Showing $(1-\frac{1}{n}) - (1-\frac{1}{n})^n$ is an increasing sequence?

Question

I've come across a familiar math expression in my research, and I want to formally prove the following.

$s_{n+1} - s_{n} > 0$ holds for integer $n \geq 2$, where $s_n := (1-\frac{1}{n}) - (1-\frac{1}{n})^n$.

I've plotted this using software, and it seems it's true. But I've been struggling to prove this. Does anyone have clues?

Thanks.

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Edited: I'm adding information as to roughly map out how I've tried. Please see below.

First, based on basic calculus.

I tried to think about functions rather than sequences, and show that both functions increase, but the former at a higher rate. To elaborate more,

$f(x) := 1 - \frac{1}{x}$ and $g(x) := (1 - \frac{1}{x})^x$

Showing $f(x)$ is increasing is easy. Took the derivative of it, and it is always positive for all $x$. The value of $f(x)$ is $\frac{1}{2}$ and its slope $f'(x)$ is $\frac{1}{4}$ at $x=2$.

What remains is to show (1) the value of $g(x)$ is less than or equal to $f(0.5) = 0.5$; (2) the slope of $g(x)$ is less than or equal to $f'(0.5) = 0.25$, and (3) the rate at which the slope increases is less than $f''(x)$. $g(0.5) = 0.25 < f(0.5) = 0.5$ (1) is done. Now (2) and (3) to go.

Showing $g(x)$ is increasing is hard. Took the logarithm of it, and tried to take the derivative of it, but to no avail.

This is pretty much it so far.

Second, based on induction.

Now, I'm trying to prove the claim, using the standard approach, but have no real progress yet.

Answer 1

If you change variable to $u=\frac1n$, then letting $n\to\infty$ is the same as letting $u\to 0^+$, and $$ (1-\tfrac1n)-(1-\tfrac1n)^n = 1-u - e^{\frac{\log(1-u)}{u}} $$ The exponent $\frac{\log(1-u)}{u}$ has a removable singularity at $u=0$; if we plug that (by dividing the series expansion of $\log(1-u)$ by $u$ term for term) we find that the right-hand-side above is nice and differentiable at $u=0$, and has derivative $\frac1{2e}-1$, which is negative.

Thus, as $u$ approaches $0$ from above, the RHS will, at least eventually, approach $1-\frac1e$ from below.

(Actually, graphing the function shows that it keeps decreasing for $u=[0,\frac12]$, so the original sequence is increasing right from the beginning at $n=2$).

Answer 2

I thought it might be instructive to prove this inequality using only Bernoulli's Inequality and some straightforward arithmetic. To that end, we proceed.

Let $s_n=\left(1-\frac1n\right)-\left(1-\frac1n\right)^n$. Then, the forward first difference of $s_n$ is given by

$$s_{n+1}-s_n=\frac{1}{n(n+1)}+\left(\left(1-\frac1n\right)^n-\left(1-\frac1{n+1}\right)^{n+1}\right) \tag 1$$

Now, we can write the second term on the right-hand side of $(1)$ as

$$\begin{align} \left(1-\frac1n\right)^n-\left(1-\frac1{n+1}\right)^{n+1}&=\left(1-\frac1{n+1}\right)^{n+1}\left(\frac{\left(1-\frac1n\right)^n}{\left(1-\frac1{n+1}\right)^{n+1}}-1\right) \tag 2\\\\ &=\left(1-\frac1{n+1}\right)^{n+1}\left(\frac{1}{\left(1-\frac1{n+1}\right)}\left(1-\frac{1}{n^2}\right)^n-1\right) \tag 3\\\\ &\ge \left(1-\frac1{n+1}\right)^{n+1}\left(\frac{1}{\left(1-\frac1{n+1}\right)}\left(1-\frac{1}{n}\right)-1\right) \tag 4\\\\ &=-\frac1{n^2}\left(1-\frac1{n+1}\right)^{n+1}\tag 5\\\\ &=-\frac{1}{n(n+1)}\left(1-\frac1{n+1}\right)^{n}\tag 6\\\\ &\ge -\frac{1}{n(n+1)}\tag 7 \end{align}$$

Therefore, we have the expected inequality

$$s_{n+1}-s_n\ge 0$$

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NOTES:

In arriving at $(2)$, we factor out the term $\left(1-\frac{1}{n+1}\right)^{n+1}$.

In going from $(2)$ to $(3)$, we noted that $\frac{\left(1-\frac1n\right)^n}{\left(1-\frac1{n+1}\right)^{n+1}}=\frac{1}{\left(1-\frac1{n+1}\right)}\left(1-\frac1{n^2}\right)^n$

In arriving at $(4)$ we used Bernoulli's Inequality.

In going from $(4)$ to $(5)$ we simplified the expression in large parentheses.

In going from $(5)$ to $(6)$ we used the equality $1-\frac{1}{n+1}=\frac{n}{n+1}$

In arriving at $(7)$, we noted that $\left(1-\frac{1}{n+1}\right)^{n}\le 1$