Question

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

$\sum _{i=0}^{\infty }\frac{i!}{(i+1)!}$

Short answer

The five terms of the series are$1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}$

Step-by-step solution

Step 1. Given information: 

$\sum _{i=0}^{\infty }\frac{i!}{(i+1)!}$

Step 2. Finding the first term of the series:

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

$\frac{i!}{(i+1)!}=\frac{0!}{(0+1)!}$(Substituting)

$$\begin{gathered} =\frac{0!}{1!} \\ =1(\text{Because}0!=1) \end{gathered}$$

The first term of the series $\sum _{i=0}^{\infty }\frac{i!}{(i+1)!}$is 1 .

Step 3. Finding the second term of the series:

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

$\frac{i!}{(i+1)!}=\frac{1!}{(1+1)!}$(Substituting)

$$\begin{gathered} =\frac{1!}{2!} \\ =\frac{1}{2} \end{gathered}$$

The зccond term of the series $\sum _{i=0}^{\infty }\frac{i!}{(i+1)!}$is$\frac{1}{2}$.

Step 4. Finding the third term of the series:

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

$\frac{i!}{(i+1)!}=\frac{2!}{(2+1)!}$(Substituting)

$=\frac{2!}{3!}=\frac{1}{3}$

The third term of the series $\sum _{i=0}^{\infty }\frac{i!}{(i+1)!}$is $\frac{1}{3}$.

Step 5. Finding the fourth term of the series:

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

$\frac{i!}{(i+1)!}=\frac{3!}{(3+1)!}$(Substituting)

$=\frac{3!}{4!}=\frac{1}{4}$

The fourth term of the series $\sum _{i=0}^{\infty }\frac{i!}{(i+1)!}$is $\frac{1}{4}$.

Step 6. Finding the fifth term of the series:

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

$\frac{i!}{(i+1)!}=\frac{4!}{(4+1)!}$(Substituting)

$=\frac{4!}{5!}=\frac{1}{5}$

The fifth term of the series $\sum _{i=0}^{\infty }\frac{i!}{(i+1)!}$is $\frac{1}{5}$.