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Using induction, I proved the base case and then proceeded to prove: $$(1 + \frac{1}{n+1}) ^ {n+1} \ge 2$$ given $$(1 + \frac{1}{n}) ^ n \ge 2$$ However, I'm stuck at this point and have no clue how to go about it. Other than induction, I tried simple algebraic transformations but couldn't prove this inequality. Any pointers on how to prove this will be appreciated.

[PS: This is my first question on stackexchange, so I'm sorry if there's anything wrong with this post and will be happy to edit if needed].

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    $\begingroup$ One way to happily "edit" the question would be not to roll back edits that did no harm. The "induction" tag certainly fits and parentheses that are the correct height are easier to read, though the syntax for them ("\left(" and "\right)") is obscure. $\endgroup$
    – David K
    Commented Mar 11, 2015 at 3:01
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    $\begingroup$ @DavidK - I didn't realize what I was doing, will keep that in mind. $\endgroup$
    – Nishant S
    Commented Mar 11, 2015 at 3:05
  • $\begingroup$ OK, these things happen. (Personally, I find the "rollback" controls not the most intuitive, and have gotten in a little trouble with them once or twice myself.) $\endgroup$
    – David K
    Commented Mar 11, 2015 at 3:10

3 Answers 3

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Hint Prove by induction the more general statement:

If $x >0$ then $$(1+x)^n \geq 1+nx $$

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  • $\begingroup$ Chose this as answer because of its general nature and wider applicability. $\endgroup$
    – Nishant S
    Commented Mar 11, 2015 at 3:41
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    $\begingroup$ @NishantS BTW, this is called the Bernoulli inequality and it actually holds for $x \geq -1$. You need $x \geq-1$ to make sure in the inductive step that multiplication by $(1+x)$ doesn't change the sign of the inequality. $\endgroup$
    – N. S.
    Commented Mar 11, 2015 at 3:50
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$$\left(1 + \frac{1}{n}\right)^{n} = 1 + {n \choose 1} \frac{1}{n} + \cdots \geq 1 + 1$$

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  • $\begingroup$ This approach to proving is based on induction? $\endgroup$
    – Nishant S
    Commented Mar 11, 2015 at 3:07
  • $\begingroup$ Nope. This is a direct proof using binomial expansion. $\endgroup$
    – TenaliRaman
    Commented Mar 11, 2015 at 3:23
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Assuming $\left ( 1+\dfrac{1}{n}\right)^n \geq 2$, we want to prove $\left ( 1+\dfrac{1}{n+1} \right)^{n+1}\geq 2$.

You may start by saying: $\left ( 1+\dfrac{1}{n+1}\right)^{n+1} = \left ( 1+\dfrac{1}{n+1} \right)^n\left ( 1+\dfrac{1}{n+1}\right).$

But if $\left( 1+\dfrac{1}{n}\right)^n\geq 2, $ then $\left( 1+\dfrac{1}{n+1}\right)^n\geq 2$.

$\left( 1+\dfrac{1}{n+1}\right)^n\left ( 1+\dfrac{1}{n+1}\right)\geq 2\left ( 1+\dfrac{1}{n+1}\right)$.

It is easy to see that $\left ( 1+\dfrac{1}{n+1}\right) \geq 1$, therefore conclusion follows.

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    $\begingroup$ The third line is not valid. We know that $\left(1+\frac{1}{n+1}\right)^n < \left(1+\frac{1}{n}\right)^n$ for $n > 0$; therefore the mere fact $2 \leq \left(1+\frac{1}{n}\right)^n$ tells us nothing about $\left(1+\frac{1}{n+1}\right)^n$. It would be a very different matter if the first $<$ were $\geq$ instead, but it isn't. $\endgroup$
    – David K
    Commented Mar 11, 2015 at 2:51

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