Q. 27 - Calculus

Question

In Exercises 21–28 provide the first five terms of the series.

$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$

Step-by-Step Solution

Answer

Ans: The five terms of the series are $1,-\frac{1}{2},\frac{1}{24},-\frac{1}{720},\frac{1}{40320}$

Step 1. Given information:

$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$

Step 2. Finding the first term of the series:

The first term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is obtained by substituting $n=0$ in $\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=0$ is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^0}{(2\times 0)!}$ (Substituting)

=1(Because 0 !=1 )

The first term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is 1 .

Step 3. Finding the second term of the series:

The second term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is obtained by substituting $n=1$ in $\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=1$ is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^1}{(2\times 2)!}$ (Substituting)

$=\frac{-1}{2!}=-\frac{1}{2}$

The second term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is $-\frac{1}{2}$.

Step 4. Finding the third term of the series:

The third term of the series$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is obtained by substituting $n=2$ in$\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=2$ is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^2}{(2\times 2)!}$ (Substituting)

$=\frac{1}{4!}=\frac{1}{24}$

The third term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is $\frac{1}{24}$.

Step 5. Finding the fourth term of the series:

The fourth term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is obtained by substituting $n=3$ in $\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=3$ is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^3}{(2\times 3)!}$( Substituting )

$=\frac{-1}{6!}=\frac{-1}{6\times 5\times 4\times 3\times 2\times 1}=-\frac{1}{720}$

The fourth term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is $-\frac{1}{720}$.

Step 6. Finding the fifth term of the series:

The fifth term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is obtained by substituting $n=4$ in $$. Therefore, the value at $n=4$ is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^4}{(2\times 4)!}$( Substituting )

$=\frac{1}{8!}=\frac{-1}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}=\frac{1}{40320}$

The fifth term of the series$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$ is $\frac{1}{40320}$.$\frac{(-1{)}^n}{\left(2n\right)!}$