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How can we show that the function $n \mapsto e_n$, where $e_n$ is the $n$-th digit in the decimal expansion of $e$, is computable?

I have some idea in terms of Cantor's diag. argument, but I need to think along the lines of writing a series expansion, and Church's thesis.

Can someone produce the series as discussed below? thanks

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Do you really want $n$ to be the $n$th digit? or do you mean for $e_n$ to be the $n$th digit? –  Gerry Myerson Feb 9 '13 at 11:17
    
en is the nth digit in e. –  mary Feb 9 '13 at 20:57
    
I believe the series to which @Carl refers is $e=\sum_0^{\infty}(n!)^{-1}$. –  Gerry Myerson Feb 9 '13 at 22:36
    
Sure, but I dont know how to fit it in with Carl's hint below. Can you please add in the detail below? –  mary Feb 9 '13 at 22:58
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Can you do some work on your own? Can you even cut'n'paste "Taylor series with remainder" into Google to see what I'm talking about? Anyone throwing around terms like "Cantor's diagonal argument", "series expansion", and "Church's thesis" ought to be able to do a little bit of mathematics on his/her own. –  Gerry Myerson Feb 10 '13 at 2:56

1 Answer 1

Hint: because $e$ is not a rational number, for each rational number $r$ you can produce a sufficiently good approximation to $e$ to tell whether $e$ is greater than $r$ or less than $r$. The way to make the approximation is to use a series for $e$ for which you can estimate the error for each partial sum.

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