2
$\begingroup$

Let $a_1,a_2,\dots,a_n$ be any $n$ positive real numbers. Show that $$\lim_{t \to 0^+}\left[\frac1n \sum_{i=1}^{n}a_i^t\right]^{1/t}$$ is the geometric mean of $a_1,a_2,\dots,a_n$.

I know $\text{GM}= \left[\displaystyle\prod_{i=1}^{n} a_i\right]^{1/n}$

$\displaystyle\lim_{t\to 0^+} a_i^t\approx 1 ~\forall~ i$ so AM=GM

So, $$\lim_{t \to 0^+}\left[\left(\displaystyle\prod_{i=1}^{n} a_i^t\right)^{1/n}\right]^{1/t}=\lim_{t \to 0^+} \left[\displaystyle\prod_{i=1}^{n} a_i\right]^{\displaystyle\frac tn\times \frac 1t}=\text{GM}$$

Am I correct? I can't give a solid argument. Can this problem solve in any other fashion.

$\endgroup$
1
  • $\begingroup$ Hint: Take logs and apply L'Hopital. $\endgroup$
    – whuber
    Dec 1 '14 at 17:09
3
$\begingroup$

The logarithm of our expression is equal to $$\frac{\log(\frac{1}{n}\sum a_i^t)}{t}.$$ Using L'Hospital's Rule we find that the limit is $$\lim_{t\to 0^+} \frac{\frac{1}{n} \sum(\log a_i) a_i^t }{\frac{1}{n}\sum a_i^t}.$$ This limit is $\frac{1}{n}\sum \log a_i$. Exponentiate.

Remark: The AM-GM argument is nice. However, some estimates would be needed to verify the intuition that we are "nearly" in the equality case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.