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I want to prove that $\left(1+\frac{1}{n}\right)^x>1+\frac{x}{n+1}$ $\forall x \ge1$ and $n \in \mathbb{N}$.

I first expanded $\left(1+\frac{1}{n}\right)^x=1+\frac{x}{n}+\frac{x(x-1)}{2n^2}+...$. So I wanted to truncate the expansion but I think the terms start becoming negative later on in the series. So I do not know how to do it. If anyone can give any hints it would be great. Thanks.

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    $\begingroup$ Don't you have that that $1+\frac{x}{n}> 1+\frac{x}{n+1}$, and you've just shown that $\left(1+\frac{1}{n}\right)^x \geq 1+\frac{x}{n}$... $\endgroup$
    – Thomas Andrews
    Jan 8, 2015 at 5:30
  • $\begingroup$ @ThomasAndrews I am getting negative terms later that is why I am not sure $\endgroup$
    – happymath
    Jan 8, 2015 at 5:32
  • $\begingroup$ (Oh, wait,that argument only works for integer $x$, since terms can theoretically be negative for $x$ real.) $\endgroup$
    – Thomas Andrews
    Jan 8, 2015 at 5:32
  • $\begingroup$ @ThomasAndrews exactly that is where I am having a problem $\endgroup$
    – happymath
    Jan 8, 2015 at 5:33
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    $\begingroup$ Check the generalised Bernoulli Inequality: en.wikipedia.org/wiki/Bernoulli's_inequality#Generalization $\endgroup$
    – Macavity
    Jan 8, 2015 at 5:34

4 Answers 4

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This is direct using Generalised Bernoulli Inequality.

If you are not familiar with that, here is a quick proof using Jensen's inequality and strict concavity of $\log$ function: $$\frac1x \log\left(1+\frac{x}n\right)+\left(1-\frac1x\right)\log 1 < \log \left(1+\frac1n\right)$$ $$\iff \left(1+\frac1{n}\right)^x> 1+\frac{x}n >1+\frac{x}{n+1}$$

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Hint: Prove instead the more attractive $(1+t)^x \ge 1+tx$ for $x\ge 1$ and $t\ge 0$. This implies the desired result.

Let $f(t)=(1+t)^x - (1+tx)$. Then $f(0)=0$. Show that $f'(t)\ge 0$ for $t\ge 0$.

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  • $\begingroup$ Nicolas thanks for this great answer and sad that I cannot accept more than one $\endgroup$
    – happymath
    Jan 8, 2015 at 5:57
  • $\begingroup$ You are welcome. The technique I used comes up a lot. $\endgroup$
    – André Nicolas
    Jan 8, 2015 at 6:01
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Let $f(x)=(1+\frac{1}{n})^{x}$, let $g(x)=1+\frac{x}{n}$. It is clear that when $x=1$ we have $$ f(1)=g(1)=1+\frac{1}{n} $$ And for all $x\ge 1$ we have $$ f'(x)=f(x)\log(1+\frac{1}{n})=(1+\frac{1}{n})^{x}\log(1+\frac{1}{n}), g'(x)=\frac{1}{n} $$ I claim that $f'(x)\ge f'(1)\ge g'(x)$ for all $x\ge 1, n\ge 1$. In other words we want to show that $$ (1+y)\log(1+y)-y\ge 0,\forall 0\le y\le 1 $$ But this can be proved by differentiating again with respect to $y$, after which we have $$ 1+\log(1+y)-1=\log(1+y)\ge 0 $$ and it is clear that for $y=0$ we obtain an equality. So the original statement is proved, because $g(x)> \frac{x}{n+1}$ at all times.

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Here are some hints. Look at $1+\frac{x}{n}+\frac{x(x-1)}{2n^{2}}+\cdots$ and compare this infinite sum to $1+\frac{x}{n+1}$. Do we know which of these two sum is bigger? Also, use the generalized binomial theorem, which can be found about halfway down on this page: http://en.wikipedia.org/wiki/Binomial_theorem.

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  • $\begingroup$ $x$ is real though. See the comets under the question. $\endgroup$
    – Karolis Juodelė
    Jan 8, 2015 at 5:36
  • $\begingroup$ @Soothsaer If you take integers I can see that but what about non integers? $\endgroup$
    – happymath
    Jan 8, 2015 at 5:42
  • $\begingroup$ Good catch. I added in the comment about using the generalized binomial theorem. $\endgroup$
    – Soothsaer
    Jan 8, 2015 at 5:44

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