Chapter 10.5, Problem 54E

(a)

To determine

Show whether the infinite series converges.

Answer to Problem 54E

The infinite series diverges by the Limit Comparison Test.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{k=1}^{\infty }\frac{1}{\sqrt{2k+7}}}$

To use with Limit Comparison Test, find a series $a_k$ :

  $a_k=\frac{1}{\sqrt{2k}}=\frac{1}{\sqrt{2}}\cdot \frac{1}{k^{1/2}}$

Thus,

  $a_k$ diverges by the P − Series Test $\left(p=1/2<1\right)$ .

Then

  $b_k=\frac{1}{\sqrt{2k+7}}$

Apply the limit comparison test:

  $\underset{k\to \infty }{\lim }\frac{\frac{1}{\sqrt{2k}}}{\frac{1}{\sqrt{2k+7}}}=\underset{k\to \infty }{\lim }\sqrt{\frac{2k+7}{2k}}=\sqrt{1}=1$

Since the limit is a constant,

  ${\displaystyle {\sum }_{k=1}^{\infty }{a}_k}$ and ${\displaystyle {\sum }_{k=1}^{\infty }{b}_k}$ either both converge or both diverge.

But

We have

Series ${\displaystyle \sum a_k}$ diverges.

Then

  ${\displaystyle \sum b_k}$ must also diverge.

Therefore,

By Limit Comparison Test, the infinite series diverges.

(b)

To determine

Explain whether the infinite series converges.

Answer to Problem 54E

The infinite series diverges.

Explanation of Solution

Given information:

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

We have

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

Use nth − term test:

Since k approaches infinity, we will take the limit of the original series.

  $\underset{k\to \infty }{\lim }{\left(1+\frac{1}{k}\right)}^k$

Then simplify:

  $\underset{k\to \infty }{\lim }{\left(1+\frac{1}{k}\right)}^k={\left(1+\frac{1}{\infty }\right)}^{\infty }={\left(1+0\right)}^{\infty }=1^{\infty }=e$

The series does not approach zero.

Therefore,

The infinite series diverges.

(c)

To determine

Discuss whether the infinite series converges.

Answer to Problem 54E

The infinite series absolutely converges by the Direct Comparison Test.

Explanation of Solution

Given information:

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

To use with Limit Comparison Test, find a series $c_n$ :

  $c_n=\frac{1}{k^2}$

Thus,

  ${\displaystyle \sum c_n}$ converges by the P − Series Test.

Since $\cos k$ can be either positive or negative,

Consider the absolute value of the terms:

  $\left|a_k\right|=\left|\frac{\cos k}{k^2+\sqrt{k}}\right|$

Now,

Perform the Direct Comparison Test:

Compare the numerators and denominators.

Smaller numerator divided by larger denominators equals the smaller number.

  $\frac{\left|\cos k\right|<1}{\left|k^2+\sqrt{k}\right|>k^2}$

Since

  $\left|a_k\right|<c_k$

Then

  ${\displaystyle \sum \left|a_k\right|}$ converges.

According to the definition of Absolute Convergence,

  ${\displaystyle \sum \left|a_k\right|}$ absolutely converges by the Direct Comparison Test.

(d)

To determine

Decide whether the infinite series converges.

Answer to Problem 54E

The infinite series diverges by the Integral Test.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{k=3}^{\infty }\frac{18}{k\left(\ln k\right)}}$

We have

  ${\displaystyle \sum _{k=3}^{\infty }\frac{18}{k\left(\ln k\right)}}$

Use the Integral Test:

  ${\displaystyle {\int }_3^{\infty }\frac{18}{x\left(\ln x\right)}dx}$

Take the anti-derivative:

  $\underset{b\to \infty }{\lim }{\left[18\ln \left|\ln x\right|\right]}_3^b$

Solve the integral:

  $\underset{b\to \infty }{\lim }{\left[18\ln \left|\ln x\right|\right]}_3^b=18\ln \left|\ln b\right|-18\ln \left|\ln 3\right|=18\ln \left|\ln \infty \right|-0=\infty -0=\infty$

The integral comes to infinity.

Therefore,

The series diverges by the Integral Test.