Question

In Exercises 29–36 provide the first five terms of the sequence of partial sums for the given series. You may find it useful to refer to your answers to Exercises 21–28.

$\sum _{k=1}^{\infty }\frac{1}{k^2+2}$

Short answer

Ans: The first five terms of partial sums for the given series$\left\{\frac{1}{3},\frac{1}{2},\frac{13}{22},\frac{64}{99},\frac{203}{297}\right\}$

Step-by-step solution

Step 1. Given information: 

$\sum _{k=1}^{\infty }\frac{1}{k^2+2}$

Step 2. Finding the first term of the series:

The first term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is obtained by substituting $k=1$in $\frac{1}{k^2+2}$. Therefore, the value at $k=1$ is:

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

$=\frac{1}{3}$

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

Step 3. Finding the second term of the series:

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

$\frac{1}{(2{)}^2+2}=\frac{1}{4+2}$(Substituting)

$=\frac{1}{6}$

The second term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is $\frac{1}{6}$.

Step 4. Finding the third term of the series:

The third term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is obtained by substituting $k=3$in $\frac{1}{k^2+2}$. Therefore, the value at $k=3$ is:

$\frac{1}{(3{)}^2+2}=\frac{1}{9+2}$(Substituting)

$=\frac{1}{11}$

The third term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is $\frac{1}{11}$.

Step 5. Finding the fourth term of the series:

The fourth term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is obtained by substituting $k=4$in $\frac{1}{k^2+2}$. Therefore, the value at $k=4$ is:

$\frac{1}{(4{)}^2+2}=\frac{1}{16+2}$ (Substituting)

$=\frac{1}{18}$

The fourth term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is $\frac{1}{18}$.

Step 6. Finding the fifth term of the series:

The fifth term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is obtained by substituting $k=5$in $\frac{1}{k^2+2}$. Therefore, the value at $k=5$ is:

$\frac{1}{(5{)}^2+2}=\frac{1}{25+2}$ (Substituting)

$=\frac{1}{27}$

The fifth term of the series $\sum _{k=1}^{\infty }\frac{1}{k^2+2}$is $\frac{1}{27}$.

Step 7.  The first five terms of the sequence of partial sums :

The first five terms in the sequence of partial sums are:

$$\begin{gathered} S_1=\frac{1}{3} \\ {S}_2=S_1+a_2 \\ =\frac{1}{3}+\frac{1}{6}\text{(Substitution)} \\ =\frac{1}{2} \\ {S}_3=S_2+a_3 \\ =\frac{1}{2}+\frac{1}{11}\text{(Substitution)} \\ =\frac{13}{22} \end{gathered}$$

Step 2.

$$\begin{gathered} S_4=S_3+a_4 \\ =\frac{13}{22}+\frac{1}{18}\text{(Substitution)} \\ =\frac{117+11}{198} \\ =\frac{64}{99} \\ {S}_5=S_4+a_5 \\ =\frac{1}{27}+\frac{64}{99}\text{(Substitution)} \\ =\frac{203}{297} \end{gathered}$$

Therefore, first five terms of partial sums for the given series is $\left\{\frac{1}{3},\frac{1}{2},\frac{13}{22},\frac{64}{99},\frac{203}{297}\right\}$.