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How would you prove that $$\displaystyle \prod_{k=1}^\infty \left(1+\dfrac{1}{2^k}\right) \lt e ?$$

Wolfram|Alpha shows that the product evaluates to $2.384231 \dots$ but is there a nice way to write this number?

A hint about solving the problem was given but I don't know how to prove the lemma.

Lemma : Let, $a_1,a_2,a_3, \ldots,a_n$ be positive numbers and let $s=a_1+a_2+a_3+\cdots+a_n$ then $$(1+a_1)(1+a_2)(1+a_3)\cdots(1+a_n)$$ $$\le 1+s+\dfrac{s^2}{2!}+\dfrac{s^3}{3!}+\cdots+\dfrac{s^n}{n!}$$

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HINT

Taking $\log$ both sides the statement is equivalent to prove that

$$\sum_{k=1}^\infty \log \left(1+\dfrac{1}{2^k}\right) \lt 1$$

then use $\log(1+x)<x$.

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It can be easily shown that for any $x\in(0,1)$ we have

$$ \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9765}}{1+x+\frac{x^2}{3}+\frac{x^3}{21}+\frac{x^4}{315}+\frac{x^5}{9765}} <1+x < \frac{1+2x+\frac{4x^2}{3}+\frac{8x^3}{21}+\frac{16 x^4}{315}+\frac{32 x^5}{9450}}{1+x+\frac{x^2}{3}+\frac{x^3}{21}+\frac{x^4}{315}+\frac{x^5}{9450}} $$ and in general, by setting $D_m=\prod_{k=1}^{m}(2^k-1)$ and $p_n(x)=1+\sum_{k=1}^{n}\frac{x^k}{D_k}$, $$ \frac{p_n(2x)+\frac{2^{n+1}x^{n+1}}{D_{n+1}}}{p_n(x)+\frac{x^{n+1}}{D_{n+1}}}<1+x< \frac{p_n(2x)+\frac{2^{n+1}x^{n+1}}{D_{n}(2^{n+1}-2)}}{p_n(x)+\frac{x^{n+1}}{D_{n}(2^{n+1}-2)}}.$$ By telescoping it follows that $$ p_n(1)+\frac{1}{D_{n+1}}<\prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)<p_n(1)+\frac{1}{D_n(2^{n+1}-2)} $$ and by picking $n=4$ we have $$ \frac{3326}{1395}<\prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)<\frac{22531}{9450} $$ such that the first figures of the middle term are $\color{green}{2.3842}$.
By picking $n=10$ we get that the middle term is $\color{green}{2.384231029\ldots}$.
As a continued fraction $$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right)=\left[2; 2, 1, 1, 1, 1, 14, 1, 3, 1, 1, 6, 9, 18, 7, 1, 27,\ldots\right]$$ while $e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots]$.

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Proof of the Lemma by Induction. Let $P(n): (1+a_1)(1+a_2)...(1+a_n) ≤ 1 + s_n + ... + \frac {s_n^n}{n!}$ For$n=1$ the result is obvious. Let $P(m): (1+a_1)(1+a_2)...(1+a_m) ≤ 1 + s_m + ... + \frac {s_m^m}{m!}$ be true. Then, $$1 + s_{m+1} + ... + \frac {s_{m+1}^{m+1}}{(m+1)!}$$ $$= 1 + (s_m + a_{m+1}) + ... + \frac {(s_m + a_{m+1})^{m+1}}{(m+1)!}$$ $$= 1 + s_m + ... + \frac {s_m^m}{m!} + Q , Q>0$$ Thus we have, $$(1 + s_{m+1} + ... + \frac {s_{m+1}^{m+1}}{(m+1)!}) - Q = 1 + s_m + ... + \frac {s_m^m}{m!} ≥ (1+a_1)(1+a_2)...(1+a_m) < (1+a_1)(1+a_2)...(1+a_m)(1+a_{m+1})$$ Hence $P(m+1)$ is true whenever $P(m)$ is true. Therefore $P(n)$ is true in general.

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  • $\begingroup$ Induction doesn't give much insight for the proof. $\endgroup$ – AtiqRahman Oct 22 '18 at 8:49

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