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\(e-2=1\)

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:

\(\begin{align}S & = \frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots \\ & = \frac 1{2!} + \frac {3-2}{3!} +\frac {4\times2-7}{4!}+\frac {5\times7-34}{5!}+\frac {6\times34-203}{6!}+\cdots \\ & = \frac 1{2!}+\frac 1{2!}-\frac 2{3!}+\frac 2{3!}-\frac 7{4!}+\frac 7{4!}-\frac {34}{5!}+\frac {34}{5!}-\frac {203}{6!}+\cdots \\ & = 1 \end{align} \)

Please indicate my mistake

Note by Mrigank Shekhar Pathak
1 week, 4 days ago

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\(\begin{align}S &= \frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots \\ &= \frac 1{2!} + \frac {3-2}{3!} +\frac {4\times2-7}{4!}+\frac {5\times7-34}{5!}+\frac {6\times34-203}{6!}+\cdots \\ & = \frac 1{2!}+\frac 1{2!}-\frac 2{3!}+\frac 2{3!}-\frac 7{4!}+\frac 7{4!}-\frac {34}{5!}+\frac {34}{5!}-\frac {203}{6!}+\cdots \\ & = 1-\lim_{n\rightarrow \large \infty} \dfrac{a_{n}}{n!}\\\\ \text{Where } &a_{n} \text{ is given by the reccurance,}\\ a_{n}&=na_{n-1}-1\\\\ \text{I was } &\text{able to simplify } a_{n} \text{ as,}\\ a_n&=3(n!)-\sum_{m=0}^n \dfrac{n!}{(n-m)!}\\\\ \implies S&=1-\lim_{n\rightarrow \large \infty} \dfrac{3(n!)-\sum_{m=0}^n \dfrac{n!}{(n-m)!}}{n!}\\\\ &=1-3+\lim_{n\rightarrow \large \infty} \dfrac{\sum_{m=0}^n \dfrac{n!}{(n-m)!}}{n!}\\\\ &=-2+\lim_{n\rightarrow \large\infty}\sum_{m=0}^n \dfrac{1}{(n-m)!}\\ &\text{setting } (n-m) =t\\ &=-2+\lim_{n\rightarrow \large\infty}\sum_{t=0}^n \dfrac{1}{(t)!}\hspace{5mm} \color{blue}\text{t varies from } 0 \text{ to } n\\ &=\color{orange}\boxed{\color{black}e-2}\end{align}\)

Anirudh Sreekumar - 1 week, 3 days ago

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

Mrigank Shekhar Pathak - 1 week, 2 days ago

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

James Wilson - 1 week, 4 days ago

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I used the recurrence relation \(a_{n+1}=na_n-1\), \(a_3=2\) 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 \(3-e\).

James Wilson - 1 week, 4 days ago

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Although you would have to show \(e\neq 3\).

James Wilson - 1 week, 4 days ago

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

James Wilson - 1 week, 4 days ago

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

Mrigank Shekhar Pathak - 1 week, 2 days ago

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

Chew-Seong Cheong - 1 week, 4 days 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 \(3!\) term and split the remaining terms in the same way then I am getting the same answer (i.e., 1)

Mrigank Shekhar Pathak - 1 week, 4 days 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 \(4!\) term you will get \(\dfrac 1{2!} + \dfrac 1{3!} + \dfrac 1{3!} = \dfrac 56\).

Chew-Seong Cheong - 1 week, 4 days ago

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

Mrigank Shekhar Pathak - 1 week, 4 days ago

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@Mrigank Shekhar Pathak No. It is tough.

Chew-Seong Cheong - 1 week, 4 days ago

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