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

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

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

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:

$$\begin{gathered} \frac{1}{k^2+2}=\frac{1}{1^2+2} \\ =\frac{1}{3} \end{gathered}$$(Substituting)

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

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:

$$\begin{gathered} \frac{1}{k^2+2}=\frac{1}{2^2+2} \\ =\frac{1}{6} \end{gathered}$$(Substituting)

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

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:

$$\begin{gathered} \frac{1}{k^2+2}=\frac{1}{3^2+2} \\ =\frac{1}{11} \end{gathered}$$(Substituting)

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

The fourthterm 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:

$$\begin{gathered} \frac{1}{k^2+2}=\frac{1}{4^2+2} \\ =\frac{1}{18} \end{gathered}$$(Substituting)

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

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:

$$\begin{gathered} \frac{1}{k^2+2}=\frac{1}{5^2+2} \\ =\frac{1}{27} \end{gathered}$$(Substituting)

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