Prove/disprove convergence of two series.

**DenJansen** · October 17th 2011, 08:38 AM

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

Consider the positive sequences $\{a_{n}\}$ and $\{b_{n}\}$ with
$b_{n}=\sqrt[n]{a_{1}\cdot a_{2}\cdot\ldots\cdot a_{n}}.$

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

I attempted to prove this as follows:

Since

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

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

Therefore

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

Thus

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

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:

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

$\delta>0$

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

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

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

**girdav** · October 17th 2011, 10:57 AM

Do you Cesaro mean? You can use the results about that taking the logarithm.

**DenJansen** · October 18th 2011, 11:26 PM

Originally Posted by girdav

Do you Cesaro mean? You can use the results about that taking the logarithm.

Figures. Reading that, it sounds like it's exactly what I need, but we skipped the section that describes it. Thanks for your help!

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