Q. 21 - Calculus

Question

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

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

Step-by-Step Solution

Answer

Ans: The five terms of the series are $\frac{1}{3},\frac{1}{6},\frac{1}{11},\frac{1}{18},\frac{1}{27}$

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:

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

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

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:

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

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

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:

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

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

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:

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

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:

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

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