This is similar to something I posted yesterday, but I can't figure it out.
What's the limit of the following as n goes to infinity,
[(a^(1/n) + b^(1/n))/2]^n for a, b > 0
?
Could someone show me how to do this with algebra?
Thanks in advance.
This is similar to something I posted yesterday, but I can't figure it out.
What's the limit of the following as n goes to infinity,
[(a^(1/n) + b^(1/n))/2]^n for a, b > 0
?
Could someone show me how to do this with algebra?
Thanks in advance.
This is the case $\displaystyle 1^{\infty}$.
$\displaystyle \displaystyle\lim_{n\to\infty}\left(\frac{a^{\frac {1}{n}}+b^{\frac{1}{n}}}{2}\right)^n=\lim_{n\to\in fty}\left(1+\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}-2}{2}\right)^n=e^{\displaystyle\lim_{n\to\infty}n\ left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}-2}{2}\right)}=$
$\displaystyle \displaystyle =e^{\displaystyle\frac{1}{2}\lim_{n\to\infty}\left (\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}}+\frac{b^{\frac{1}{n}}-1}{\frac{1}{n}}\right)}=e^{\frac{1}{2}(\ln a+\ln b)}=e^{\ln\sqrt{ab}}=\sqrt{ab}$
Use the fact that,
$\displaystyle \left( 1 + \frac{s_n}{n} \right)^n \to e^s$ where $\displaystyle s_n \to s$.
So here, $\displaystyle 2s_n = na^{1/n} + nb^{1/n} - 2n$
This means, $\displaystyle 2s_n = n(a^{1/n} - 1) + n(b^{1/n} - 1)$.
Thus,
$\displaystyle 2s = \log a + \log b = \log ab$.
It is related to This