Math 126B

Worksheet #2

Spring 2003

Names:

(0) Once again, let's spend a moment to get to know each other.

Which member of the group has the largest number of brother and sisters?

Which member has traveled to the most number of countries?

This work sheet focuses on the Comparison Test (Section 11.4). One of the challenging stages of

the Comparison Test for new students is determining which series is the comparison. The following

is an attempt to address this issue.

(1) Suppose you have the series ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{a}}_{\mathit{n}}$. To determine whether the series converges, you want to

compare it to ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{b}}_{\mathit{n}}$.

(a) If your intuition says that ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{a}}_{\mathit{n}}$ converges, then you need to Iind ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{b}}_{\mathit{n}}$ where ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{b}}_{\mathit{n}}$

converges and each term of is larger than each term of

(b) If your intuition says that ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{a}}_{\mathit{n}}$ diverges, then you need to Iind ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{b}}_{\mathit{n}}$ where ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{\mathit{b}}_{\mathit{n}}$

diverges and each term of is larger than each term of

(2) The key point to the Comparison Test is to have some ready made functions that you have

information about. Two series that are commonly used for comparison are the geometric series

and the $\mathit{p}$-series. In your own words, indicate when a geometric series and $\mathit{p}$ series converges

and when it diverges. Then, give a basic example that converges and one that diverges for

each series.

Math 126B

Worksheet #2

Spring 2003

(a) Geometric series:

Converge: Diverge:

(b) p-series:

Converge: Diverge:

(3) Okay, now let's look at a somewhat challenging problem. We want to show that ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{2}}^{\mathrm{\infty }}\frac{\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}}{{\mathit{n}}^{\mathrm{2}}}$

converges.

(a) First off, intuition suggests that we want to compare our series to a convergent p-series.

With regards to the placement of the variable, $\mathit{n}$, explain why a comparison to a p-series

is better than a comparison to a geometric series.

(b) Next, it would have been great if we could have used $\frac{\mathrm{1}}{{\mathit{n}}^{\mathrm{2}}}$ because $\mathit{p}\mathrm{=}\mathrm{2}$. But, we can't

because even a wanna-be rock star knows that for $\mathit{n}\underset{\mathrm{‾}}{\mathrm{>}}\mathrm{1}\mathrm{0}$, $\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}\underset{\mathrm{‾}}{\mathrm{>}}\mathrm{1}$. The inequalities

are working the wrong way. So, $\mathrm{w}\mathrm{e}\mathrm{\text{'}}\mathrm{v}\mathrm{e}$ got to come up with a function that's greater than

$\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}$, but produces a power larger than 1 in the denominator. How about, $\sqrt{\mathit{n}}\mathrm{?}\mathrm{!}$

Recall the induction process covered in Worksheet #1. Well, $\mathrm{w}\mathrm{e}\mathrm{\text{'}}\mathrm{1}\mathrm{1}$ use a variation of it

to show that $\sqrt{\mathit{n}}\underset{\mathrm{‾}}{\mathrm{>}}\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}$ for $\mathit{n}$ $\mathrm{>}\mathit{N}$.

Math 126B

Worksheet #2

Spring 2003

$\mathrm{i}$. Select a number, ${\mathit{n}}^{\mathrm{*}}$, and show that $\sqrt{{\mathit{n}}^{\mathrm{*}}}\underset{\mathrm{‾}}{\mathrm{>}}\ln \mathrm{\left(}{\mathit{n}}^{\mathrm{*}}\mathrm{\right)}$. (This is like showing the initial

value works.)

$\mathrm{i}\mathrm{i}$. Rather than showing that the $\mathit{n}$ $\mathrm{+}\mathrm{1}$ case works, we are going to use the derivatives.

Determine whether the derivative of $\sqrt{\mathit{n}}$ is greater than the derivative of $\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}$ for

$\mathit{n}\underset{\mathrm{‾}}{\mathrm{>}}{\mathit{n}}^{\mathrm{*}}$

$\mathrm{i}\mathrm{i}\mathrm{i}$. Explain how the two previous results show that $\sqrt{\mathit{n}}\underset{\mathrm{‾}}{\mathrm{>}}\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}$ for $\mathit{n}$ $\mathrm{>}{\mathit{n}}^{\mathrm{*}}$

Now, we have all of the pieces. We have $\frac{\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}}{{\mathit{n}}^{\mathrm{2}}}\underset{\mathrm{‾}}{\mathrm{<}}\frac{\sqrt{\mathit{n}}}{{\mathit{n}}^{\mathrm{2}}}$ for $\mathit{n}\underset{\mathrm{‾}}{\mathrm{>}}{\mathit{n}}^{\mathrm{*}}$ Also, we know that

${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{2}}^{\mathrm{\infty }}\frac{\sqrt{\mathit{n}}}{{\mathit{n}}^{\mathrm{2}}}\mathrm{=}{\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{2}}^{\mathrm{\infty }}\frac{\mathrm{1}}{{\mathit{n}}^{\mathrm{3}\mathrm{/}\mathrm{2}}}$ and it converges because $\mathit{p}\mathrm{=}\mathrm{3}\mathrm{/}\mathrm{2}\mathrm{>}\mathrm{1}$. Thus, by the Comparison Test,

${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{2}}^{\mathrm{\infty }}\frac{\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}}{{\mathit{n}}^{\mathrm{2}}}$ converges as well.

Math 126B

Worksheet #2

Spring 2003

(4) Using the process above, determine a series that you can use to show that ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{2}}^{\mathrm{\infty }}\frac{\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}}{{\mathit{n}}^{\mathrm{3}\mathrm{/}\mathrm{2}}}$ converges.

(5) Does ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{2}}^{\mathrm{\infty }}\frac{\ln \mathrm{\left(}\mathit{n}\mathrm{\right)}}{\mathit{n}}$ converge or diverge?

(6) If there is time, determine whether the following series converge or diverge?

(a) ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}\frac{\mathrm{1}}{\sqrt{{\mathit{n}}^{\mathrm{3}}\mathrm{+}\mathrm{4}\mathit{n}\mathrm{+}\mathrm{1}}}$

(b) ${\displaystyle \sum }_{\mathit{n}\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}\frac{\mathrm{1}\mathrm{+}{\mathrm{2}}^{\mathit{n}}}{\mathrm{1}\mathrm{+}{\mathrm{3}}^{\mathit{n}}}$

(c) $\frac{\mathrm{1}}{\mathrm{2}}\mathrm{+}\frac{\mathrm{1}}{{\mathrm{2}}^{\mathrm{3}}}\mathrm{+}\frac{\mathrm{1}}{\mathrm{2}\mathrm{ċ}\mathrm{3}\mathrm{2}}\mathrm{+}\frac{\mathrm{1}}{{\mathrm{4}}^{\mathrm{3}}}\mathrm{+}\frac{\mathrm{1}}{\mathrm{2}\mathrm{ċ}\mathrm{5}\mathrm{2}}\mathrm{+}\frac{\mathrm{1}}{{\mathrm{6}}^{\mathrm{3}}}\mathrm{+}\mathrm{\text{...}}$