mathimatikoi.org

a) Prove that $$\displaystyle\frac{1}{2}\cdot\frac{3}{4}\cdot\ldots\cdot\frac{99}{100}<\frac{1}{10}\,.$$
b) Examine if, for every \(n\in\mathbb{N}\,,\) holds $$\displaystyle\mathop{\prod}\limits_{k=1}^{n}\frac{2k-1}{2k}<\frac{1}{\sqrt{2n}}\,.$$

Taking logarithms, it suffices to show that \[-\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k}\right)>\frac{1}{2}\ln 2n\,.\] But \[-\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k}\right)=\sum_{k=1}^{n}\sum_{m=1}^{+\infty}\frac{(2k)^{-m}}{m}>\sum_{k=1}^{n}\sum_{m=1}^{2}\frac{(2k)^{-m}}{m}=\frac{H_{n}}{2}+\frac{1}{4}\sum_{k=1}^{n}\frac{1}{k^2}\] so it suffices to show that \[\frac{H_{n}}{2}+\frac{1}{8}\sum_{k=1}^{n}\frac{1}{k^2}>\frac{1}{2}\ln 2n\,.\] Now, setting \[a_n:=\frac{H_{n}}{2}+\frac{1}{8}\sum_{k=1}^{n}\frac{1}{k^2}-\frac{1}{2}\ln 2n\] we have that \[a_{n+1}<a_n\] (this is straightforward by studying \(a_{x+1}-a_x\)) and \[a_n\to\frac{\gamma-\ln2}{2}+\frac{\pi^2}{48}>0\] (since \(H_n=\gamma+\ln n+\mathcal O(n^{-1})\)). These mean that \(a_n>0\) for \(n\geq1\) and we are done.

Trying to prove the result directly by induction does not seem to work but actually if we are asked to prove the stronger result \[ \prod_{k=1}^n \frac{2k-1}{2k} < \frac{1}{\sqrt{2n+1}}\] then it becomes a rather easy exercise on mathematical induction:

The result is immediate if \(n=1\). Moreover if the result is true for \(n=r\) then \[ \prod_{k=1}^{r+1} \frac{2k-1}{2k} = \frac{2r+1}{2r+2}\prod_{k=1}^{r} \frac{2k-1}{2k} < \frac{2r+1}{2r+2} \cdot \frac{1}{\sqrt{2r+1}} = \frac{\sqrt{2r+1}}{2r+2}< \frac{1}{\sqrt{2r+3}}\] so the result is true for \(n=r+1\) completing the induction step and thus the proof.

A solution by Foteini Kaldi:

\[\dfrac{1}{2}\cdot \dfrac{2}{3}\cdot \ldots \cdot\dfrac{2n}{2n+1}=\dfrac{1}{2n+1}\] but \[\dfrac{1}{2} <\dfrac{2}{3}\quad \Rightarrow \quad \left(\dfrac{1}{2}\right)^2 <\dfrac{1}{2}\cdot \dfrac{2}{3}\]
\[\dfrac{3}{4} <\dfrac{4}{5}\quad \Rightarrow \quad \left(\dfrac{3}{4}\right)^2 <\dfrac{3}{4}\cdot \dfrac{4}{5}\]
\[\vdots\]
\[\dfrac{2n-1}{2n} <\dfrac{2n}{2n+1}\quad \Rightarrow \quad \left(\dfrac{2n-1}{2n}\right)^2 <\dfrac{2n-1}{2n}\cdot\dfrac{2n}{2n+1}\,.\]
The above give \[\left(\dfrac{1}{2}\cdot \dfrac{3}{4}\cdot \ldots \dfrac{2n-1}{2n}\right)^2 < \dfrac{1}{2n+1}\Rightarrow\dfrac{1}{2}\cdot \dfrac{3}{4}\cdot \ldots \dfrac{2n-1}{2n}<\dfrac{1}{\sqrt{2n+1}}<\dfrac{1}{\sqrt{2n}}\,.\]