## 175 – Quiz 7

Happy Thanksgiving!

Here is quiz 7.

Problem 1 asks to write out the first few terms of the following series to show how the series starts, and then find the sum of the series:

$\displaystyle \sum_{n=0}^\infty(-1)^n\frac5{4^n}.$

Let ${S_n}$ denote the sum of the first ${n}$ terms of the series. Then, for example,

$\displaystyle S_0=\frac{(-1)^0 5}{4^0}=5,$

$\displaystyle S_1=\frac{(-1)^0 5}{4^0}+\frac{(-1)^1 5}{4^1}=5-\frac54=\frac{15}4,$

$\displaystyle S_2=\frac{(-1)^0 5}{4^0}+\frac{(-1)^1 5}{4^1}+\frac{(-1)^2 5}{4^2}=5-\frac54+\frac5{16}=\frac{65}{16},$

and

$\displaystyle S_3=\frac{(-1)^0 5}{4^0}+\frac{(-1)^1 5}{4^1}+\frac{(-1)^2 5}{4^2}+\frac{(-1)^3 5}{4^3}$ $\displaystyle =5-\frac54+\frac5{16}-\frac5{64}=\frac{255}{64}.$

The series ${\displaystyle \sum_{n=0}^\infty(-1)^n\frac5{4^n}}$ is geometric, i.e., it has the form ${\sum_{n=0}^\infty ar^n.}$ In this case, ${a=5}$ and ${r=-1/4.}$ Since ${|r|=1/4<1,}$ the series converges, and adds up to

$\displaystyle \frac{5}{1-\left(-\frac14\right)}=4.$

Problem 2 asks to explain why the following series converges or diverges, and if it converges, find its sum:

$\displaystyle \sum_{n=1}^\infty\ln\left(\frac n{2n+1}\right).$

The series diverges. Perhaps the easiest way of checking this is by using the ${n}$-th term test: If a series ${\sum a_n}$ converges, then ${\lim_n a_n=0.}$ So, if ${\lim_n a_n\ne0,}$ then ${\sum a_n}$ diverges. In this case,

$\displaystyle a_n=\ln\left(\frac n{2n+1}\right)=\ln\left(\frac 1{2+\frac1n}\right)\rightarrow\ln\left(\frac12\right)\ne0,$

so the series diverges.

Problem 3 asks to express the following number as the ratio of two integers:

$\displaystyle 1.\overline{414}=1.414\,414\,414\dots$

To do this, simply note that

$\displaystyle 1.\overline{414}=1+\frac{414}{1000}+\frac{414}{1000^2}+\frac{414}{1000^3}+\dots,$

and that (except for the first term) this is a geometric series,

$\displaystyle \frac{414}{1000}+\frac{414}{1000^2}+\frac{414}{1000^3}+\dots=\sum_{n=0}^\infty ar^n,$

with ${\displaystyle a=\frac{414}{1000}}$ and ${\displaystyle r=\frac1{1000}.}$ Hence, the series adds up to

$\displaystyle 1+\frac{\frac{414}{1000}}{1-\frac1{1000}}=1+\frac{414}{999}=\frac{1413}{999}=\frac{157}{111}.$

Typeset using LaTeX2WP. Here is a printable version of this post.