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More fully, if $n\ge 2$ is an integer and $0 < x < 1$, prove that $(1+\frac{x}{n})^n < \frac1{1-x}$.

In addition, if $c > 1$ and $0 < x \le \frac{c-1}{c}$, prove that $(1+\frac{x}{n})^n < 1+cx$.

Proofs by elementary means (no calculus or limits) are particularly sought.

As an example of the utility of this result, set $x = \frac12$. Then this shows that $2 > (1+\frac1{2n})^n$ or $2^{1/n} > 1+\frac1{2n}$ .

This is an example of what I call a contra-Bernoulli inequality (CBI) which gives an upper bound to $(1+y)^n$ as opposed to Bernoulli's inequality, which gives a lower bound to $(1+y)^n$ of $1+ny$.

Note that any CBI of the form $(1+y)^n < 1+c y$ for $n \ge 2$ requires that $y$ is bounded, since $(1+y)^n > 1+y^n$ so $1+cy > 1+y^n$ or $cy > y^n$ or $y < c^{1/(n-1)}$.

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\begin{align} \left ( 1+\frac{x}{n} \right )^n &=1+n\cdot \frac{x}{n}+\frac{n(n-1)}{2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}\\ &<1+x+x^2+\cdots\\ &=\frac{1}{1-x} \end{align} for $n\geq 2$, $0<x<1$.

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For the first part, we only need the binomial theorem:

$$\left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k} = \sum_{k=0}^n \frac{\prod_{j=1}^k(n+1-j)}{k!n^k}x^k \leqslant \sum_{k=0}^n \frac{x^k}{k!} \leqslant \sum_{k=0}^n x^k < \frac{1}{1-x}.$$

For the second, we observe that

$$\frac{1}{1-x} \leqslant 1 + cx$$

for $0 < x < \frac{c-1}{c}$, since

$$(1-x)(1+cx) -1 = (c-1)x - cx^2$$

has zeros in $x = 0$ and $x = \frac{c-1}{c}$, and is positive between the zeros, since the coefficient of $x^2$ is negative.

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In response to "more fully...", but not necessarily "in addition...": Note that $\lim_{n\rightarrow \infty} (1+(x/n))^n = e^x$ by definition and that, for any finite $n$, $(1+(x/n))^n \leq e^x$ for $0 \leq x \leq 1$ (can check this on graph, or use Taylor series). Now consider the end points of the interval: $e^{0+} < 1/(1-(0+))$ and $e^1 < 1/(1-1)$. Since both $e^x$ and $1/(1-x)$ are strictly monotonic in $]0,1]$ this means that $e^x < 1/(1-x)$ throughout this interval (no crossings). A fortiori, because of the previous part, $(1+(x/n))^n < 1/(1-x)$.

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  • $\begingroup$ That $f$ and $g$ are strictly increasing and such that $f(0)\lt g(0)$ and $f(1)\lt g(1)$ does not prevent the equation $f(x)=g(x)$ to have solutions $x$ in $(0,1)$. $\endgroup$
    – Did
    Jul 8, 2013 at 15:40
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Here's my elementary proof by induction that $(1+x/n)^n < 1/(1-x)$.

Let $y = x/n$. This becomes $(1+y)^n < 1/(1-ny)$ when $0 < y < 1/n$.

For $n=1$, this is $1+y < 1/(1-y)$ or $1-y^2 < 1$ which is true.

Suppose it it true for $n$, so that $(1+y)^n < \frac1{1-ny}$.

Then, if $0 < y < \frac1{n+1}$,

$\begin{align} (1-(n+1)y)(1+y)^{n+1} &=(1-(n+1)y)(1+y)(1+y)^{n}\\ &<\frac{(1-(n+1)y)(1+y)}{1-ny}\\ &=\frac{1-ny-(n+1)y^2}{1-ny}\\ &=\frac{1-ny}{1-ny}\\ &=1 \end{align} $

so $(1+y)^{n+1} < \frac1{1-(n+1)y}$.

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Because $$ (1 + \frac{x}{n})^n < e^x $$ We can rewrite the expression as $$ e^x < \frac{1}{1-x} $$ Replacing the terms with their infinite sumation $$ \sum_{k=0}^\infty \frac{x^k}{k!} < \sum_{k=0}^\infty x^k $$ After trivial manipulations we get $$ \sum_{k=2}^\infty \frac{x^k (1 - k!)}{k!} < 0 $$ The inequality is trivially proved

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