# Thread: Prove/disprove convergence of two series.

1. ## Prove/disprove convergence of two series.

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 $\displaystyle a_{n}\geq b_{n}$

2. ## Re: Prove/disprove convergence of two series.

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

3. ## Re: Prove/disprove convergence of two series.

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!