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

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## Comments

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

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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, \(S=e-2<1\)

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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 \(e-3\).

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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/Recurrence

relation#Solvingfirst-ordernon-homogeneousrecurrencerelationswithvariablecoefficients) to get the formula. The limit of the absolute value of the terms ultimately simplifies to \(3-e\).Log in to reply

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

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

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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.

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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)

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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 \(\dfrac 1{2!} + \dfrac 1{3!} + \dfrac 1{3!} = \dfrac 56\).

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