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

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?
