How to prove $\frac{1}{n} > \frac{1}{n+1}$?

Question

I understand this fact is very simple, but how would I go about it from a fundamental perspective only know that $n$ is less than $n+1$?

Answer 1

You could also subtract.

$$\frac1n - \frac{1}{n+1} = \frac{1}{n(n+1)} > 0.$$

Since the difference is positive, $\frac1n$ must be the larger one.

Answer 2

You can use that $a>b>0$ implies $\frac{1}{b}>\frac{1}{a}>0$. You have $n+1>n$ so $\frac{1}{n}>\frac{1}{n+1}.$

Answer 3

Comparing $\frac 1n$ and $\frac 1{n+1}$ we get,Numerator of both are same but denominator of the second is greater than the first (assuming $n$ to be positive).So,$1$ divided by a greater number is less than $1$ divided by a smaller number(from definition of fractions).Hence we get the result.

Answer 4

Since $n < n+1$, we can divide both sides by $n$ to get $$ 1 < \frac{n+1}{n}. $$ Then divide both sides by $n + 1$ to get $$ \frac{1}{n+1} < \frac{1}{n}. $$ The above argument is only for $n$ and $n+1$ are both positive. However, it is important to note that this only holds when $n$ and $n+1$ are both the same signs.

Answer 5

The long form of this argument.

The general approach to defining inequalities is to assume[*] there is a set of positive numbers with the properties that they are closed under addition and multiplication, and such that for every $x$, exactly one of the following is true:

1. $x$ is positive
2. $-x$ is positive
3. $x=0$

Definition: We say $a>b$ if $a+(-b)$ is positive.

Lemma 1: 1 is positive.

Proof: We know $1\neq 0$. If $(-1)$ is positive, then, by closure, $(-1)\cdot(-1)=1$ is positive. This violates our last condition- only one of $1$ and $-1$ is positive.

Lemma 2: If $a$ is positive and $ab$ is positive, then $b$ is positive.

Proof: We can quickly see that $b\neq 0$, because otherwise $ab=0$ and we know that $0$ is not positive.

If $b$ is not positive, then $-b$ is positive, so $a(-b)=-(ab)$ is positive. This contradicts that $ab$ is positive.

So $b$ must be positive.

Theorem: If $n$ is positive, then $\frac{1}{n}>\frac{1}{n+1}$.

Proof: Since $n$ is positive, and $1$ is positive, $n+1$ is positive, and in turn $n(n+1)$ is positive.

Since $n(n+1)\cdot \frac{1}{n(n+1)}=1$ is positive, and $n(n+1)$ is positive, we have that $\frac{1}{n(n+1)}$ is positive.

Finally, we have that $$\frac{1}{n} + \left(-\frac{1}{n+1}\right)=\frac{1}{n(n+1)}$$

So, by definition, $\frac{1}{n}>\frac{1}{n+1}$.

[*] Technically, you could prove such a set of positive numbers exists, but that heavily depends on what set of numbers you are using (reals, rationals?) and how you defined them.

Answer 6

Yet another argument. You know, I hope, that multiplying an inequality by a positive number preserves the direction of the inequality. Your question makes sense only when both integers $n$ and $n+1$ are nonzero, so that $n(n+1)$ is always positive. Then start with the inequality of unknown direction $1/n\,?\,1/(n+1)$, and multiply by the positive number $n(n+1)$, to get the no longer unknown inequality $n+1 \,?\,n$. That is, the question mark represents a “$>$” in both its occurrences.