What does $\frac{1}{n}$ converge to?

Question

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?

Answer 1

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.

Answer 2

It converges to $0$ since for any $\epsilon > 0$, we can find $N$ such that $1/N < \epsilon$. So for $n > N$ we have $|1/n - 0| < 1/N < \epsilon$.

Answer 3

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 $

Answer 4

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 $\frac {1}{n}\to 0$.

Answer 5

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 $n$.