Q. 35 - Calculus

Question

Provide the first five terms of the sequence of partial sums for the given series.

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

Step-by-Step Solution

Answer

The first five terms of the partial sums for the series $\sum _{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}$ are, $\left\{1,\frac{1}{2},\frac{13}{24},\frac{389}{720},\frac{21785}{40320}\right\}$.

Step 1 . Given Information

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

Step 2 . Find the First term

Substitute $n=0$ in the given series.

$\begin{array}{l}\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^0}{(2\times 0)!}\\ =\frac{1}{0!}\\ =1\end{array}$

The first term is, $1.$

Step 3 . Find the Second term

Substitute $n=1$ in the given series.

$\begin{array}{l}\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^1}{(2\times 1)!}\\ =\frac{-1}{2!}\\ =-\frac{1}{2}\end{array}$

The second term is, $-\frac{1}{2}$.

Step 4 . Find the third term.

Substitute $n=2$ in the given series.

$$\begin{gathered} \frac{{\left(-1\right)}^n}{\left(2n\right)!}=\frac{{\left(-1\right)}^2}{\left(2\left(2\right)\right)!} \\ =\frac{\left(1\right)}{\left(4\right)!} \\ =\frac{1}{24} \end{gathered}$$

The third term is,$\frac{1}{24}$.

Step 5 . Find the fourth term.

Substitute $n=3$ in the given series.

$$\begin{gathered} \frac{{\left(-1\right)}^n}{\left(2n\right)!}=\frac{{\left(-1\right)}^3}{\left(2\left(3\right)\right)!} \\ =-\frac{1}{720} \end{gathered}$$

The fourth term is, $-\frac{1}{720}$.

Step 6 . Find the fifth term.

Substitute $n=4$ in the given series.

$$\begin{gathered} \frac{{\left(-1\right)}^n}{\left(2n\right)!}=\frac{{\left(-1\right)}^4}{\left(2\left(4\right)\right)!} \\ =\frac{1}{40,320} \end{gathered}$$

The fifth term is, $\frac{1}{40,320}$.

Step 7 . The first five terms in the sequence of partial sums are,

$$\begin{gathered} S_1=1 \\ {S}_2=S_1+a_2 \\ =1-\frac{1}{2} \\ =\frac{1}{2} \\ {S}_3=S_2+a_3 \\ =\frac{1}{2}+\frac{1}{24} \\ =\frac{13}{24} \\ {S}_4=S_3+a_4 \\ =\frac{13}{24}-\frac{1}{720} \\ =\frac{389}{720} \\ {S}_5=S_4+a_5 \\ =\frac{389}{720}+\frac{1}{40,320} \\ =\frac{21785}{40,320} \end{gathered}$$