I want to prove $\prod^{k-1}_{i=0}(1+\frac{1}{k+2i})<2$ for all integers $k\ge2$.
I can see that this is true by testing many values, but I would like to prove it. I have tried proof by induction. My problem is that the induction hypothesis for (n-1) does not tell me the amount by which $k=(n-1)$ and $2$ differ. Is there a straightforward induction proof? Any other good methods?
Denote the product by $p_k$ and check (for $k>1$) that $p_{k+1}/p_{k-1}=(9k^2-4)/(9k^2-1)$.
Thus, $p_{k+1}<p_{k-1}$ for $k>1$. Since $p_1=2$ and $p_2=15/8<2$, the result follows.
Products like these can be expressed in a closed form:
See
$\displaystyle{\prod\limits^{k - 1}_{i=0}{\dfrac{k + 2i + 1}{k + 2i}}}=\dfrac{k + 1}{k}\cdot\dfrac{k + 3}{k + 2}\cdot\dfrac{k + 5}{k + 4}\cdot \ldots \cdot \dfrac{3k - 1}{3k - 2}=\displaystyle{\dfrac{3\cdot\left(\frac{k}{2}\right)!\left(\frac{3k -1}{2}\right)!}{\left(\frac{3k}{2}\right)!\left(\frac{k-1}{2}\right)!}}$
Now, use Stirling's approximation to calculate the limit as $k - 1 \to \infty$
Let $f(n) = \sqrt{2\pi n} \cdot \displaystyle{\left(\frac{n}{e}\right)^n} \approx n! \newcommand{\Stirling}[1]{ \sqrt{2\pi \cdot #1} \cdot \displaystyle{\left(#1 \cdot \dfrac{1}{e}\right)^#1}}$
Therefore, $$\displaystyle{\lim\limits_{k \to \infty}{\prod\limits^{k - 1}_{i=1}{\dfrac{k + 2i + 1}{k + 2i}}} = \lim\limits_{k \to \infty}{\dfrac{3\cdot f\left(\dfrac{k}{2}\right) \cdot f\left(\dfrac{3k - 1}{2}\right)}{f\left(\dfrac{3k}{2}\right) \cdot f\left(\dfrac{k-1}{2}\right)}} \\ \quad \\}$$
$$\displaystyle{= \lim\limits_{k \to \infty}{\dfrac{3\cdot \Stirling{\dfrac{k}{2}} \cdot \Stirling{\dfrac{3k - 1}{2}}}{\Stirling{\dfrac{3k}{2}} \cdot \Stirling{\dfrac{k - 1}{2}}}} \\ \quad \\}$$
$$\displaystyle{= \lim\limits_{k \to \infty} { \require{\cancel} \underbrace{\dfrac{3\cdot \Stirling{\dfrac{k}{2}}}{\Stirling{\dfrac{3k}{2}}}}_{\dfrac{\sqrt{3} \cdot \cancel{\sqrt{2\pi \cdot \dfrac{k}{2}}} \cdot \cancel{\displaystyle{\left(\dfrac{k}{2} \cdot \dfrac{1}{e}\right)^\dfrac{k}{2}}}}{3^\cfrac{3k}{2} \cdot \cancel{\sqrt{2\pi \cdot \dfrac{k}{2}}} \cdot \displaystyle{\left(\dfrac{k}{2} \cdot \dfrac{1}{e}\right)^{\large{k}}}}} \space \cdot \space \underbrace{\dfrac{\Stirling{\dfrac{3k - 1}{2}}}{\Stirling{\dfrac{k - 1}{2}}} }_{\dfrac{\left(\dfrac{3k - 1}{2}\right)^\dfrac{3k}{2}}{\left(\dfrac{k - 1}{2}\right)^\dfrac{k}{2}} \large{\space \cdot \space e^{-k}}}}}$$
$$\displaystyle{\Rightarrow \lim\limits_{k \to \infty}{\dfrac{\sqrt{3} \cdot \left(\dfrac{3k - 1}{2}\right)^\dfrac{3k}{2}}{3^\cfrac{3k}{2} \cdot \left(\dfrac{k}{2}\right)^{\large{k}} \cdot \left(\dfrac{k - 1}{2}\right)^\dfrac{k}{2}}} \cdot e^{-k} \cdot e^{k} }$$
$$\displaystyle{\Rightarrow \lim\limits_{k \to \infty}{\big(k - 1\big)^\cfrac{-k}{2} \cdot \left(\frac{3k - 1}{3}\right)^\cfrac{3k}{2} \cdot k^{-k} \cdot \sqrt{3}}}$$
$$\displaystyle{\Rightarrow \sqrt{3} \cdot \lim\limits_{k \to \infty}{\big(k^3 - k^2\big)^\cfrac{-k}{2} \cdot \left(\left(\frac{3k - 1}{3}\right)^3\right)^\cfrac{k}{2}}}$$
$$\displaystyle{\Rightarrow \sqrt{3} \cdot \lim\limits_{k \to \infty}{\left(\frac{\left(\frac{3k - 1}{3}\right)^3}{k^3 - k^2}\right)^\cfrac{k}{2}}}$$
$$\displaystyle{\Rightarrow \sqrt{3} \cdot \lim\limits_{k \to \infty}{\left(\frac{27k^3 - 27k^2 + 9k - 3}{27k^2\left(k - 1\right)}\right)^\cfrac{k}{2}}}$$
$$\displaystyle{\Rightarrow \sqrt{3} \cdot \lim\limits_{k \to \infty}{\left(1 + \frac{9k - 3}{27k^2\left(k - 1\right)}\right)^\cfrac{k}{2}}}$$
$$\displaystyle{ = \bbox[yellow, 5px,border:2px solid black] {\sqrt{3} \cdot 1 < 2}}$$
