Chapter 8, Problem 10P

To determine

To find: The sum of the series ${\displaystyle \sum _{n\to 0}^{\infty }{a}_n}$ .

Answer to Problem 10P

The sum of ${\displaystyle \sum _{n\to 0}^{\infty }{a}_n}$ is $e$ .

Explanation of Solution

Given:

The sequence $\left\{a_n\right\}$ is defined recursively by the equations $a_0=a_1=1$ , $n\left(n-1\right)a_n=\left(n-1\right)\left(n-2\right)a_{n-1}-{\left(n-3\right)}_{a_{n-2}}$ .

Calculation:

Rewrite the above recurrence relation as follows.

  $\begin{array}{c}n\left(n-1\right)a_n=\left(n-1\right)\left(n-2\right)a_{n-1}-{\left(n-3\right)}_{a_{n-2}}\\ a_n=\frac{\left(n-1\right)\left(n-2\right)a_{n-1}-{\left(n-3\right)}_{a_{n-2}}}{n\left(n-1\right)}\end{array}$

Let $n=2$

Substitute 2 for $n$ in the above equation.

  $\begin{array}{c}{a}_2=\frac{\left(2-1\right)\left(2-2\right)a_{2-1}-{\left(2-3\right)}_{a_{2-2}}}{2\left(2-1\right)}\\ =\frac{1{\left(0\right)}_{a_{2-1}}-{\left(2-3\right)}_{a_{2-2}}}{2\left\{2-1\right\}}\\ =\frac{a_0}{2}\\ =\frac{1}{2}\end{array}$

Therefore, the value of $a_2=\frac{1}{2!}$ .

Let $n=3$

Substitute 3 for $n$ in the above equation.

  $\begin{array}{c}{a}_3=\frac{\left(3-1\right)\left(3-2\right)a_{3-1}-{\left(3-3\right)}_{a_{3-2}}}{3\left(3-1\right)}\\ =\frac{2{\left(1\right)}_{a_2}-{\left(0\right)}_{a_1}}{3\left(2\right)}\\ =\frac{a_2}{3}\\ =\frac{1}{6}\end{array}$

Therefore, the value of $a_3=\frac{1}{3!}$ .

Let $n=4$

Substitute 4 for $n$ in the above equation.

  $\begin{array}{c}{a}_4=\frac{\left(4-1\right)\left(4-2\right)a_{4-1}-{\left(4-3\right)}_{a_{4-2}}}{4\left(4-1\right)}\\ =\frac{3{\left(2\right)}_{a_3}-{\left(1\right)}_{a_2}}{4\left(3\right)}\\ =\frac{6a_3-a_2}{12}\\ =\frac{1}{24}\end{array}$

Therefore, the value of $a_4=\frac{1}{4!}$ .

Repeating like this it will get $a_n=\frac{1}{n!}$ .

Now find the sum of ${\displaystyle \sum a_n}={\displaystyle \sum \frac{1}{n!}}$ .

  $\begin{array}{c}{\displaystyle \sum a_n}={\displaystyle \sum \frac{1}{n!}}\\ =1+\frac{1}{2!}+\frac{1}{3!}+\cdots \\ =e\end{array}$

Therefore, the sum of ${\displaystyle \sum _{n\to 0}^{\infty }{a}_n}$ is $e$ .