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

  • 1
    The first two terms of your rearranged series sum to $1$. After that the terms cancel in pairs. To say that correctly you should look at the sequence of partial sums. – Ethan Bolker Dec 6 '17 at 17:00
  • 10
    Are you sure that your rearrangement results in a convergent series? What's the general formula for the numerators? They seem to grow rather quickly. – Ian Dec 6 '17 at 17:02
  • 1
    I'm voting to close this question as off-topic because it seems to be a way to silently challenge us while the OP knows the answer. – Yves Daoust Dec 7 '17 at 14:23
  • @YvesDaoust Please do not misunderstand me , I really didn't knew initially that the procedure I used can't be applied and I have learnt a lot from all the answers below. I don't know why you misunderstood me. – Mrigank Shekhar Pathak Dec 7 '17 at 17:36
  • @MrigankShekharPathak: I removed my vote. – Yves Daoust Dec 7 '17 at 18:33
up vote 11 down vote accepted

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

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

  • 3
    The example in the OP is a little bit more interesting because the "adjustments" go to zero, but not "summably fast". Your "adjustments" don't go to zero. – Ian Dec 6 '17 at 17:20
  • @Ian I think that the adjustments go to $3-e$ not $0$. – Mark Bennet Dec 6 '17 at 18:19
  • 1
    Oh, you're right, what the OP did is actually somewhat more dramatic than I thought. +1 for pointing that out. – Ian Dec 6 '17 at 18:26

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

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.

  • You're misunderstanding the meaning of the ...; the next term in the sum is $+\frac{203}{6!}$. The point is that it is set up so that after the first two terms, the terms come in $\pm$ pairs. – Ian Dec 7 '17 at 16:28
  • Yeah, I m sorry. I thought the summation ends there. I didn't read it properly – user166465 Dec 7 '17 at 16:30

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