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.

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

Short answer

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

Step-by-step solution

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∞ ak and ∑k=1∞ bk 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">∑k=1∞bkdiverges, then$\sum _{k=1}^{\infty }{a}_k$diverges.

Step 3. The terms of the series ∑k=1∞ 1k-π2 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 localid="1649179701987" $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^2}$is convergent by $p-$series test.

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

Hence., the given series is convergent.