# Does limit of $x_n=(1+\frac{1}{2})*(1+\frac{1}{4})*…*(1+\frac{1}{2^n})$ exist?

I tried to show that the sequence is increasing and limited, but couldn't find the limit. I also tried with squeeze theorem, but $(1+\frac{1}{2^n})^n<=x_n<=(1+\frac{1}{2})^n$ is not helping and I ran out of ideas.

• – John Dvorak Nov 4 '14 at 12:25
• Try taking the ln of $x_n$, and use : $ln(1+u) \leq u$ you'll get : $a_n = ln(1+\frac{1}{2^n}) \leq \frac{1}{2^n} = v_n$ ; $\sum v_n$ converges, hence $\sum a_n$ does as well, and your sequence converges – mvggz Nov 4 '14 at 12:26
• You're right, I meant limited. Edited first post. – duke Nov 4 '14 at 12:26
• Use the AM/GM inequality to provide a bound? – Mark Bennet Nov 4 '14 at 12:29
• – Lucian Nov 4 '14 at 19:06

A handy fact about products like this is $$1+\sum_{k=1}^n a_k \le \prod_{k=1}^n (1+a_k) \le \exp\Big(\sum_{k=1}^n a_k\Big)$$ (The first inequality is a generalization of Bernoulli's inequality; the second is $1+x\le e^x$ used $n$ times.) So you can check the convergence of this kind of product by checking the convergence of a related series, for which we have a bunch of standard techniques. In this case, the related series is a familiar one.

• Thank You. Technically I still hasn't had series and series convergency at uni, but this seems reasonable. – duke Nov 4 '14 at 12:49

$$A_N=\prod_{n=1}^{N}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n=1}^{N}\log\left(1+\frac{1}{2^n}\right)\leq\exp\sum_{n=1}^{N}\frac{1}{2^n}\leq e.$$ Since the LHS is increasing and bounded, the limit $$\lim_{N\to +\infty} A_N = \prod_{n=1}^{+\infty}\left(1+\frac{1}{2^n}\right)$$ exists.

I'll write it this way:

Let : $S_n = ln(x_n) = \sum_{k=0}^n ln(1 + \frac{1}{2^k})$

As I said: $a_n = ln(1+\frac{1}{2^n}) \leq \frac{1}{2^n} = v_n$

You get : $S_n \leq \sum_{k=1}^n v_k \leq \sum_{k=1}^{+\infty} v_k = 1$

$(S_n)$ is increasing, bounded so it converges.