FOR BIG EPSILON TOO! The definition states $\forall\epsilon>0$ - so pick any number - any number you like, it must work for that, you can pick 0.00005 or 5,000,000 and for any of these epsilon values YOU MUST BE ABLE TO FIND an $N$ such that whenever n>N you have the "limit" and the $n^\text{th}$ term being AT Most epsilon far apart. $\endgroup$ – Alec Teal Dec 25 '14 at 0:02
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    $\begingroup$ @gbox for bonus points, to show that something doesn't converge you literally negate all the definitions, so rather than (in full) $\forall\epsilon>0\exists N\forall n>N$ we have $|a_n-a|<\epsilon$, negating it: $\exists\epsilon>0\forall N\exists n>N:|a_n-a|\ge\epsilon$ and this is asserting "There exists an epsilon such that for ANY N such that we we can find an n > N where $a_n$ and $a$ are more than epsilon apart. $\endgroup$ – Alec Teal Dec 25 '14 at 0:11

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