$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $

Question

I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.

The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives $$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$

If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.

The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing; However this also does not seem to help too much.

Any ideas?

Answer 1

For $x>1$, $\frac{1}{x-1}=\frac{1}{x}\frac{1}{1-1/x}$ is a convergent geometric series. Let's return to your title inequality. The left-hand side is $\sum_{n\ge 0}\frac{1}{n!x^n}$; the right-hand side is $\sum_{n\ge 0}\frac{1}{x^n}$.

Answer 2

Prove instead that, for $0<x<1$, $$ e^x<1+\frac{1}{\frac{1}{x}-1}=1+\frac{x}{1-x}=\frac{1}{1-x} $$ which is the same as proving that, for $0<x<1$, $$ e^{1-x}<\frac{1}{x} $$ that is, $e^x>ex$. Now differentiating is easier, isn't it? If $f(x)=e^x-ex$, $$ f'(x)=e^x-e $$ Thus $f'$ is negative for $0<x<1$. Since $f(1)=0$, the inequality is proved.

Answer 3

Let $t=1/x$, then $e^{1/x} < 1 + \frac{1}{x-1}$ for $x>1$ is equivalent to

$e^t -1< +\frac{t}{1-t}$ for $0<t<1$.

We have $e^t-1=t+ \frac{t^2}{2!}+\frac{t^3}{3!}+.....$ and $\frac{t}{1-t}=t+t^2+t^3+...$for $|t|<1.$

Can you proceed ?

Answer 4

$$e^{\frac{1}{x}}<\frac{x}{x-1}$$

$$e^{\frac{1}{x}}<\frac{1}{1-\frac{1}{x}}$$ $$e^{-\frac{1}{x}}<\frac{1}{1+\frac{1}{x}}$$ $$\frac{1}{e^{\frac{1}{x}}}<\frac{1}{1+\frac{1}{x}}$$ $$\frac{1}{1+\frac{1}{x}+\frac{1}{2x^2}+...}<\frac{1}{1+\frac{1}{x}}$$

Answer 5

HINT:: $$\begin{align} e^{\frac{1}{x}} &< 1 + \frac{1}{x-1} \implies e < \left(\frac{x}{x-1}\right)^x\\ \end{align}$$ And

$$\lim_{x\to\pm\infty}\left(\frac{x}{x-1}\right)^x = \lim _{x\to \pm\infty }\left(e^{x\ln \left(\frac{x}{x-1}\right)}\right) = e$$

Answer 6

Prove: $e^{\frac{1}{x}}<1+\frac{1}{x-1}$ , $\:$ $\forall >1$ : $x\in\mathbb{R}$

Let $\delta = 1+\epsilon $ , $\:$where $0<\epsilon$
Suppose $x=\delta$:
$\Rightarrow$ the proof becomes:

Prove $e^{\frac{1}{\delta}}<1+\frac{1}{\delta-1}$

This is equivalent to:
$\frac{1}{\delta}<\ln(1+\frac{1}{\delta-1})$
$\Rightarrow \frac{1}{\delta}<\ln(\frac{\delta}{\delta-1})$
$\Rightarrow \frac{1}{\delta}<\ln(\delta)-\ln(\delta-1)$
$\Rightarrow 1<\delta\ln(\delta)-\delta\ln(\delta-1)$

Suppose $y=x\ln(x)-x\ln(x-1)$
If we can prove $y<1$, $\forall x$, then everything else yields.

The next part consists of me being lazy, and I will leave all the formal work to the reader.

It is known that $y$ has no turning points; hence no minima values - (via setting the first dirivative to 0).

It is also known of the following:

$\lim_{x\rightarrow\infty}(y)=1$ ^ $\lim_{x\rightarrow 0^{+}}(y)\rightarrow\infty$

$\Rightarrow y>1$ for all $x\in\mathbb{R}$
Hence: $1<\delta\ln(\delta)-\delta\ln(\delta-1)$ is proven.
$\therefore$ $\:$ $e^{\frac{1}{x}}<1+\frac{1}{x-1}$ for all $x>1$.

Answer 7

For this question I proved that for all real $t, e^t \ge 1 + t $ and that equality holds only for $t=0.$

From this it follows that, for $t>0,$ $$\exp\left(\frac {-1}{1+t}\right)\gt 1-\frac 1 {1+t}=\frac t {1+t},$$

which implies (taking reciprocal of both sides, reversing the order) $$\exp\left(\frac {1}{1+t}\right)\lt \frac {1+t} {t}$$ so (setting $x=1+t$) $$\exp\left(\frac {1}{x}\right)\lt \frac {x} {x-1} = 1 + \frac 1 {x-1}$$ for $x>1$ as desired.