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

$\sum _{j=0}^{\infty }{\left(-1\right)}^{j+1}$

$\sum _{j=0}^{\infty }{\left(-1\right)}^{j+1}$

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

$(-1{)}^{0+1}=(-1{)}^1$(Substituting)

$=-1$

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

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

$(-1{)}^{1+1}=(-1{)}^2$(Substituting)

=1

The second term of the series $\sum _{j=0}^{\infty }(-1{)}^{j+1}$is 1 .

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

$(-1{)}^{2+1}=(-1{)}^3$(Substituting)

=-1

The third term of the series $\sum _{j=0}^{\infty }(-1{)}^{j+1}$is -1.

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

$(-1{)}^{3+1}=(-1{)}^4$( Substituting )

=1

The fourth term of the series $\sum _{j=0}^{\infty }(-1{)}^{j+1}$is 1 .

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

$(-1{)}^{4+1}=(-1{)}^5$(Substituting)

=-1

The fifth term of the series $\sum _{j=0}^{\infty }(-1{)}^{j+1}$is -1.