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How many terms are required in the series $e =\sum^∞_{k=0}{1\over k!}$ to give $e$ with an error of at most ${6\over 10}$ unit in the $20$th decimal place?

Here is what I have:

$$e\approx \sum_{k=0}^n {1\over k!}$$ where the remainder term is $$R=\sum_{k=n+1}^\infty {1\over k!}={1\over (n+1)!} \left[1+{(n+1)!\over (n+2)!}+{(n+1)!\over (n+3)!}+\cdots\right] \leq {1\over(n + 1)!} \left[1 + {1\over 2} + {1\over 4}+{1\over 8} + \cdots \right] = {2\over (n + 1)!}.$$ I'm not sure how to do this and give the error of at most $6\over 10$ unit in the $20$th decimal place. Any solutions or hints are greatly appreciated.

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  • $\begingroup$ Note that one can do a tiny bit sharper, $\frac{1}{n+2}$ rather than $\frac{1}{2}$. Then just fool around with a calculator, to get the suitable $n$. $\endgroup$ – André Nicolas Aug 28 '15 at 19:21
  • $\begingroup$ What do you mean by "a tiny bit sharper, ${1\over {n+2}}$ rather than ${1\over 2}$"? $\endgroup$ – 1233dfv Sep 6 '15 at 18:48
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Every tail of this series is bounded above by a geometric series. For example: \begin{align} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1 {7!} + \frac 1 {8!} + \frac 1 {9!} + \frac 1 {10!} + \cdots} \\[10pt] \le {} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1 {7!} + \frac 1 {7!\cdot 8} + \frac 1 {7!\cdot 8^2} + \frac 1 {7!\cdot 8^3} + \cdots} \\[10pt] = {} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \frac{1/7!}{1 - 1/8} \\[10pt] = {} & \frac{1957}{720} + \frac 1 {630}. \end{align} So the problem is: how many terms are needed to make the upper bound in place of $1/630$ as small as what you need?

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  • $\begingroup$ Sorry, forgive me, but I don't understand. I understand the math you typed, but now I don't understand what is meant "by to give $e$ with an error of at most ${6\over 10}$ unit in the $20$th decimal place? $\endgroup$ – Mike Aug 28 '15 at 19:51
  • $\begingroup$ I would guess this means the 20th digit after the decimal point. Every time the 20th digit increases by $1$, the number whose digit it is increases by $10^{-20}$. So I would take that to mean the error should be less than $\frac 6 {10}\times 10^{-20}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 28 '15 at 19:54

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