$e-2\approx0.71828$, but I got $1$ We know that:
$$\frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots =e-2\approx0.71828$$
But I am getting the above sum as $1,$ as shown below:
\begin{align}
S & = \frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots \\[10pt]
& = \frac 1{2!} + \frac {3-2}{3!} +\frac {4\times2-7}{4!}+\frac {5\times7-34}{5!}+\frac {6\times34-203}{6!}+\cdots \\[10pt]
& = \frac 1{2!}+\frac 1{2!}-\frac 2{3!}+\frac 2{3!}-\frac 7{4!}+\frac 7{4!}-\frac {34}{5!}+\frac {34}{5!}-\frac {203}{6!}+\cdots \\[10pt]
& = 1
\end{align}
Please indicate my mistake
 A: You can do this with any series.
You have a sum $a_1+a_2+a_3+a_4+\cdots$. 
You can rewrite it as $1+(a_1-1)-(a_1-1)+(a_2+a_1-1)-(a_2+a_1-1)+(a_3+a_2+a_1-1)-\cdots$
The partial sums are $1, a_1, 1, a_2+a_1, 1, a_3+a_2+a_1, \dots$
Clearly splitting the elements in this way does not help any to sum the original series.

In response to the comments what OP has done in the question is rather split as $$a_1+(1-a_1)-(1-a_1-a_2)+(1-a_1-a_2)-\dots$$ with partial sums $a_1, 1, a_1+a_2, 1 \dots$
Whichever is used the "adjustment" term is of the form $\pm \left(1-\sum_{r=1}^n a_r\right)$. Here it tends in absolute value to $1-(e-2)=3-e$.
A: You are only shifting the difference at the infinity!
EG
note that $$1-\frac {203}{6!}=1-\frac {203}{720}=0.71805\ldots$$

More in general, it can be easily shown that  the remainder is solution of the following recurrence equation:
$$a_k=a_{k-1}-\frac{1}{k!}$$
$$a_3=\frac13$$
which quickly converges to the value: 0.281718... = 1-e

https://www.wolframalpha.com/input/?i=a(k)%3Da(k-1)-1%2Fk!,+a(3)%3D1%2F3
A: Consider the numerators of the negative terms. The first one, call it $a_3$ for ease of notation, is $2$. Then they follow the recursion $a_{n+1}=(n+1)a_n-1$. What's the behavior of the sequence $\frac{a_n}{n!}$? Cases:


*

*It doesn't go to zero. In this case, your series doesn't even converge.

*It goes to zero but isn't summable. In this case, Riemann's theorem on conditional vs. absolute convergence will warn you that such a rearrangement may (or may not) change the value of the sum, or even cause it to fail to converge at all.

*It is summable. In this case such a rearrangement cannot change the value of the sum.


Also note that you are only considering the sequence of partial sums at even indices (i.e. the sum of 2 terms, 4 terms, ...). This can cause a sum that doesn't converge to look like it does; consider for a simpler example $\sum_{n=0}^\infty (-1)^n$.
A: After cancelling some of the terms, you are finally left with
{1}/{2!} +{1}/{2!} - some fraction.
I think you should recheck if the series you have written is converging. 
Although the numerators don't increase as fast as the denominators, I am not sure if you can completely ignore the further terms.
