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Prove by induction that $$(1+\frac{1}{n})^n \lt n$$ for all $n \ge 3$. So I have the smallest case done where $n=3$: $$(1+\frac{1}{3})^3 < 3$$ $$2.37 <3 $$ Now I have the $n+1$ case set up like this: $$(1+\frac{1}{m+1})^{m+1}<m+1$$ but I'm unsure where to go from here. Should I split up the left hand side into $$(1+\frac{1}{m+1})^m(1+\frac{1}{m+1})$$

Bryn Burns
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    Use $\displaystyle \frac{1}{m+1}<\frac{1}{m}$ and so, use the induction hypothesis. – Lucas Oct 23 '18 at 00:42
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    Keep in mind you're trying to show that $(1+\frac{1}{m+1})^{m+1}<m+1$ is true. So your "n+1 case" shouldn't be saying anything. You are just starting with the left hand side, manipulating it, and hoping to get $<m+1$. – J. Moeller Oct 23 '18 at 00:43
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    @LucasCorrêa where does $\frac{1}{m+1} < \frac{1}{m}$ come from? I thought the induction hypothesis was $1+\frac{1}{m} < m$ – Bryn Burns Oct 23 '18 at 00:49
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    @BrynBurns, $m < m+1 \Longrightarrow \frac{1}{m+1} < \frac{1}{m}$ – Lucas Oct 23 '18 at 00:51
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    I am going to show you kind of what I am thinking. You are doing the problem correct. You want to let $n\geq 3$ for your $n+1$ case too. So then $(1+\frac{1}{m+1})^m<m$ by your inductive hypotheses. So, then you have some positive number $a$ where $a<m$ and you want to show $a(1+\frac{1}{m+1})<m+1$. In other words, show $a+\frac{a}{m+1}<m+1$. Why would $\frac{a}{m+1}<1$ given that $a<m$ and $a$ and $m$ are both positive numbers? – W. G. Oct 23 '18 at 00:59
  • Similar with https://math.stackexchange.com/questions/2953479 and https://math.stackexchange.com/questions/582459 – Nosrati Oct 18 '22 at 09:42

2 Answers2

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$$(1+\frac{1}{m+1})^m(1+\frac{1}{m+1}) \le \color{blue}{(1+\frac{1}{m})^m}(1+\frac{1}{m+1}) \le m(1+\frac{1}{m+1})$$

Try to expand and conclude.

Siong Thye Goh
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Try this: Show by induction that $$\left(1 + {1\over n}\right)^n < 3 - {1\over n}$$ for $n \ge 1.$

ncmathsadist
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