Q. 22 - 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.

$\sum _{k=1}^{\infty } \frac{3}{{\left(1+2k\right)}^2}$.

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty } \frac{3}{{\left(1+2k\right)}^2}$ is convergent.

Step 1 . Given information

$\sum _{k=1}^{\infty } \frac{3}{{\left(1+2k\right)}^2}$.

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

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

Step 3 . Find 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role="math" localid="1649149649494" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/df813edc-6697-4bb0-95da-7fa873bb36cd.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220405%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220405T095114Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=ac3d505e4383bef0aa6a8d908e33435917815b29b9a27e60cade3ecc7a756d9b" ∑ k = 1 ∞   b k for the given series.

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

Next find $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}$ for the given series.

$$\begin{gathered} \underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\underset{k\to \infty }{\lim } \frac{{\displaystyle \frac{3}{{\left(1+2k\right)}^2}}}{{\displaystyle \frac{1}{k^2}}} \\ =\underset{k\to \infty }{\lim } \frac{3k^2}{{\left(1+2k\right)}^2} \\ =\underset{k\to \infty }{\lim } \frac{3k^2}{k^2{\left({\displaystyle \frac{1}{k}}+2\right)}^2} \\ =\underset{k\to \infty }{\lim } \frac{3}{{\left({\displaystyle \frac{1}{k}+2}\right)}^2} \\ =\frac{3}{4} \end{gathered}$$

Step 4 . From the obtained values,

The value of $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\frac{3}{4}$ which is a non zero finite number.

The value of $\sum _{k=1}^{\infty } b_k=\sum _{k=1}^{\infty } \frac{1}{k^2}$ is convergent by the $p-$series test.

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

Hence, the given series is convergent.