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I want to find an infinite sequence that conerges to e so that the kth term of the sequence is less than 10^-k away from e. Obviously, I've considered the Taylor series, but asymptotic bounds on the truncation error don't tell you what the actual upper bound for a specific index is.

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The error of $\sum_{k=0}^n\frac1{k!}$ is approximately the next term $\frac1{(n+1)!}$ and an upper bound for the error is obtained from $$ \sum_{k=n+1}^\infty\frac1{k!}< \frac1{n!}\sum_{k=1}^\infty\frac1{(n+1)^k}=\frac1{n\cdot n!}.$$ As $n\cdot n!>10^n$ for $n$ sufficiently big, this should help you.

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