Prove that: $((1+a)^{(1/n)}-1)n$ is decreasing

Question

I want to prove that: $((1+a)^{(1/n)}-1)n$ is decreasing with respect to $n\in\mathbb N$. Also $a>0$.

I have tried both derivating and taking the difference, but no results. Perhaps there is a trick I am missing.

The derivative is:

$$\frac{(n-\text{log}⁡(1+a))(1+a)^{(1/n)}⁡-n)}{n}$$

The difference is:

$$(1+a)^{\frac{1}{n}}n-(1+a)^{\frac{1}{n+1}}n-(1+a)^{\frac{1}{n+1}}+1$$

So I want to show that the dervative is negative, or that the dfference is positive. Or both...

Answer 1

We can write the derivative as \begin{equation} f'(n) = \overbrace{-\dfrac{\left(a+1\right)^\frac{1}{n}\ln\left(a+1\right)}{n}+\left(a+1\right)^\frac{1}{n}}^A-1 \end{equation} Second derivative is \begin{equation} f''(n) = \dfrac{\left(a+1\right)^\frac{1}{n}\ln^2\left(a+1\right)}{n^3}>0 \end{equation} Hence $f'(n)$ is increasing. Hence $f'(n)$ maximum value is at infinity, we can say \begin{equation} f'(n) < \lim_{n \rightarrow \infty} f'(n) = 0 \end{equation} Hence $f'(n) < 0$. So $f(n)$ is decreasing for any $a>0$.

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Why is $\lim_{n \rightarrow \infty} f'(n) = 0$

Let's study $A$ that appears as $f'(n) = A-1$. \begin{equation} \ln A =\frac{1}{n}\ln(a+1) + \ln(1- \frac{(a+1)^{\frac{1}{n}}}{n}) \end{equation} Both terms go to zero so $A$ has to go to $e^0 = 1$, hence $f'(n) \rightarrow 1-1 = 0$

Answer 2

After some algebraic manipulations, the derivative becomes $$(a+1)^{1/n}\left(1-\frac{1}{n}\log(a+1)\right)-1.$$ Let $$x := \frac{1}{n}\log(a+1).$$ If $x \ge 1$, then the first term is negative, so the derivative is negative. If $x < 1$, then taking the logarithm of the first term and doing some algebraic manipulations yields $$ \log\left((a+1)^{1/n}\left(1-\frac{1}{n}\log(a+1)\right)\right) = x + \log(1-x), $$ which is negative since the logarithm is convex and $-x$ is a tangent line to $\log(1-x)$.

Answer 3

We may set $1+a=e^k$ and $\frac{1}{n}=x$, then prove that $$ \frac{e^{kx}-1}{x} $$ is increasing for $x>0$. But, well, this is trivial since $$ \frac{e^{kx}-1}{x}=\sum_{n\geq 1}\frac{k^n}{n!}x^{n-1} $$ is a convergent series of increasing functions on $\mathbb{R}^+$ (notice that $a>0$ ensures $k>0$).