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I have to prove the following inequality: $$\prod\limits_{r=1}^{\infty}\left(1+\frac{1}{2^r}\right)<\frac{5}{2}$$ and I tried to prove by induction that $\prod_{r=1}^{R}\left(1+\frac{1}{2^r}\right)<\frac{5}{2}-\frac{2}{2^R}$ but I haven't managed to.

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  • $\begingroup$ But anyhow, I hope you could take a look at that useful trick in my answer. Such an evaluation trick is most useful at times. $\endgroup$ – Vim Jan 19 '15 at 11:44
  • $\begingroup$ Creative telescoping is perfectly suited for the job. $\endgroup$ – Jack D'Aurizio Aug 24 '18 at 19:43
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If we start at $r=1$, then the result is true.

Using $1+x\le e^x$, we get $$ \begin{align} \prod_{r=1}^\infty\left(1+\frac1{2^r}\right) &\le\frac32\prod_{r=2}^\infty e^{1/2^r}\\ &=\frac32e^{1/2}\\ &\lt\frac52 \end{align} $$ since $e\lt\frac{25}9$.


Another Approach

Lemma: if $0\le a_k\le1$, then $$ \prod_{k=1}^n(1-a_k)\ge1-\sum_{k=1}^na_k\tag{1} $$ Proof: Suppose $(1)$ is true for $n$, then $$ \begin{align} \prod_{k=1}^{n+1}(1-a_k) &=(1-a_{n+1})\prod_{k=1}^n(1-a_k)\\ &\ge(1-a_{n+1})\left(1-\sum_{k=1}^na_k\right)\\ &=1-a_{n+1}-\sum_{k=1}^na_k+a_{n+1}\sum_{k=1}^na_k\\ &\ge1-\sum_{k=1}^{n+1}a_k\tag{2} \end{align} $$ Thus, $(1)$ is true for $n+1$. Since $(1)$ is trivial for $n=0$, we see that $(1)$ is true for all integer $n\ge0$. $$\square$$ Corollary: Taking the limit of $(1)$ yields $$ \prod_{k=1}^\infty(1-a_k)\ge1-\sum_{k=1}^\infty a_k\tag{3} $$ Notice that for $0\lt a_k\lt1$, $(1+a_k)(1-a_k)=1-a_k^2\lt1$. Therefore, $$ 1+a_k\lt\frac1{1-a_k}\tag{4} $$ The Corollary implies $$ \begin{align} \prod_{r=3}^\infty\left(1-\frac1{2^r}\right) &\ge1-\frac14\\ &=\frac34\tag{5} \end{align} $$ Combining $(4)$ and $(5)$ gives $$ \begin{align} \prod_{r=1}^\infty\left(1+\frac1{2^r}\right) &=\frac32\cdot\frac54\prod_{r=3}^\infty\left(1+\frac1{2^r}\right)\\ &\lt\frac{15}8\prod_{r=3}^\infty\left(1-\frac1{2^r}\right)^{-1}\\ &\le\frac{15}8\cdot\frac43\\ &=\frac52\tag{6} \end{align} $$

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Yet another approach is creative telescoping. Here it is simpler to apply it to produce tight lower bounds. We may check that for any $x\in(0,1)$ we have $$ 1+x > \left(\frac{1+\frac{2x}{3}}{1+\frac{x}{3}}\right)^3 $$ hence it follows that $$ \prod_{r\geq 1}\left(1+\frac{1}{2^r}\right) > \frac{3}{2}\cdot\frac{5}{4}\cdot\prod_{r\geq 3}\left(\frac{1+\frac{1}{3\cdot 2^{r-1}}}{1+\frac{1}{3\cdot 2^r}}\right)^3=\frac{15}{8}\left(1+\frac{1}{12}\right)^3 > \frac{50}{21}.$$ Similarly $$ 1+x < \frac{1+2x+\frac{4}{3}x^2+\frac{8}{7}x^3}{1+x+\frac{1}{3}x^2+\frac{1}{7}x^3} $$ implies $$ \prod_{r\geq 1}\left(1+\frac{1}{2^r}\right)<\frac{15}{8}\left[1+2x+\frac{4}{3}x^2+\frac{8}{7}x^3\right]_{x=1/8}<\frac{74}{31}. $$ The difference between the upper bound and the lower bound is already less than $7\cdot 10^{-3}$.


