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If $d(a,b)=$ largest $n$ such that $a$ and $b$ agree on all digits upto $n$. Eg. $d(\pi,3.14)=3$, $d(0.1234667,0.1234669)=7$. What is the asymptotics of $d(\pi/4,1-1/3+1/5-1/7+\cdots(\pm)1/m)$ as $m\rightarrow\infty$?

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So $d(a,b)=\lceil-\log_{10}|a-b|\rceil$? –  J. M. Oct 27 '11 at 2:56
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This is a complicated way of asking about the rate of convergence of the Gregory series.. –  Ragib Zaman Oct 27 '11 at 3:00
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No, there can be problem if there are a long sequence of 9's in decimal representation of $\pi$, you can see it here, section 10: ics.org.ru/doc?pdf=440&dir=e –  Nurdin Takenov Oct 27 '11 at 3:21
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up vote 4 down vote accepted

Using the standard estimates for an alternating series of decreasing terms [1], we know the error $$ \frac{\pi}{4} - \sum_{k=1}^m \frac{(-1)^{k+1} }{2k-1} = \sum_{k=m+1}^{\infty} \frac{(-1)^{k+1} }{2k-1}$$

has magnitude $\sim \displaystyle \frac{1}{2m}.$ The number of agreed digits is thus asymptotic to the number of leading decimal zeros in the decimal expansion of $ \displaystyle \frac{1}{2m},$ which is $ \sim \log_{10} 2m .$

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