Is there a better way to estimate $(1-1/(1+e^{\epsilon/k}))^k$ from above? I am trying to prove an inequality that would imply my algorithm satisfies $\epsilon$-differential privacy for $k_i$ being a parameter. The inequality is
$$
\left(1-\frac{1}{1+e^{\epsilon/k_i}}\right)^{k_i} < 1 - \frac{1-\frac{1}{2^{k_i}}}{e^\epsilon} ,
$$
for $\epsilon > 0$ and $k_i \in \{1,2,...\}$.
I've been trying for weeks, and I've got a rather large proof. First by showing that the two sides are never equal when $\epsilon > 0$, and using the Intermediate Value theorem to show that if at a given $\epsilon^\prime$ and $k_i^\prime$ the RHS is less than the LHS, then this also holds for all $\epsilon>0$ for the same $k_i^\prime$. After that I show the base  case at $k_i=1$ and proceed by induction on $k_i$ to show it holds for all $k_i$. 
The resulting proof are rather long (5 pages with lots of hyperbolic tangents and friends). I was just wondering if there is a faster and more elegant way to prove the same result.
 A: Call $x=\mathrm{e}^{\epsilon/k_i}$ hence $x>1$ is a real number, $n=k_i$ hence $n\ge1$ is an integer, and $z_n=1-2^{-n}$. You want to prove
$$
\left(1-\frac{1}{1+x}\right)^n<1-\frac{z_n}{x^n}.
$$
This is equivalent to
$$
x^n\left(1-\left(\frac{x}{1+x}\right)^n\right)>z_n.
$$
Call $u_n(x)$ the LHS. Then $u_n(1)=z_n$, hence if $u_n$ is increasing, you are done. But the derivative of $u_n$ is
$$
u'_n(x)=nx^{n-1}\left(1-\left(\frac{x}{1+x}\right)^n\right)-nx^n\left(\frac{x}{1+x}\right)^{n-1}\frac{1}{(1+x)^2}=nx^{n-1}v_n(x),
$$
with
$$
v_n(x)=1-\left(\frac{x}{1+x}\right)^n\frac{2+x}{1+x}.
$$
Since $n\ge1$, $v_n(x)\ge v_1(x)=1/(1+x)^2>0$ and the proof is complete.
This also shows that the constant $z_n=1-2^{-n}$ is optimal in the sense that one cannot replace it by any greater value and still hope the inequality to hold for every positive $\epsilon$ (that is, except if $\epsilon$ is restricted to $\epsilon\ge\epsilon_0$ for a given positive $\epsilon_0$).
A: First, for positive integer $k$, write your inequality as
$$
\bigg(\frac{{1+e^{\varepsilon /k} -1}}{{1 + e^{\varepsilon /k} }}\bigg)^k  < \frac{{e^\varepsilon   - 1 + 2^{ - k} }}{{e^\varepsilon  }},
$$
or
$$
e^{2\varepsilon }  < (1 + e^{\varepsilon /k} )^k (e^\varepsilon   - 1 + 2^{ - k} ).
$$
Next note that
$$
(1 + e^{\varepsilon /k} )^k  \ge 1 + (e^{\varepsilon /k} )^k  = 1 + e^\varepsilon  .
$$
Hence 
$$
(1 + e^{\varepsilon /k} )^k (e^\varepsilon   - 1 + 2^{ - k} ) \ge (1 + e^\varepsilon  )(e^\varepsilon   - 1) + \bigg(\frac{{1 + e^{\varepsilon /k} }}{2}\bigg)^k  > (e^{2\varepsilon }  - 1) + 1 = e^{2\varepsilon } ,
$$
and so we are done.
