Help with the convergence of $[\gamma \sqrt[n]{a} + (1-\gamma)\sqrt[n]{b} ]^n$ [closed]

Question

Let be $a,b> 0$ and $\gamma \in (0,1) $. Set $x_n = (\gamma \sqrt[n]{a} + (1-\gamma)\sqrt[n]{b} )^n$ for each $n\in\mathbb{N}$.

My question is if this sequence $(x_n)$ is convergent and, if it so, which is its limit. I tried some ideas but none of them work. I guess I am stuck now, but it seems not a difficult sequence, so I'm sure somebody can give me a hint or a good idea here.

Thank you in advance.

Answer 1

Let us write $p = \gamma, q = 1 - \gamma$ so that $p,q > 0$ and $p+q = 1$. Without loss of generality, we may assume $a > b$ (the case when $a = b$ is considerably easier). Thus $$(pa^{1/n} + qb^{1/n})^n = b (p(a/b)^{1/n} + q)^n, $$ so it suffices to consider the limit $$\lim_{n \rightarrow \infty} (p u^{1/n} + q)^n$$

for $u = a/b > 1$. Now put $t = 1/n$, so the above limit turns into

$$\exp \left(\lim_{t \rightarrow 0^+} \frac{\log(pu^t + q)}{t} \right).$$

Applying L'Hospital's rule we obtain

$$\exp \left(\lim_{t \rightarrow 0^+}\frac{p \log(u) u^t}{pu^t + q} \right) = \exp \left(p \log u\right) = \left(\frac{a}{b} \right)^\gamma.$$

Multiplying by the factor of $b$ from the very beginning gives $a^\gamma b^{1 - \gamma}$.

Answer 2

We have $$\gamma \sqrt[n]{a} + (1-\gamma)\sqrt[n]{b}=\gamma a^{1/n} + (1-\gamma)b^{1/n} \\ =\gamma e^{(1/n)\ln a} + (1-\gamma)e^{(1/n)\ln b} \\ =\gamma\Big(1+\frac1n\,\ln a+O(1/n^2)\Big) + (1-\gamma)\Big(1+\frac1n\,\ln b+O(1/n^2)\Big) \\ =1+\frac1n\,\ln(a^\gamma b^{1-\gamma})+O(1/n^2)\\ =\exp\Big\{\frac1n\,\ln(a^\gamma b^{1-\gamma})+O(1/n^2)\Big\},$$ whence the limit is $a^\gamma b^{1-\gamma}$.

Answer 3

Here's a very unclever proof:

You're wondering if $x_n =[\gamma a^{\frac{1}{n}} + (1-\gamma)b^{\frac{1}{n}}]^n$ converges.

$y_n = \log(x_n) = \frac{1}{\frac{1}{n}}\log(\gamma a^{\frac{1}{n}} + (1-\gamma)b^{\frac{1}{n}})$ is a good candidate for l'Hopital's Rule, and is easily shown to be equal to $\frac{\gamma a^{\frac{1}{n}}\log(a) + (1-\gamma) b^{\frac{1}{n}}\log(b)}{\gamma a^{\frac{1}{n}} + (1-\gamma)b^{\frac{1}{n}}}$, leading to $\lim_{n\to\infty} x_n = e^{\gamma\log(a) + (1-\gamma) \log(b)}$.

Answer 4

Power means of order $1/n$ converge to the geometric mean when $n$ tends to $+\infty$. You ask about the weighted version of this fact, when we have the numbers $a, b$ with weights $\gamma$, $1-\gamma$.