Chapter 9.1, Problem 85E

To determine

Torepresent:The sum of the series $\frac{1^2}{2}+\frac{2^2}{6}+\frac{3^2}{24}+\frac{4^2}{120}+\mathrm{...}+\frac{7^2}{40,320}$ in the form ofsigma notation.

Answer to Problem 85E

The sum of the series $\frac{1^2}{2}+\frac{2^2}{6}+\frac{3^2}{24}+\frac{4^2}{120}+\mathrm{...}+\frac{7^2}{40,320}$ is ${\displaystyle \sum _{i=1}^7\frac{i^2}{\left(1+i\right)!}}$ .

Explanation of Solution

Given information:

The given series is $\frac{1^2}{2}+\frac{2^2}{6}+\frac{3^2}{24}+\frac{4^2}{120}+\mathrm{...}+\frac{7^2}{40,320}$ .

Calculation:

Let’s a sequence has $n$ terms and it can be represented as follows.

  ${\displaystyle \sum _{i=1}^n{a}_i}=a_1+a_2+a_3+a_4+\mathrm{...}+a_n$

Here, the index of summation is $i$ , the lower limit of summation is 1 and the upper limit of summation is $n$ .

Simplify the given sequence and find the sum of the series.

  $\begin{array}{c}\frac{1^2}{2}+\frac{2^2}{6}+\frac{3^2}{24}+\frac{4^2}{120}+\mathrm{...}+\frac{7^2}{40,320}=\left[\begin{array}{l}\left(\frac{1^2}{2\times 1}\right)+\left(\frac{2^2}{3\times 2\times 1}\right)+\left(\frac{3^2}{4\times 3\times 2\times 1}\right)+\\ \left(\frac{4^2}{5\times 4\times 3\times 2\times 1}\right)+\mathrm{...}+\left(\frac{7^2}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\right)\end{array}\right]\\ =\frac{1^2}{2!}+\frac{2^2}{3!}+\frac{3^2}{4!}+\frac{4^2}{5!}+\mathrm{...}+\frac{7^2}{8!}\\ =\frac{1^2}{\left(1+1\right)!}+\frac{2^2}{\left(1+2\right)!}+\frac{3^2}{\left(1+3\right)!}+\frac{4^2}{\left(1+4\right)!}+\mathrm{...}+\frac{7^2}{\left(1+7\right)!}\\ ={\displaystyle \sum _{i=1}^7\frac{i^2}{\left(1+i\right)!}}\end{array}$

Therefore, the sum of the series $\frac{1^2}{2}+\frac{2^2}{6}+\frac{3^2}{24}+\frac{4^2}{120}+\mathrm{...}+\frac{7^2}{40,320}$ in the form of sigma notation is ${\displaystyle \sum _{i=1}^7\frac{i^2}{\left(1+i\right)!}}$ .