$e-2\approx0.71828$, but I got $1$

Question

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

Answer 1

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 $\sum_{n=0}^\infty (-1)^n$.

Answer 2

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.

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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 $1-(e-2)=3-e$.

Answer 3

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

Answer 4

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.