Okay, this problem has been kicking my butt for several hours now.

Consider the positive sequences $\displaystyle \{a_{n}\}$ and $\displaystyle \{b_{n}\}$ with

$\displaystyle b_{n}=\sqrt[n]{a_{1}\cdot a_{2}\cdot\ldots\cdot a_{n}}.$

Prove or disprove: If $\displaystyle \{a_{n}\}$ converges to $\displaystyle A$, then $\displaystyle \{b_{n}\}$

converges to $\displaystyle A$.

I attempted to prove this as follows:

Since

$\displaystyle \lim_{n\rightarrow\infty}a_{n}=A,$

$\displaystyle \lim_{n\rightarrow\infty}a_{n+1}=A,$

Therefore

$\displaystyle b_{n+1}=\sqrt[n+1]{a_{1}\cdot a_{2}\cdot\ldots\cdot A\cdot A}.$

Thus

$\displaystyle \lim_{n\rightarrow\infty}b_{n}=\sqrt[n]{A^{n}}=A$

Did I even remotely do that right? I feel like I should be using the

triangle inequality to prove this, but I haven't been able to produce

a useful result with it. The best I have is:

$\displaystyle \varepsilon=a_{n}+b_{n}>0$

$\displaystyle \delta>0$

$\displaystyle \left|b_{n}+A\right|\leq\left|b_{n}\right|+\left|A \right|$

$\displaystyle \left|\sqrt[n]{a_{1}\cdot a_{2}\cdot\ldots\cdot a_{n}}+A\right|\leq\left|\sqrt[n]{a_{1}\cdot a_{2}\cdot\ldots\cdot a_{n}}\right|+\left|A\right|$

I could probably get a more useful result if $\displaystyle \varepsilon=a_{n}-b_{n}>0$, but I can't find a single effective way to prove that $\displaystyle a_{n}\geq b_{n}$