Is this sequence converging monotonically?

Question

Consider the sequence $a_n = \frac{ a^{1/n}-1}{1/n}$ for a>0 . Is it monotone?

I know that $a_n \to \log a$ as $ n \to \infty$, and I think that $\{a_n\}$ is monotonically decreasing, but I'm not able to prove it.

Answer 1

For $a > 0$ the function $$ f(x) = a^x = e^{x \log a} \quad (x \ge 0) $$ is convex ($f''(x) \ge 0$), therefore for $0 < x < y$ $$ f(x) \le \frac{y-x}{y-0} \, f(0) + \frac{x-0}{y-0} \, f(y) \\ \Longrightarrow \frac{f(x) - f(0)}{x} \le \frac{f(y) - f(0)}{y} \, , $$ i.e. the function $$ g(x) = \frac{f(x)-f(0)}{x-0} = \frac{a^x-1}{x} $$ is increasing, so that $a_n = g(\frac 1n)$ is a decreasing sequence.

If $a \ne 1$ then $f$ is strictly convex and therefore $a_n$ strictly decreasing.

Answer 2

Assuming $a<1$:$$\frac{\frac{ a^{1/n}-1}{1/n}}{\frac{ a^{1/(n+1)}-1}{1/(n+1)}}=\frac{a^{1/n}-1}{a^{1/(n+1)}-1}\frac{1/(n+1)}{1/n}=\frac{a^{1/n}-1}{a^{1/(n+1)}-1}\overbrace{\frac{n}{n+1}}^{<1}$$ Now let's check if $\frac{a^{1/n}-1}{a^{1/(n+1)}-1}<1:$ $$\frac{a^{1/n}-1}{a^{1/(n+1)}-1}<1\iff a^{1/n}-1<a^{1/(n+1)}-1\iff a^{1/n}<a^{1/(n+1)}\iff 1/n>1/(n+1)$$

So $a_n/a_{n+1}<1\implies |a_n|<|a_{n+1}|\implies\begin{cases}a_n<a_{n+1}&\forall n:a_n>0\\a_n>a_{n+1}&\forall n:a_n<0\end{cases}$

Because we are talking about $a<1$ I can see that $a_n<0$

Hence for $a<1$ it is monotonically decreasing.