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How can we prove the following inequality: $$\frac{n}{n+1} < \frac{n+1}{n+2}$$

I understand how to do proof by inductions and contradictions, but I am getting stuck on this question.

My proof would start out with its base case. $n= 0$; $0/1 < 1/2$ holds true;

I.H.: this following equality holds true for a natural number $n$. We must now show that it also holds true for $n+1$:

$$\frac{n+1}{n+2} < \frac{n+2}{n+3}$$

From here I am stuck. We know the first term is greater than $n/n+1$ from our IH, but it gives no evidence that $\frac{n+2}{n+3}$ is greater than $\frac{n+1}{n+2}$ I see no correlation on how to prove this using the IH. Am I missing something?

please delete me
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user177061
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  • By the way, the induction hypothesis does not say "Assume that the [property] holds for all natural numbers $n$"; it says "Assume that it holds for some $n \in \mathbb{N}$". There is a subtle but important difference. – beep-boop Sep 18 '14 at 17:56
  • Very true. Thank you. – user177061 Sep 18 '14 at 17:57

4 Answers4

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Replace n+1 with m. Then it boils down to showing

$$\frac{m-1}{m} < \frac{m}{m+1} \to m^2 - 1 < m^2$$

QED

Gautam Shenoy
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As an alternative to induction, note that $$ \frac{n}{n+1}=1-\frac{1}{n+1}<1-\frac{1}{n+2}=\frac{n+1}{n+2}. $$

Kim Jong Un
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Note that for $n\ge 0$ $$\frac{n+1}{n+2}-\frac{n}{n+1}=\frac{(n+1)^2-n(n+2)}{(n+2)(n+1)}=\frac{1}{(n+2)(n+1)}\gt 0.$$

mathlove
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Another proof "without induction": $$ n(n + 2) = n^2 + 2n < n^2 + 2n + 1 = (n+1)^2.$$ Divide both sides by $(n + 1)(n + 2)$: $$ \frac{n}{n + 1} < \frac{n+1}{n+2}.$$

If you want to be a stickler for detail, you could use induction to prove the first inequality.

David K
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