clovnerie0q

2022-10-03

Having trouble understanding why the $r$-th mean tends to the geometric mean as $r$ tends to zero
I am having trouble understanding the proof of Theorem 3 in "Inequalities" by Hardy, Littlewood and Pólya. This theorem states that the $r$-th mean approaches the geometric mean as $r$ approaches zero.
I have seen the following post which makes things a little clearer (albeit using $o\left(r\right)$ instead of $O\left({r}^{2}\right)$:
Why is the 0th power mean defined to be the geometric mean?
However, I still cannot determine why:
(a): ${a}^{r}=1+r.log\left(a\right)+o\left(r\right)$ as $r$ tends to zero,
and
(b): $\underset{r\to 0}{lim}\left(1+rx+o\left(r\right){\right)}^{1/r}={e}^{x}$
I have a pretty solid grasp of limits, as well as the log and exp functions, but I have never really been taught anything substantial on big/little-O notation, in particular as the variable approaches zero. Could somebody point me towards a suitable proof of (a) and (b) above please.

### Answer & Explanation

Mckenna Friedman

a) Use the definition of ${a}^{r}$ and the well known Taylor series for ${e}^{x}$
${a}^{r}=\left({e}^{\mathrm{log}a}{\right)}^{r}={e}^{r\mathrm{log}a}=1+r\mathrm{log}a+{r}^{2}\left(\mathrm{log}a{\right)}^{2}\dots$
b) Hint: Do you remember this formula?
$\underset{n\to \mathrm{\infty }}{lim}\left(1+\frac{x}{n}{\right)}^{n}={e}^{x}$

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