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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

$$A_N=\prod_{n=1}^{N}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n=1}^{N}\log\left(1+\frac{1}{2^n}\right)\leq\exp\sum_{n=1}^{N}\frac{1}{2^n}\leq e.$$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.