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Suppose you have this sum: $1- \frac1{2!} + \frac1{3!}-\cdots-\frac1{52!}$

How do you show that this is equal to approximately $1- e^{-1}$?

Attempt: I know the formula for $e^{-x}$ but am not sure where to proceed now.

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    It should be "approximately equal"2012-01-21

1 Answers 1

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You know that $e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+-\dots\;,$ so $\begin{align*} e^{-1}&=1-\frac1{1!}+\frac1{2!}-\frac1{3!}+-\dots\\ &=\frac1{2!}-\frac1{3!}+\frac1{4!}-\frac1{5!}+-\dots\;. \end{align*}$

If you add this to the infinite series $1- \frac1{2!} + \frac1{3!}-+\dots\;,$ you obviously get $1$, so $1-e^{-1}=1- \frac1{2!} + \frac1{3!}-+\dots\;.$ This is an alternating series with strictly decreasing terms, so the error when you truncate it at $1/52!$ is less in magnitude than the first missing term, namely, $1/53!$. That’s certainly a very small error.

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    $1/53! \approx 2 \times 10^{-70}$2012-01-21