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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?

$$ a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right) $$

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    $\begingroup$ What about taking the logarithm ? $\endgroup$
    – user65203
    Feb 10 '19 at 16:54
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    $\begingroup$ The sequence is monotone. Is it bounded? $\endgroup$
    – rtybase
    Feb 10 '19 at 16:55
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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.

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  • $\begingroup$ How did you get from $ \log{a_n} = \sum_{n=1}^{\infty} \log{1+\frac{1}{3^n}} $ to $ \log{a_n} \leq \sum_{n=1}^{\infty} \frac{1}{3^n} $ ? I don't understand that part $\endgroup$
    – leller
    Feb 10 '19 at 17:10
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    $\begingroup$ $\log{x} \leq x-1$ is the same as $\log{(1+x)}\leq x$, so $\log{(1+3^{-n})} \leq 3^{-n}$ $\endgroup$
    – JoseSquare
    Feb 10 '19 at 17:12
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    $\begingroup$ Could you please use brackets? I don't know if you mean $ log(1)+x $ or $log(1+x)$ $\endgroup$
    – leller
    Feb 10 '19 at 17:14
  • $\begingroup$ Already edit, I meant $\log{(1+x)}$ $\endgroup$
    – JoseSquare
    Feb 10 '19 at 17:16
  • $\begingroup$ Thank you very much. Could you please show me the path from $ log(x) \leq x - 1 $ to $ log(1+x) \leq x $? $\endgroup$
    – leller
    Feb 10 '19 at 17:19
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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.

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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 $.

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