Jtdylktuy

Using the series $displaystyle sum_{k=0}^{infty}frac{(-1)^{k}}{k!}=frac{1}{e}$, evaluate the first $3$ decimal digits of $1/e$.

Attempt. In alternating series $displaystyle sum_{k=0}^{infty}(-1)^{k+1}alpha_n$, where $alpha_n searrow 0$, if $alpha$ is the sum of the series then $$|s_n-alpha|leq alpha_{n+1}.$$ So, in our case we need to find $n$
such that $|s_n-1/e|<0.001$, where $displaystyle s_n=sum_{k=0}^{n-1}frac{(-1)^{k}}{k!}$ and it is enough to find $n$ such that $dfrac{1}{n!}<0.001$, so $ngeq 7$. Therefore:

$$s_7=sum_{k=0}^{6}frac{(-1)^{k}}{k!}=0.36805ldots$$

so I would expect $dfrac{1}{e}=0.368ldots$. But:
$dfrac{1}{e}=0.36787944ldots$.

Where am I missing something?

Thanks in advance.

To know the first three digits in the decimal expansion of $e^{-1}$ you don't want $n$ such that
$$|s_n-e^{-1}|<0.001,$$
but in stead you want $n$ such that $|s_n-e^{-1}|$ is strictly less than the part of $e^{-1}$ after the first three digits. Denoting the fractional part of a real number $x$ by ${x}$, this can be formally described as
$$|s_n-e^{-1}|<left|s_n-frac{{10^3s_n}}{10^3}right|,$$
so it suffices to find $n$ such that
$$frac{1}{n!}<left|s_n-frac{{10^3s_n}}{10^3}right|.$$
As your computations show $n=7$ does not suffice; you've found that
$$s_7=0.3680555...qquadtext{ but }qquad 0.0000555...<frac{1}{7!}.$$ A little more work shows that $n=8$ does suffice; indeed
$$s_8=0.3678571...qquadtext{ and }qquad 0.0001428...>frac{1}{8!}.$$

Where am I missing something?

Um... nowhere?

$|s_7 - frac 1e| = |0.36805.... - 0.36787944| approx 0.00017056... < 0.001$

And $|0.368 - frac 1e| = .00012055882855767840447622983853913..... < 0.001$

So... why do you think you are missing something?

so we know:
$$S=sum_{n=0}^inftyfrac{(-1)^n}{n!}=frac{1}{e}$$
and we want to know this $frac1e$. To do this we simply evaluate terms of the sequence until it is consistant to 3 decimal decimal places.
$$a_0=frac11=1$$
$$a_0+a_1=1+frac{-1}1=0$$
$$a_0+a_1+a_2=0+frac{1}{2}=frac{1}{2}=0.5$$
$$a_0+a_1+a_2+a_3=frac{1}{2}+frac{-1}{6}=frac{1}{3}approx0.333$$
$$a_0...+a_4=frac{1}{3}+frac{1}{24}=frac{3}{8}=0.375$$
$$a_0...+a_5=frac{3}{8}+frac{-1}{120}approx0.367$$
If you continue this in the way that you did then you do indeed end up with $0.368$ which is correct to 3 decimal places, so you have missed nothing

Making the problem more general, you want to compute $p$ such that
$$S_p=sum_{k=0}^{p}frac{(-1)^{k}}{k!}$$ such that
$$frac{1}{(p+1)!} le 10^{-n}$$ $n$ being the number of significant decimal places you need.

If you look at this question of mine, you will see a magnificent approximation by @robjohn for the inverse of the factorial function.

Applied to your case $(a=1)$, this would give
$$color{blue}{p=frac{n log (100)-log left({2 pi }right)}{2 Wleft(frac{n log
(100)-log left({2 pi }right)}{2 e}right)}-frac 32}$$ where appears Lambert function that you can evaluate using the expansions given in the linked Wikipedia page.

For $n=3$, this would give, as a real number, $p=5.17$. So, you need to use $p=6$. Let us check : $frac 1 {7!}=frac{1}{5040}approx 0.00020$ while $frac 1 {6!}=frac{1}{720}approx 0.00139$.

Note
$$bigg|sum_{k=0}^{n}frac{(-1)^{k}}{k!}-frac{1}{e}bigg|=bigg|sum_{k=n+1}^{infty}frac{(-1)^{k}}{k!}bigg|lesum_{k=n+1}^inftyfrac{1}{3^{k}}lefrac{1}{2cdot3^{n}}.$$
Let $frac{1}{2cdot3^{n}}<0.001$ and then $n>frac{ln500}{ln 3}approx5.65678$. Now set $n=6$ and then
$$ bigg|sum_{k=0}^{6}frac{(-1)^{k}}{k!}-frac{1}{e}bigg|<0.001. $$
In fact, it is easy to see
$$ bigg|sum_{k=0}^{5}frac{(-1)^{k}}{k!}-frac{1}{e}bigg|=0.001212774505,bigg|sum_{k=0}^{6}frac{(-1)^{k}}{k!}-frac{1}{e}bigg|=0.0001761143841<0.001. $$
So $n=6$ is the small number such that $|s_n-frac1e|<0.001$.