Why is $\frac{1}{10}\left(\frac{9}{10}+\frac{8}{9}+\dots+\frac{1}{2}\right)\approx \frac{\sqrt{2}}{2}$?

Question

When doing a stochastics problem recently I noticed that \begin{equation*} \frac{1}{10}\sum\limits_{k=1}^9 \frac{k}{k+1}=\frac{1}{10}\left(\frac{9}{10}+\frac{8}{9}+\dots+\frac{1}{2}\right)=0.7071031746 \end{equation*} while \begin{equation*} \frac{\sqrt{2}}{2}=0.7071067812\dots \end{equation*} These two quantities are amazingly close to each other (in fact, the discrepancy is only about $5\cdot 10^{-4}\%$) and thus, I wondered if there is any deeper reason behind this.

Is there some relationship between those two quantities that explains why they are so close to each other or is it just "pure luck"?

Answer 1

Converted to an answer at @martycohen's suggestion:

For $n:=10$ your sum is$$\frac1n\sum_{k=0}^{n-1}\frac{k}{k+1}=1-\frac1n\sum_{k=0}^{n-1}\frac{1}{k+1}=1-\frac1nH_n\approx1-\frac1n(\ln n+\gamma).$$Since the large-$n$ behaviour is to approximate $1$, it looks like a coincidence that boils down to asking which $n$ will make this work best.

Answer 2

A more detailed approach:

$ \frac{9}{10}+\frac{8}{9}+\frac{7}{8}+ . . . +\frac{1}{2}=(1-\frac{1}{10})+(1-\frac{1}{9})+(1-\frac{1}{8})+ . . . +(1-\frac{1}{2})=9-\big(\frac {1}{10}+\frac{1}{9}+\frac{1}{8}+ . . . + \frac {1}{2}\big )$

$s=\big(\frac {1}{10}+\frac{1}{9}+\frac{1}{8}+ . . . + \frac {1}{2}\big ) =\int ^{10}_2 \frac{dx}{x}= \big[ Ln (x)\big]^{10}_2$

⇒ $S=\frac{1}{10}\big(\frac{9}{10}+\frac{8}{9}+\frac{7}{8}+ . . . +\frac{1}{2}\big)=\frac{9-(Ln (10)- Ln (2)}{10}≈ 0.739056$

$\frac{\sqrt 2}{2}=0.707106$

$0.739056-0.707106=0.03195≈0.1-0.06805$

$0.06805≈ \gamma/10$

Where $\gamma$ is Euler constant.