Q. 24 - Calculus

Question

In Exercises $21-30$ use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.

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

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty } \frac{\sqrt{k}}{1+k^2}$ is convergent.

Step 1 . Given information

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

Step 2 . The comparison test states that for ∑ k = 1 ∞   a k and ∑ k = 1 ∞   b k be two series with positive terms then,

1. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=L$, where $L$ is any positive real number then either both converge or both diverge.
2. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=0$ and $\sum _{k=1}^{\infty } b_k$ converges, then $\sum _{k=1}^{\infty } a_k$ converges.
3. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\infty$ and $\sum _{k=1}^{\infty } b_k$ diverges, then $\sum _{k=1}^{\infty } a_k$ diverges.

Step 3 . The terms of the series ∑ k = 1 ∞   k 1 + k 2 are positive.

Find $\sum _{k=1}^{\infty } b_k$ for the given series.

$$\begin{gathered} \sum _{k=1}^{\infty } b_k=\sum _{k=1}^{\infty } \frac{k^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{k^2} \\ =\sum _{k=1}^{\infty } \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

Step 4 . Next find lim k → ∞   a k b k for the given series.

$$\begin{gathered} \underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\underset{k\to \infty }{\lim } \frac{{\displaystyle \frac{\sqrt{k}}{1+k^2}}}{{\displaystyle \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ =\underset{k\to \infty }{\lim } \frac{k^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{k}}{1+k^2} \\ =\underset{k\to \infty }{\lim } \frac{k^2}{1+k^2} \\ =\underset{k\to \infty }{\lim } \frac{k^2}{k^2\left(1+{\displaystyle \frac{1}{k^2}}\right)} \\ =\underset{k\to \infty }{\lim } \frac{1}{\left(1+{\displaystyle \frac{1}{k^2}}\right)} \\ =1 \end{gathered}$$

Step 5 . From the obtained values,

The value of $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=1$ which is a finite non zero number.

The value of $\sum _{k=1}^{\infty } b_k= \sum _{k=1}^{\infty } \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ is convergent by using $p-$series test.

Therefore, $\sum _{k=1}^{\infty } a_k$ is convergent.

Hence, the given series is convergent.