# Prove that $\frac{e^{-a}}{1-e^{-a}}<\frac{1}{a}$ for positive $a$

The question is to prove the following inequality: $$\frac{e^{-a}}{1-e^{-a}}<\frac{1}{a}$$ for positive $$a$$.

I have tried to make the RHS into the following form: $$\frac{1}{1-(1-a)} = 1 + (1-a) + (1-a)^2 +...$$, and the LHS similarily. But this doesn't help to solve the problem. Could anyone provide some hints?

• Do you know that $e^a=1+a+\dfrac{a^2}2+...$? Commented Aug 9, 2019 at 1:32
• You should put the entire problem in the body, not just the title; addendum: thanks for doing so Commented Aug 9, 2019 at 1:44

It's just $$e^a>1+a,$$ which is smooth:

Let $$f(a)=e^a-1-a.$$

Thus, $$f'(a)=e^a-1>0,$$ which gives $$f(a)>f(0)=0.$$

You want to show $$\dfrac{e^{-a}}{1-e^{-a}}<\dfrac1a$$ for every $$a>0$$.

This is equivalent to showing:

$$\dfrac{1-e^{-a}}{e^{-a}}> a$$ for every $$a>0$$.

This simplifies to $$e^a-1>a\Leftrightarrow e^a> a+1$$

And $$e^a=\sum_{k=0}^\infty \frac{a^k}{k!}>1+a+\frac{a^2}{2}>a+1$$

Sometimes, it can be useful to try reverse engineering a solution to an inequality. This means manipulating the inequality into something you know is true, and reversing the steps, if possible.

Suppose $$x > 0$$. Then from \begin{align}\frac{e^{-x}}{1-e^{-x}} < \frac 1x, \end{align} by multiplying both sides by $$x$$, and then by $$1-e^{-x}$$, we get \begin{align} xe^{-x} < 1 - e^{-x}. \end{align} Notice that we haven't flipped the direction of the inequality, because $$x$$ is positive, and so is $$1 - e^{-x}$$ (convince yourself of this!). For $$x > 0$$, we also have $$e^x > 0$$, and so by multiplying both sides by $$e^x$$, \begin{align} x < e^x - 1, \end{align} which we can arrange to get \begin{align} x + 1 < e^x, \ x > 0. \end{align} How can we convince ourselves that the above is true? We start with claim that the line defined by the equation $$y = x + 1$$ is tangent to the graph of $$y = e^x$$, at $$x = 0$$.

The slope of the graph of $$y = e^x$$ at $$x = 0$$ is $$e^0 = 1$$, which follows from the fact that for each real number $$a$$, the slope of the graph of $$y = e^x$$ at $$x = a$$ is $$e^a$$. Furthermore, at $$x = 0$$, the $$y$$-value of the graph of $$y = e^x$$ is $$e^0 = 1$$.

At $$x = 0$$, we also have $$x + 1 = 0 + 1 = 1$$, and the slope of the line $$y = x + 1$$ is $$1$$.

Thus, the slopes and values of $$y = x + 1$$ and $$y = e^x$$ coincide at $$x = 0$$, and so $$y = x + 1$$ is tangent to $$y = e^x$$ at $$x = 0$$.

Now, the slope of the graph of $$y = e^x$$ is always increasing as $$x$$ increases (never stays the same, or decreases with increasing $$x$$). However, the slope of $$y = x + 1$$ remains the same. This means that $$y = e^x$$ will always lie above its tangent line $$y = x + 1$$ (at $$x = 0$$), for $$x > 0$$.

In other words, \begin{align} x + 1 < e^x, \ \text{when} \ x > 0. \end{align} The goal now is to start with this inequality which we now know is true, and reverse the steps to obtain the truth of the desired inequality.

There are no steps in the rearrangement of the inequality given in the question, to the "tangent line" inequality, which are irreversible. I leave you to do this!

Edit: To visualise the situation, you could try graphing $$y = e^x$$, and $$y = x+1$$, for example on the desmos graphing calculator. You can also plot $$y = \frac{e^{-x}}{1-e^{-x}}$$ and $$y = \frac 1x$$ and see if $$y = \frac 1x$$ lies below $$y = \frac{e^{-x}}{1-e^{-x}}$$ for $$x > 0$$.

For $$a>0$$, $$e^a=1+a+\dfrac {a^2}2+...>1+a>1,$$

so $$e^a-1>a$$

and

$$\dfrac1a>\dfrac1{e^a-1}=\dfrac{e^{-a}}{1-e^{-a}}.$$

Here is a way using the fact that

• $$(\star):\log y = \int_1^y\frac{1}{x}\,dx$$

You may substitute $$\boxed{-a = \log t}$$ with $$\boxed{0

Then, your inequality is equivalent to $$\begin{eqnarray*} \frac{e^{-a}}{1-e^{-a}}<\frac{1}{a} & \stackrel{-a = \log t}{\Longleftrightarrow} & \frac{t}{1-t} < -\frac{1}{\log t} \\ & \stackrel{-\log t = \log \frac{1}{t}>0}{\Longleftrightarrow} & \boxed{\log \frac{1}{t} <\frac{1}{t}-1} \end{eqnarray*}$$

The last inequality follows now directly by $$(\star)$$:

$$\color{blue}{\log \frac{1}{t}} \stackrel{(\star)}{=} \int_1^{\frac{1}{t}}\frac{1}{x}\,dx \stackrel{1\leq x \leq \frac{1}{t}}{\color{blue}{<}}\int_1^{\frac{1}{t}}1\,dx = \color{blue}{\frac{1}{t}-1}$$

Since $$x \mapsto e^{-x}$$ is strictly decreasing, for positive $$a > 0$$ we have

$$e^{-a} = \int_{0}^{1} e^{-a} \, \mathrm{d}x < \int_{0}^{1} e^{-ax} \, \mathrm{d}x = \frac{1 - e^{-a}}{a}.$$

Rearranging this inequality yields the desired conclusion.

Let $$S=1+x+x^2+...+x^n$$ Then $$xS=x+x^2+x^3+...+x^{n+1}=S-1+x^{n+1}$$ $$xS=S-1+x^{n+1}$$ $$S=\frac{1-x^{n+1}}{1-x}$$ If $$|x|<1$$ and $$n\rightarrow\infty$$ $$S=\frac{1}{1-x}$$ Substitute $$x=1-a$$ and you get what you want

• I think OP wants proof that $$\frac{e^{-a}}{1-e^{-a}}<\frac{1}{a}$$ Commented Aug 9, 2019 at 1:51

Multiply by $$a(1-e^{-a})$$ on both sides to get $$ae^{-a} < 1-e^{-a}$$ Add $$e^{-a}$$ on both sides to get $$(a+1)e^{-a}<1$$ From here, find the maximum of $$(a+1)e^{-a}$$ by differentiating. You'll find that the maximum is $$1$$, occuring only at $$x = 0$$.