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What does the sequence of $n=1$ to infinity converge to for $\dfrac{1}{n}$ and how do I prove this?

I understand that as $n$ gets bigger, the fraction gets smaller, but how do I find the exact value it converges to?

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    $\begingroup$ It goes to 0. Just use the epsilon definition to prove it. $\endgroup$ – user60887 Dec 11 '13 at 1:39
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    $\begingroup$ Why the downvotes? The question is valid and it seems that the asker needs basic help. $\endgroup$ – robjohn Dec 11 '13 at 1:41
  • $\begingroup$ This question is equivalent to "what is the exact value of $n$ as $n \to \inf$ ?" $\endgroup$ – AHH Dec 22 '13 at 10:32
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Remember a sequence $\{ x_n \}$ converges to $x$, $x_n \to x$ if given any $\epsilon > 0$, then we can find an $N \in \mathbb{N}$ such that if $n > N$, then $|x_n -x | < \epsilon$.

Now, as for your problem, the claim is that $x_n = \frac{1}{n} \to 0 $. To show this, suppose $\epsilon > 0$. Now, we want to find $N$ such taht if $n > N $, then $|\frac{1}{n} - 0| < \epsilon $. well, notice

$$ |\frac{1}{n} - 0| = |\frac{1}{n}| = \frac{1}{n} < \epsilon \iff n > \frac{1}{\epsilon} $$

So, it is evident we should take $N > \frac{1}{\epsilon} $. With this choice, then

if $n > N$, $\frac{1}{n} < \epsilon \implies |\frac{1}{n}| < \epsilon$.

So, by definition, $\frac{1}{n} \to 0 $

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It's not clear how much formal-proof experience you have, so it might help to present an informal (no epsilons!) argument.

For every element $\frac 1 n$ in the sequence, $\frac 1 n > 0$. So if the sequence converges it must converge to a number $\geq 0$. Now pick any number $0 < x \leq 1$. For high enough n $x > \frac 1 n$. So if the sequence converges, it (1) can't converge to any number $< 0$ and (2) can't converge to any number $> 0$. So it converges to 0.

A teacher might not be completely satisfied with this and go "wait how do you know it converges at all", but it's pretty simple to show that it converges to something from what we already have.

EDIT: MJD brought up in the comments that this doesn't exclude it converging to a negative number- the argument as it stands holds if you replace $0$ with, say, $-17$. We can patch the argument like this: Pick an arbitrary negative number $-s$. Since every number in the sequence is positive, the distance between the $nth$ element and $-s$ is $|s + \frac{1}{n}| > \frac{s}{2} \forall s,n$. If the sequence never comes within $s/2$ of $-s$, it can't converge to $-s$. Since $-s$ was arbitrary, this holds true for all negative numbers.

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  • $\begingroup$ Making the clever substitution $x \mapsto \epsilon$ we see this is the standard $\epsilon$ proof in disguise! $\endgroup$ – Deven Ware Dec 11 '13 at 2:21
  • $\begingroup$ @DevenWare curses you found me out! $\endgroup$ – Hovercouch Dec 11 '13 at 18:50
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    $\begingroup$ I don't like how you sneakily changed the $>$ sign to a $\ge$ sign with no explanation. Without something that your argument is missing, you could just as easily have begun like this: “For every element $\frac 1n$ in the sequence, $\frac1n > -17$. So if the sequence converges it must converge to a number $\ge -17$… $\endgroup$ – MJD Dec 18 '13 at 5:45
  • $\begingroup$ @mjd you're right, that would have to be elaborated on to make the proof rigorous. I'll edit when I'm not on mobile. $\endgroup$ – Hovercouch Dec 21 '13 at 21:25
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It converges to $0$ since for any $\epsilon > 0$, we can find $N$ such that $1/N < \epsilon$. So for $n > N$ we have $|1/n - 0| < 1/N < \epsilon$.

