Show, that the sequence $ a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right) $ converges. I need to show, that the following sequence converges.
I think I can somehow do it by Riemann Integral, but I cannot figure out a way to extract $ \frac{1}{n} $ from it. I also cannot find two sequences which could let me show that it converges by the sandwich theorem. How to approach it?
$$ a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right) $$

If you are ok with using GM-AM (inequality between geometric and arithmetic mean) and with using the well known limit


*

*$\left(1+ \frac{x}{n}\right)^n\stackrel{n\to \infty}{\longrightarrow}e^x$
then you can reason directly as follows:
\begin{eqnarray*} \prod_{k=1}^n \left(1+\frac{1}{3^k}\right)
 & \leq & \left(\frac{\sum_{k=1}^n \left(1+\frac{1}{3^k}\right)
}{n} \right)^n\\
& = & \left(\frac{n + \sum_{k=1}^n \frac{1}{3^k}}{n} \right)^n\\
 & \leq &  \left(1 + \frac{\sum_{k=1}^{\infty} \frac{1}{3^k}}{n} \right)^n \\
 & =  &  \left(1 + \frac{\frac{1}{2}}{n} \right)^n \\
 & \stackrel{n \to \infty}{\longrightarrow}  &  \sqrt{e} \\
\end{eqnarray*}
Since $a_n$ is obviously increasing and according to above calculation also bounded, it follows that $a_n$ is also convergent.
