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?

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

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

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.

• 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 Feb 10 '19 at 17:10
• $\log{x} \leq x-1$ is the same as $\log{(1+x)}\leq x$, so $\log{(1+3^{-n})} \leq 3^{-n}$ Feb 10 '19 at 17:12
• Could you please use brackets? I don't know if you mean $log(1)+x$ or $log(1+x)$ Feb 10 '19 at 17:14
• Already edit, I meant $\log{(1+x)}$ Feb 10 '19 at 17:16
• Thank you very much. Could you please show me the path from $log(x) \leq x - 1$ to $log(1+x) \leq x$? Feb 10 '19 at 17:19

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.

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 $$x > 0$$:

$$\ln(1+x) =\int_1^{1+x}\dfrac{dt}{t} =\int_0^{x}\dfrac{dt}{1+t} \lt\int_0^{x}dt =x$$.