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}.

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This entry was posted on Wednesday, November 25th, 2009 at 2:45 pm and is filed under 175: Calculus II. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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