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Assume $c \geq 1$. I want to show that the function $$f(n) = \left(1-\frac{1}{n}\right)^{n-1} \cdot \left(1+\frac{1}{n-c}\right)^{n-c}$$ satisfies $f(n) \leq 1$ for all $n > c$.

My attempt: I think the fastest way to show this is by extending $f$ to $\mathbb R$ and show the result using derivatives. Notice that $\lim_{x \rightarrow \infty} f(x) = e^{-1} \cdot e =1$, so it suffices to show that $f(x)$ is increasing for $x > c$.

After computing the derivative, one notices that it suffices to show $$\log\left( \frac{x-1}{x} \cdot \frac{x-c+1}{x-c} \right) \geq \frac{c-1}{x(x-c+1)}.$$

I am stuck here.

thewatcher
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2 Answers2

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$$ \log f(x) = (x-1) \log \left( 1 - \frac 1x \right) + (x-c) \log \left( 1 + \frac {1}{x-c} \right) $$ has the derivative $$ \log \left( 1 - \frac 1x \right) + \frac 1x + \log\left( 1 + \frac {1}{x-c} \right) - \frac{1}{x-c+1} \\ = \log\left( \frac{x-1}{x} \cdot \frac{x-c+1}{x-c} \right) - \frac{c-1}{x(x-c+1)}. $$ Using the estimate $\log (u) \ge (u-1)/u$ (see for example here) this is $$ \ge \frac{c-1}{(x-1)(x-c+1)} - \frac{c-1}{x(x-c+1)} = \frac{c-1}{x(x-1)(x-c+1)} \ge 0 $$ for $x > c \ge 1$, so that $f$ is indeed increasing on $(c, \infty)$.

Martin R
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Alternative proof.

By Bernoulli inequality $(1 + u)^r \le 1 + ur$ for all $0 < r \le 1$ and $u \ge 0$, we have $$\left(1+\frac{1}{n-c}\right)^{\frac{n-c}{n-1}} \le 1 + \frac{1}{n-c}\cdot \frac{n-c}{n-1} = 1 + \frac{1}{n-1}.$$

Thus, we have $$\left(1-\frac{1}{n}\right)^{n-1} \cdot \left(1+\frac{1}{n-c}\right)^{n-c} \le \left(1-\frac{1}{n}\right)^{n-1} \cdot \left(1+\frac{1}{n-1}\right)^{n-1} = 1.$$

River Li
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    $\left(1+\frac{1}{n-c}\right)^{n-c} \le \left(1+\frac{1}{n-1}\right)^{n-1}$ also follows from the (well-known) fact that $x \mapsto \left(1+\frac{1}{x}\right)^x$ is increasing in $x$. – Martin R Oct 21 '24 at 07:21