91Ó°ÊÓ

$\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n}{\left(2n\right)!}$

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

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

=1(Because 0 !=1 )

The first term of the series $\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n}{\left(2n\right)!}$is 1 .

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

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

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

The second term of the series $\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n}{\left(2n\right)!}$is $-\frac{1}{2}$.

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

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

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

The third term of the series $\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n}{\left(2n\right)!}$is $\frac{1}{24}$.

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

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

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

The fourth term of the series $\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n}{\left(2n\right)!}$is $-\frac{1}{720}$.

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

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

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

The fifth term of the series$\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n}{\left(2n\right)!}$is $\frac{1}{40320}$.$\frac{(-1{)}^n}{\left(2n\right)!}$