# $$e-2=1$$

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}

Note by Mrigank Shekhar Pathak
1 year, 4 months 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 $$e-3$$.

- 1 year, 4 months ago

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 $$3-e$$.

- 1 year, 4 months ago

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

- 1 year, 4 months ago

But I think the flaw should be clear...

- 1 year, 4 months ago

\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}

- 1 year, 4 months ago

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, $$S=e-2<1$$

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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)

- 1 year, 4 months ago

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

- 1 year, 4 months ago

could you please show with some test that the procedure does not converge

- 1 year, 4 months ago

No. It is tough.

- 1 year, 4 months ago

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

- 1 year, 4 months ago