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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.

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  • $\begingroup$ an increasing function is always monotonic... $\endgroup$ – John Dvorak Nov 4 '14 at 12:25
  • $\begingroup$ 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 $\endgroup$ – mvggz Nov 4 '14 at 12:26
  • $\begingroup$ You're right, I meant limited. Edited first post. $\endgroup$ – duke Nov 4 '14 at 12:26
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    $\begingroup$ Use the AM/GM inequality to provide a bound? $\endgroup$ – Mark Bennet Nov 4 '14 at 12:29
  • $\begingroup$ See Infinite product: Convergence criteria. $\endgroup$ – Lucian Nov 4 '14 at 19:06
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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.

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  • $\begingroup$ Thank You. Technically I still hasn't had series and series convergency at uni, but this seems reasonable. $\endgroup$ – duke Nov 4 '14 at 12:49
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$$A_N=\prod_{n=1}^{N}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n=1}^{N}\log\left(1+\frac{1}{2^n}\right)\leq\exp\sum_{n=1}^{N}\frac{1}{2^n}\leq e.$$ 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.

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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.

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