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?
$$ a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right) $$
 A: If you aplied logarithm you get that $\log{a_n} = \sum_{n=1}^{\infty} \log{(1+\frac{1}{3^n})}$. Now using that $\log{x} \leq x-1$ for $x>0$ you get that $\log{a_n} \leq 
 \sum_{n=1}^{\infty} \frac{1}{3^n}$ that is a geometric series that converges, so as $\log{a_n}$ converges, then $a_n$ converges.
A: 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.
A: More generally, let
$a_n
=\prod_{k=1}^n (1+c^k)
$
where $0 < c < 1$.
This problem is
$c = \frac13$.
Let
$b_n = \ln(a_n)
=\sum_{k=1}^n \ln(1+c^k)
$.
Then $b_n$ is increasing and,
since $\ln(1+x) < x$ for $x > 0$,
$b_n 
\lt \sum_{k=1}^n c^k
=\dfrac{c-c^{n+1}}{1-c}
\lt\dfrac{c}{1-c}
$
for all $n$.
Therefore
$b_n$ is a bounded,
monotone increasing sequence
and so converges.
Therefore $a_n = e^{b_n}$
also converges.
Another proof can be given
using Cauchy's criterion.
If $n > m$,
$b_n-b_m
=\sum_{k=m+1}^n \ln(1+c^k)
\lt\sum_{k=m+1}^n c^k
=\dfrac{c^{m+1}-c^n}{1-c}
\lt\dfrac{c^{m+1}}{1-c}
$.
This last can be made
as small as desired
by choosing $m$ large enough.
Note:
To show that
$\ln(1+x) < x$
for $ x > 0$:
$\ln(1+x)
=\int_1^{1+x}\dfrac{dt}{t}
=\int_0^{x}\dfrac{dt}{1+t}
\lt\int_0^{x}dt
=x
$.
