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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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Can you write more clearly your series? If it is alternating, the last term seems wrong... Also, what happens if you set $x=1$ in the formula for $e^{-x}$? – N. S. Jan 21 '12 at 21:51
When I added $\LaTeX$ I also turned your finite sum into an infinite series; I hope that that was what you intended it to be. – Brian M. Scott Jan 21 '12 at 21:58
I fixed the question. It's supposed to be a finite series. – lord12 Jan 21 '12 at 22:00
This can't be true if it is a finite series. – user17090 Jan 21 '12 at 22:02
If it’s a finite series, the statement is false. – Brian M. Scott Jan 21 '12 at 22:03
up vote 6 down vote accepted

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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$1/53! \approx 2 \times 10^{-70}$ – lhf Jan 21 '12 at 23:47

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