Prove that $b_n$ is decreasing and that $\lim_{n \to \infty}b_n=e$

Deduce that $e<3$.

Any help would be appreciated.

Edit: we can't use log/ln nor derive or integrate because we haven't covered that yet.

  • $\begingroup$ I tried to show by induction that its decreasing with the inequality of arithmetic and geometric means but got stuck. I tried to use Bernoulli's inequality but got stuck again... $\endgroup$ – GinKin Nov 15 '13 at 18:50
  • $\begingroup$ This problem turned out to be more difficult than what I thought. $\endgroup$ – hhsaffar Nov 16 '13 at 7:48
  • $\begingroup$ I also tried the induction method with the binomial theorem but it doesn't help. It can't be that difficult, this should be my easiest question... $\endgroup$ – GinKin Nov 16 '13 at 8:31
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    $\begingroup$ I think this is what you are looking for: math.stackexchange.com/questions/121076/proving-1-1-nn1-gt-e/… $\endgroup$ – hhsaffar Nov 16 '13 at 18:01
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    $\begingroup$ This is highly related to this question. I showed in my answer that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing; thereby allowing us to estimate $e\le\left(1+\frac15\right)^6\lt3$. $\endgroup$ – robjohn Nov 16 '13 at 19:20

Without using induction, you could take the log of $b_n$ and then the limit is much clearer : \begin{eqnarray*} \lim_{x\rightarrow \infty} log(b_n) &=& \lim_{x\rightarrow \infty}(n+1)log(1 + \frac1n)\\ &=& \lim_{x\rightarrow \infty} (n+1)(\frac1n + 0(n^{-2}))\\ &=& \lim_{x\rightarrow \infty}\frac{n+1}{n}\\ &=& 1 \end{eqnarray*}

So $\lim_{x\rightarrow \infty} b_n = e^1 = e$

To show it is decreasing, consider $\frac{b_{n+1}}{b_n}$ perhaps? Then check that for some $n_0\in \mathbb N$, $b_{n_0} <3$ which implies that $e<3$ since $b_n$ is a decreasing sequence.

  • $\begingroup$ Thanks but we can't use log, we haven't covered that yet. $\endgroup$ – GinKin Nov 15 '13 at 19:14
  • $\begingroup$ In that case you should mention it in the question and write more about what to use, as well as the work you did so far. $\endgroup$ – user88595 Nov 15 '13 at 19:16

Hint: Write $(1+\frac1n)^{n+1}=e^{(n+1)\ln(1+\frac1n)}$.

(1) What is $\lim_{n\rightarrow\infty}(n+1)\ln(1+\frac1n)$? What does this imply for the limit of the original sequence? (Hint: Continuity)

(2) What is the derivative of $e^{(x+1)\ln(1+\frac1x)}$? When is this function decreasing?

For showing $e<3$, try computing the first few terms until you get a value smaller than $3$.

  • $\begingroup$ Thanks but we can't use ln nor derive, I edited the question. $\endgroup$ – GinKin Nov 15 '13 at 19:19
  • $\begingroup$ Then I guess the question is how you define $e$? Numerically, as a limit, ...? What tools can you actually use? Monotone convergence? $\endgroup$ – Sean Clark Nov 15 '13 at 19:22
  • $\begingroup$ Yes and a few inequities among those I mentioned in the replies to the question. $\endgroup$ – GinKin Nov 15 '13 at 19:26

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