2

I am trying to prove that for $t>0$$$\frac{t}{t+1} \le 1-e^{-t}\le \frac{2t}{1+t}$$

I know that Simplest or nicest proof that $1+x \le e^x$

taking $$e^{-x} \le \frac{1}{1+x} \implies 1-e^{-x}\ge \frac{x}{x+1}$$

Now I don't know to prove the other inequality.

Guy Fsone
  • 23,903

5 Answers5

3

$f(t) = (1+t)e^{-t} + t$ is strictly increasing since $f'(t) =1-te^{-t} =e^{-t}(e^{t}-t)>0$ for all $t > 0$. This holds as $t < 1 + t \le e^t$ .

Therefore, $$1=f(0)\le f(t) =(1+t)e^{-t} + t $$ for all $t>0$ which, together with $(1+t)e^{-t}\le 1$ since $1+t \le e^t$, implies that $$0\le 1-(1+t)e^{-t} \le t$$

so adding $t$ on both sides, we get $$t\le (1+t)(1-e^{-t})\le 2t.$$

As $1+t-e^{-t}-te^{-t}=(1+t)(1-e^{-t})$, the result follows on dividing the last inequality by $t+1.$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005
2

If we set $$ g(t) = (t+1)\frac{1-e^{-t}}{t} $$ we have that $\lim_{t\to 0^+}g(t)=\lim_{t\to +\infty}g(t)=1$ and $g(t)\geq 1$ for any $t>0$. $g(t)$ attains its maximum value close to $t=2$, and such maximum value is way less than $2$, it is around $1.3$.

Indeed, the maximum value of $g(t)$ is attained at the only positive solution of $e^{-t}=\frac{1}{1+t+t^2}$.
If we introduce $h(t)=(t+1)\frac{1-\frac{1}{1+t+t^2}}{t}=1+\frac{t}{1+t+t^2}$ we may easily check that $h(t)\leq \frac{4}{3}$.

It follows that $$\boxed{ \forall t>0,\qquad \frac{t}{t+1}\leq 1-e^{-t} \leq\color{red}{\frac{4}{3}}\cdot\frac{t}{t+1}.} $$

Jack D'Aurizio
  • 353,855
2

You already proved the first part, $$ 1-e^{-t}\ge\frac{t}{t+1}. \tag{1}$$ For the second part, noting that, for $t\ge0$, $$ e^t=1+t+\frac12t^2+\cdots\ge\frac12+t+\frac12t^2=\frac12(1+t)^2$$ one has $$ e^{-t}\le \frac{2}{(1+t)^2}. $$ Integrating from $0$ to $t$, one has $$ \int_0^te^{-s}ds\le \int_0^t\frac{2}{(1+s)^2}dt $$ or $$ 1-e^{-t}\le\frac{2t}{t+1}. \tag{2}$$ From (1) and (2), one has $$ \frac{t}{t+1}\le1-e^{-t}\le\frac{2t}{t+1}. $$

xpaul
  • 44,000
2

The second inequality is trivial if $t \geqslant 1$, so we may assume that $0 < t < 1$. In fact, we only assume $0 < t < 2$. Put $u = t/2$, so that $0 < u < 1$. The curve $y = \tfrac{1}{x}$ lies above the tangent at $(1, 1)$ and below the secant from $(1-u, \tfrac{1}{1-u})$ to $(1+u, \tfrac{1}{1+u})$, therefore, calculating the areas of two trapezia, $$ 2u < \int_{1-u}^{1+u}\frac{dx}{x} < \frac{2u}{1-u^2}. $$ Using the first of these inequalities (the second was only mentioned for interest), \begin{gather*} t < \log\left(1+\frac{t}{2}\right)-\log\left(1-\frac{t}{2}\right) = \log\frac{2+t}{2-t}, \\\text{i.e.}\quad e^t < \frac{2+t}{2-t}, \quad\text{i.e.}\quad 1 - e^{-t} < \frac{2t}{2 + t}, \end{gather*} which is stronger than the required inequality (and is trivially true when $t \geqslant 2$, so holds for all $t > 0$).

Calum Gilhooley
  • 12,174
  • In the non-trivial case $t < 1$, one could also argue that $e^t = 1 + t + \tfrac{t^2}{2} + \tfrac{t^3}{6} + \tfrac{t^4}{24} + \cdots < 1 + t + \tfrac{t}{2} + \tfrac{t}{4} + \tfrac{t}{8} + \cdots < 1 + 2t$, whence $1 - e^{-t} < \tfrac{2t}{1+2t} < \tfrac{2t}{1+t}$. – Calum Gilhooley Nov 12 '17 at 07:52
  • One could also argue that $\tfrac{1}{2}\tfrac{\log(1+u)}{\log(1-u)} = u + \tfrac{u^3}{3} + \tfrac{u^5}{5} + \cdots < u + u^3 + u^5 + \cdots = \tfrac{u}{1-u^2}$ (and of course the sum is also greater than $u$, which is all we need here). – Calum Gilhooley Nov 12 '17 at 08:27
  • Correction: the third "$<$" in the first comment should be a "$=$". – Calum Gilhooley Nov 12 '17 at 08:44
  • Flogging this dead horse one more time, the series expansion of $\tfrac{2+t}{2-t} = 1 + t\left(1-\tfrac{t}{2}\right)^{-1}$ is $1 + t + \tfrac{t^2}{2} + \tfrac{t^3}{4} + \cdots$. – Calum Gilhooley Nov 14 '17 at 23:50
  • Oops! Obviously, that second comment should begin: "$\tfrac12\log\frac{1 + u}{1 - u} \ldots$". I must have been hitting the sauce that day, or something. $\ \ddot\smile$ – Calum Gilhooley Jul 12 '20 at 14:44
2

$\frac{t}{t+1} \le 1-e^{-t}\le \frac{2t}{1+t} $ is the same as $t \le (1+t)(1-e^{-t}) \le 2t $.

Since $(1+t)(1-e^{-t}) =1+t-(1+t)e^{-t} $, this is the same as $0 \le 1-(1+t)e^{-t} \le t $.

The left side is $(1+t)e^{-t} \le 1 $ or $1+t \le e^t$ which is well known.

The right side is $1-(1+t)e^{-t} \le t $ or $1-t \le (1+t)e^{-t} $.

If $t \ge 1$, this is obviously true.

If $0 \le t \lt 1$, this is

$e^t \le \dfrac{1+t}{1-t} $ and even more than this is true because $e^t \le \dfrac1{1-t} $ for $0 \le t \lt 1$ by comparing terms in the power series ($\dfrac1{k!} \le 1$).

marty cohen
  • 107,799