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.

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

Step 2. The comparison test states that for ∑k=1∞ ak and ∑k=1∞ bk 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∞ k1+k2 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 limk→∞ akbk 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 role="math" localid="1649153106872" $\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.