\(e-2=1\)

We know that:

12!+13!+14!+15!+16!+⋯=e−2≈0.71828\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:

S=12!+13!+14!+15!+16!+⋯=12!+3−23!+4×2−74!+5×7−345!+6×34−2036!+⋯=12!+12!−23!+23!−74!+74!−345!+345!−2036!+⋯=1\begin{aligned}S & = \frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots \\ & = \frac 1{2!} + \frac {3-2}{3!} +\frac {4\times2-7}{4!}+\frac {5\times7-34}{5!}+\frac {6\times34-203}{6!}+\cdots \\ & = \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 \\ & = 1 \end{aligned}

Please indicate my mistake

Note by Mrigank Shekhar Pathak
3 years, 6 months ago

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1 vote

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Comments

The absolute value of each term of the alternating series approaches 3−e3-e rather than zero. So, strangely enough, the new series does not converge. In particular, in any finite version, you are left with (∑k=4n1k!)−23!(\sum_{k=4}^n \frac{1}{k!})-\frac{2}{3!} as a final term that does not cancel with anything, which approaches e−3e-3.

James Wilson - 3 years, 6 months ago

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I used the recurrence relation an+1=nan−1a_{n+1}=na_n-1, a3=2a_3=2 to describe the numerators of the fractions. And then used the method from this wikipedia page (https://en.wikipedia.org/wiki/Recurrencerelation#Solvingfirst-ordernon-homogeneousrecurrencerelationswithvariablecoefficients) to get the formula. The limit of the absolute value of the terms ultimately simplifies to 3−e3-e.

James Wilson - 3 years, 6 months ago

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Although you would have to show e≠3e\neq 3.

James Wilson - 3 years, 6 months ago

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@James Wilson – But I think the flaw should be clear...

James Wilson - 3 years, 6 months ago

S=12!+13!+14!+15!+16!+⋯=12!+3−23!+4×2−74!+5×7−345!+6×34−2036!+⋯=12!+12!−23!+23!−74!+74!−345!+345!−2036!+⋯=1−lim⁡n→∞ann!Where an is given by the reccurance,an=nan−1−1I was able to simplify an as,an=3(n!)−∑m=0nn!(n−m)!  ⟹  S=1−lim⁡n→∞3(n!)−∑m=0nn!(n−m)!n!=1−3+lim⁡n→∞∑m=0nn!(n−m)!n!=−2+lim⁡n→∞∑m=0n1(n−m)!setting (n−m)=t=−2+lim⁡n→∞∑t=0n1(t)!t varies from 0 to n=e−2\begin{aligned}S &= \frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots \\ &= \frac 1{2!} + \frac {3-2}{3!} +\frac {4\times2-7}{4!}+\frac {5\times7-34}{5!}+\frac {6\times34-203}{6!}+\cdots \\ & = \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 \\ & = 1-\lim_{n\rightarrow \large \infty} \dfrac{a_{n}}{n!}\\\\ \text{Where } &a_{n} \text{ is given by the reccurance,}\\ a_{n}&=na_{n-1}-1\\\\ \text{I was } &\text{able to simplify } a_{n} \text{ as,}\\ a_n&=3(n!)-\sum_{m=0}^n \dfrac{n!}{(n-m)!}\\\\ \implies S&=1-\lim_{n\rightarrow \large \infty} \dfrac{3(n!)-\sum_{m=0}^n \dfrac{n!}{(n-m)!}}{n!}\\\\ &=1-3+\lim_{n\rightarrow \large \infty} \dfrac{\sum_{m=0}^n \dfrac{n!}{(n-m)!}}{n!}\\\\ &=-2+\lim_{n\rightarrow \large\infty}\sum_{m=0}^n \dfrac{1}{(n-m)!}\\ &\text{setting } (n-m) =t\\ &=-2+\lim_{n\rightarrow \large\infty}\sum_{t=0}^n \dfrac{1}{(t)!}\hspace{5mm} \color{#3D99F6}\text{t varies from } 0 \text{ to } n\\ &=\color{#EC7300}\boxed{\color{#333333}e-2}\end{aligned}

Anirudh Sreekumar - 3 years, 6 months ago

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What you have done is excellent , but it would be better if you show that 2<lim⁡n→∞∑t=0n1t!<3\displaystyle 2<\lim_{n\to \infty}\sum_{t=0}^{n}\dfrac 1{t!}<3 thus, S=e−2<1S=e-2<1

Mrigank Shekhar Pathak - 3 years, 6 months ago

Let the general term be nln−1−lnn!\dfrac {nl_{n-1}-l_n}{n!}. You are assuming that lnn!\dfrac {l_n}{n!} converges as n→∞n \to \infty. In your example you start splitting with the 3!3! term, you will get another value if you start with 4!4! term. It cannot be the answer.

Chew-Seong Cheong - 3 years, 6 months ago

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please explain in detail through some example, I can't understand why it would be wrong. Suppose I leave the 3!3! term and split the remaining terms in the same way then I am getting the same answer (i.e., 1)

Mrigank Shekhar Pathak - 3 years, 6 months ago

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You are assuming your procedure converges. It does not converge so you don't get the answer. I thought if you start splitting at 4!4! term you will get 12!+13!+13!=56\dfrac 1{2!} + \dfrac 1{3!} + \dfrac 1{3!} = \dfrac 56.

Chew-Seong Cheong - 3 years, 6 months ago

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@Chew-Seong Cheong – could you please show with some test that the procedure does not converge

Mrigank Shekhar Pathak - 3 years, 6 months ago

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@Mrigank Shekhar Pathak – No. It is tough.

Chew-Seong Cheong - 3 years, 6 months ago

You may have a look at Mathematics Stack Exchange it has got a good discussion on it.

Mrigank Shekhar Pathak - 3 years, 6 months ago
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