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.
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.
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.