A more advanced technique is to exploit the Mellin transform. The approach outlined by Marko Riedel here leads to

$$ \prod_{r \geq 1}\left(1+\frac{1}{2^r}\right) \approx 2^{-\frac{11}{24}} e^{\frac{\pi^2}{12\log 2}}=\exp\left(\frac{\pi^2}{12\log 2}-\frac{11\log 2}{24}\right)$$ where the approximation error is less than $1.1\cdot 10^{-12}$ (!!!).

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Do you actually mean that $r$ starts at 1?
My hint is, apply logarithm to break the LHS into a sum, and evaluate the sum by an integral.
After such operation, the inequality turns into such an equivalent form as: $$\displaystyle\sum_{r=1}^{\infty}\ln(1+(\frac{1}{2})^r)<\ln(\frac{5}{2})$$ Now that the value RHS is quite easy to evaluate, it is approximately $0.916291$. So all we have to do is evaluate the LHS.
For such evaluation about a sum concerning consecutive integers there is a quite useful trick (although not an exact evaluation, yet useful at most times, esp. in proving an inequality). First we need to know what the plot of $y=\ln(1+(1/2)^x)$ looks like. So I made a plot using Maple (see below). Obviously it is ever-decreasing and downward convex in $\mathbb R^{+}$. In fact, as you will later see, the only two things that matter in our evaluation is whether the function is an increasing or decreasing one and whether it is upward convex or downward convex, so we don't necessarily have to turn to Maple, calculating $y'$ and $y''$ by hand will also do. Well, that depends on whether you have a plotting software at hand.
enter image description here

Ok, so back to the subject, we are now evaluating one term, like $\ln(1+(1/2)^{k+1})$, in the LHS sum. Note that what we are to prove is a "$<$" inequality. Therefore, we naturally want to make a "$<$" evaluation. From the plot, it is very geometrically clear that $\ln(1+(1/2)^{k+1})$ just amounts to the area of a "block" with the width $1$ and the height $\ln(1+(1/2)^{k+1})$, which is exactly the leftmost block shown in the plot. And likewise, the next term $\ln(1+(1/2)^{k+2})$ amounts to the second block from the left......
Now let the "first block" be the one from $0$ to $1$, say, the "k-th" block from $(k-1)$ to $k$. Since all the blocks are strictly below the function's plot, we have $$LHS<\int_{0}^{+\infty} \ln(1+(1/2)^x)dx $$ where the RHS, calculated by Maple, is $1.18657$. Oops, that doesn't look like what we want! But don't worry, it just means that our evaluation is much too surplus. We can get rid of the "overflow" easily just by putting our evaluation some blocks forward. Try starting our "block evaluation" from $x=10$, say, from the eleventh term $\ln(1+(1/2)^{11})$ $$LHS<\displaystyle\sum_{r=1}^{10}\ln(1+(\frac{1}{2})^r)+\int_{10}^{+\infty} \ln(1+(1/2)^x)dx$$ This time we must turn to software for numerical calculation. Maple tells me that (only the first $6$ digits are shown here, in fact it tells much more) $$\displaystyle\sum_{r=1}^{10}\ln(1+(\frac{1}{2})^r) \approx 0.867900$$ $$\int_{10}^{+\infty} \ln(1+(1/2)^x)dx=0.001409$$ Put them together, and the sum, by no means, will exceed $0.9$, let alone $0.916291$. Therefore, at long last, we can claim for certain that $$LHS<0.9<RHS$$ QED.

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