$$(1 + \frac{1}{n})^n \geq 2$$
Base base: $n = 1$
LHS: $(1+ 1)^1 \leq 2$ RHS: 2
$2 \leq 2$ True.
Inductive step:
$$(1 + \frac{1}{n+1})^{n+1} \geq 2$$
$$(1 + \frac{1}{n+1})^n (1 + \frac{1}{n+1}) \geq 2$$
Im stuck
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1just use binomial expansion to prove the inequality – Vasili Sep 11 '19 at 17:33
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we barely went over that. How would that look? – K. Gibson Sep 11 '19 at 17:34
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1this might be useful $\frac{1}{n} > \frac{1}{n+1}$, – Donlans Donlans Sep 11 '19 at 17:41
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See also https://math.stackexchange.com/questions/167843/show-that-left1-dfrac1n-rightn-is-monotonically-increasing – lhf Sep 11 '19 at 17:54
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1@K.Gibson $(1+\frac{1}{n+1})^{n+1}=1^{n+1}+(n+1)1^{n}\frac{1}{n+1}+ ... + (\frac{1}{n+1})^{n+1}=2+ \ some \ positive \ number$ – Vasili Sep 11 '19 at 19:58
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I was able to use the binomial expansion – K. Gibson Sep 11 '19 at 23:02
1 Answers
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Hint: Use Bernoulli's inequality: $$ (1+x)^n \ge 1 + nx \quad \mbox{for $x\ge 0$} $$ This can be proved by induction.
lhf
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