Question

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

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

Short answer

The first five terms of the partial sums for the series$\sum _{n=1}^{\mathrm{\infty }} (-1{)}^n\frac{n^2}{n!}$is,$\left\{-1,1,-\frac{1}{2},-\frac{1}{6},-\frac{9}{24}\right\}$.

Step-by-step solution

Step 1. Given Information

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

Step 2. Find the first term.

Substitute $n=1$in the given series.

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

The first term is,$-1$.

Step 3. Find the second term.

Substitute $n=2$in the given series.

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

The second term is,$2$.

Step 4. Find the third term.

Substitute $n=3$in the given series.

$$\begin{gathered} (-1{)}^n\frac{n^2}{n!}= (-1{)}^3\frac{{\left(3\right)}^2}{3!} \\ =\left(-1\right)\left(\frac{9}{6}\right) \\ =-\frac{3}{2} \end{gathered}$$

The third term is,$-\frac{3}{2}$

Step 5. Find the fourth term.

Substitute $n=4$in the given series.

$$\begin{gathered} (-1{)}^n\frac{n^2}{n!}= (-1{)}^4\frac{{\left(4\right)}^2}{4!} \\ =1.\left(\frac{16}{24}\right) \\ =\frac{2}{3} \end{gathered}$$

The fourth term is,$\frac{2}{3}$.

Step 6. Find the fifth term.

Substitute $n=5$in the given series.

$$\begin{gathered} (-1{)}^n\frac{n^2}{n!}= (-1{)}^5\frac{{\left(5\right)}^2}{5!} \\ =\left(-1\right)\frac{25}{120} \\ =-\frac{5}{24} \end{gathered}$$

The fifth term is,$-\frac{5}{24}$.

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

$$\begin{gathered} S_1=-1 \\ {S}_2=S_1+a_2 \\ =-1+2 \\ =1 \\ {S}_3=S_2+a_3 \\ =1-\frac{3}{2} \\ =-\frac{1}{2} \\ {S}_4=S_3+a_4 \\ =-\frac{1}{2}+\frac{2}{3} \\ =-\frac{1}{6} \\ {S}_5=S_4+a_5 \\ =-\frac{1}{6}-\frac{5}{24} \\ =-\frac{9}{24} \end{gathered}$$