$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ 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?
 A: 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 $\sum_{n\ge 0}\frac{1}{x^n}$.
A: 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.
A: 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 ?
A: $$e^{\frac{1}{x}}<\frac{x}{x-1}$$
$$e^{\frac{1}{x}}<\frac{1}{1-\frac{1}{x}}$$
$$e^{-\frac{1}{x}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{e^{\frac{1}{x}}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{1+\frac{1}{x}+\frac{1}{2x^2}+...}<\frac{1}{1+\frac{1}{x}}$$
A: HINT::
$$\begin{align}
 e^{\frac{1}{x}} &< 1 + \frac{1}{x-1} \implies e < \left(\frac{x}{x-1}\right)^x\\ 
\end{align}$$
And 

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

A: 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 $x>1$.
A: 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.
