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

$\sum _{n=1}^{\infty }\frac{n+1}{n!}$

$\sum _{n=1}^{\infty }\frac{n+1}{n!}$

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

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

$=2$

The first term of the series is 2 .

The second term of the series $\sum _{n=1}^{\infty }\frac{n+1}{n!}$is obtained by substituting localid="1649385671923" $n=2$in $\sum _{n=1}^{\infty }\frac{n+1}{n!}$. Therefore, the value at $n=2$is:

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

$=\frac{3}{2}$

The second term of the series $\sum _{n=1}^{\infty }\frac{n+1}{n!}$is $\frac{3}{2}$.

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

$\frac{3+1}{3!}=\frac{4}{3\times 2\times 1}$ (Substituting)

$=\frac{2}{3}$

The third term of the series $\sum _{n=1}^{\infty }\frac{n+1}{n!}$is $\frac{2}{3}$.

The fourth term of the series $\sum _{n=1}^{\infty }\frac{n+1}{n!}$is obtained by substituting $n=4$in $\frac{n+1}{n!}$Therefore, the value at $n=4$is:

$\frac{4+1}{4!}=\frac{5}{4\times 3\times 2\times 1}$( Substituting)

$=\frac{5}{24}$

The fourth term of the series $\sum _{n=1}^{\infty }\frac{n+1}{n!}$is $\frac{5}{24}$.

The fifth term of the series$\sum _{n=1}^{\infty }\frac{n+1}{n!}$is obtained by substituting $n=5$in$\frac{n+1}{n!}$. Therefore, the value at $n=5$ is:

$\frac{5+1}{5!}=\frac{6}{5\times 4\times 3\times 2\times 1}$ (Substituting)

$=\frac{1}{20}$

The fifth term of the series $\sum _{n=1}^{\infty }\frac{n+1}{n!}$is $\frac{1}{20}$.