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  • $\begingroup$ I highly doubt the OP's class is using $\epsilon-\delta$ arguments. $\endgroup$ – user98602 Dec 11 '13 at 1:43
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    $\begingroup$ Do people cover convergence before calculus now? How do they define what it means? EDIT: Based on other posts by the OP, I assume they are in a calculus course, so $\epsilon-\delta$ should be fair game there. $\endgroup$ – zrbecker Dec 11 '13 at 1:44
  • $\begingroup$ if he is studying in U.S, math.stackexchange.com/questions/596557/… $\endgroup$ – ILoveMath Dec 11 '13 at 1:48
  • $\begingroup$ It is difficult to give a rigorous proof without using an $\epsilon$-$N$ type argument, and the OP is asking for a proof. $\endgroup$ – Stefan Smith Dec 11 '13 at 1:54
  • $\begingroup$ @DonAnselmo, I learned calculus in the US and had proofs. I think that post is overgeneralizing. $\endgroup$ – zrbecker Dec 11 '13 at 2:32
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Without $ε$ definition .

Let $x_n=\frac {1}{n}$.Then $x_n$ is strictly decreasing because $n+1>n<=>\frac {1}{n+1}<\frac {1}{n}<=>x_{n+1}<x_n$. Now,let $A=${$x_n$} be the set of sequence's terms.We can see that $0$ is a lower bound of $A$.All we need to do is to prove that $0$ is the infimum of $a$ ($infA=0$). Note that the infimum exists because $A$ is lower bounded.

Suppose (by contradiction) that there is a $x>0$ such that $x=infA$.This means that every term of the sequence $x_n$ will be larger than $x$. But :

If $x$ is a rational, then $x=\frac {m}{n}$ with $(m,n)=1$ and i can find a term $x_{mn}=\frac {1}{mn}<\frac {m}{n}<=>\frac {1}{m}<m$.

So $x$ is not a rational.

So it's an irrational number.In it's decimal writing we have that $x=0,\underbrace {000...00}a....$ for example with $a\neq 0$ with $k$ underbraced zeros.Then the floor value $$\frac {[x\cdot 10^k]}{10^k}=\frac {a}{10^k}<x$$ and the term $$x_{a\cdot 10^l}<x$$. So $x$ cannot be irrational too.

So there isn't any real number that can do the job. So $0$ is the infimum and thus $\frac {1}{n}\to 0$.

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  • $\begingroup$ @StefanSmith See here if you like $\endgroup$ – Haha Dec 11 '13 at 9:45
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One can easily show that $\frac 1n$ goes to $0$ as $n$ goes large, but it is equally possible that the sum of these fractions diverge. The following is the proof that the sum must diverge.

Consider the sets $1$, and $\frac 12$ and $\frac 13 ,\frac 14$ and $\frac 15 \frac 16 \frac 17 \frac 18$ and so on, each set ending in a power of two. Since each set consists of fractions larger than the last, the sum of each set is then greater than a half (ie $1+ \frac 12+\frac 24 + \frac 48 \dots$.

The total must then be bigger than $1+\log_2(n)/2$. For example, the sets above add to something more than $1+\frac 32 = 1+\frac{\log_2(8)}2$.

One cal likewise place an upper limit, because each set is less than the lead fraction, so we have $1+1/2+2/2+8/4\dots$ which means that it is less than $2 \log_2(n)-1/2$.

One can of course refine the limits. For example, $\frac 13+\frac 14 = \frac 7{12}$. Since half of each subsequent set is greater than $\frac 13$ we can replace $\frac 6{12}$ with $\frac 7{12}$. amd imcrease the lower limit to $\frac {11}{12}+\frac{7}{12}\log_2(n)$. Likewise the upper limit is likewise reduced because we're including $3\cdot 2^n$ in the series. The intial factor is reduced because the first set $\frac 12$ is not being subdivided into less than 3 and greater than 3. So its multiplier would be still $\frac 6{12}$, but we transfer a unit from the $1$ to do this.

The upper limit can be set by noting that the function is convex from above. Suppose you have $a \le x \le b$. Then if we show that $\frac 1x$ is less than point on the line between $\frac 1a$ and $\frac 1b$, we can use multiply the upper limit per power by $\frac 34$, and progressively floser values as more iterations are done.

It might be noted that something like $\pi$ was calculated by iterating the inscribed polygon through $2^n 3$ for successively large $n$.

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