Q. 28 - 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.

$\sum _{k=1}^{\infty } \frac{1}{{\left(k-\mathrm{\pi }\right)}^2}$.

Step-by-Step Solution

Answer

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

Step 1 . Given information

$\sum _{k=1}^{\infty } \frac{1}{{\left(k-\mathrm{\pi }\right)}^2}$.

Step 2 . The comparison test states that for ∑ k = 1 ∞   a k and ∑ k = 1 ∞   b k be two series with positive term 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 $$" width="9" style="max-width: none;" >$\sum _{k=1}^{\infty } b_k$ diverges, then $\sum _{k=1}^{\infty } a_k$ diverges.

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

Find $\sum _{k=1}^{\infty } b_k$ 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.

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

Step 4 . From the obtained values,

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

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

Therefore, the value of $\sum _{k=1}^{\infty } b_k$ convergent.

Hence., the given series is convergent.