Prove that: $((1+a)^{(1/n)}-1)n$ is decreasing 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... 
 A: 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 $a>0$.

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 $f'(n) \rightarrow 1-1 = 0$
A: 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 $k>0$).
A: 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 $\log(1-x)$.
