## [ Sums of numerical sequences and difference series] Common fractions

What is the value of the expression $\frac{1}{2!}+\frac{2}{3!}+\ldots+\frac{2000}{2001!} ?$
Can the $n$-th term of the given sum be represented as a difference $a_{n}-b_{n}$ in such a way that $b_{n}$ is equal to $a_{n+1} \cdot$

## Solution

Let's use the fact that $\frac{n}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$. Then the sum given in the condition can be written as

$\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\ldots+\left(\frac{1}{2000!}-\frac{1}{2001!}\right)$. In the resulting sum, all terms except the first and the last cancel out, so the value of this expression is $1-\frac{1}{2001!}$.

## Answer

$1-\frac{1}{2001!}$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0ce4a33d8b15522751eg-12.jpg?height=89&width=1064&top_left_y=114&top_left_x=14)

A computational machine can only perform one operation: $\boldsymbol{a} * \boldsymbol{b}=\mathbf{1} \boldsymbol{\boldsymbol { a }} \boldsymbol{\boldsymbol { d }} \boldsymbol{b}$. How can all four arithmetic operations be performed using this machine?

## Hint

The machine remembers the numbers it works with. Understand how to get $\mathbf{0}, \mathbf{1}$.

## Solution

Given two numbers $\boldsymbol{a}, \boldsymbol{b} . \mathbf{1}$ ) If we take the pair (a;a), the machine will correspond this pair to the number 0. 2) If we take the pair ( $\mathbf{( ; b )}$ ), the machine will correspond this pair to the number 1. 3) If we take the pair (1-a:b;1), the machine will correspond this pair to the number $\boldsymbol{a}: \boldsymbol{b}$. Indeed, the machine will associate the given pair with the number 1-(1-a:b):1=a:b. Thus, we have LEARNED to DIVIDE known numbers. 4) Since we have already learned to divide, we can obtain the number $\mathbf{1 : b}$, and then we can obtain the number $\boldsymbol{a}:(\mathbf{1}: \boldsymbol{b})=\boldsymbol{a} * \boldsymbol{b}$. We have LEARNED to MULTIPLY. 5) By dividing the number $\mathbf{1 - a : \boldsymbol { b }}$ by $\mathbf{1 : b}$, we get the number $\boldsymbol{b}-\boldsymbol{a}$. We have found the difference. 6) By subtracting the numbers $\boldsymbol{b}-\boldsymbol{a}$ and $\boldsymbol{b}$, we get the number - $\boldsymbol{a}$. It remains to subtract the number $\boldsymbol{b}$ from the number - $\boldsymbol{a}$, i.e., to write $\boldsymbol{b}-(-\boldsymbol{a})=\boldsymbol{b}+\boldsymbol{a}$. We have found the sum of two numbers.