# calc 3..3 questions left NEED HELP URGENTLY

• May 29th 2008, 12:14 AM
aloufy
calc 3..3 questions left NEED HELP URGENTLY
Helllo people. I have been doing this assignment and got stuck on 3 problems that I cant seem to figure out. I know its late since the assignment is due well later on today but any help will be greatly appreciated. I uploaded the section I need help with. The questions i couldnt figure out were f,k and l. For some reason my mind just hit a blank.

• May 29th 2008, 12:44 AM
notyeteuler
Hey there!

I'm not sure if you have learned the root test for analysis yet and so I will assume that you haven't

f) Notice that $\lim_{n\rightarrow\infty}\frac{3^{2n}}{2^{3n+1}}=\ frac{9}{8}.$
This cannot possibly converge as the terms do not tend to zero as n tends to infinity. The root test would give an equal result

k) Note that, as n becomes large, $\frac{1}{\sqrt{n}+\sqrt{n-1}}\rightarrow\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1} {2\sqrt{n}}$. This is known to diverge by the integral test

l) Again, this sequence tends to the limit $1$ and so, as this is not zero, the series cannot converge

Hope that helps!
• May 29th 2008, 01:02 AM
mr fantastic
Quote:

Originally Posted by notyeteuler
Hey there!

I'm not sure if you have learned the root test for analysis yet and so I will assume that you haven't

f) Notice that $\lim_{n\rightarrow\infty}\frac{3^{2n}}{2^{3n+1}}=\ frac{9}{8}.$
This cannot possibly converge as the terms do not tend to zero as n tends to infinity. The root test would give an equal result

k) Note that, as n becomes large, $\frac{1}{\sqrt{n}+\sqrt{n-1}}\rightarrow\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1} {2\sqrt{n}}$. This is known to diverge by the integral test

l) Again, this sequence tends to the limit $1$ and so, as this is not zero, the series cannot converge

Hope that helps!

The limit in f) is actually +oo, but this doesn't change the conclusion.

For k) you'll need to set up the appropriate inequality in order to run the argument ....... $\frac{1}{\sqrt{n}+\sqrt{n-1}}> \frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}}$ therefore ......
• May 29th 2008, 10:23 AM
aloufy
Quote:

Originally Posted by mr fantastic
The limit in f) is actually +oo, but this doesn't change the conclusion.

For k) you'll need to set up the appropriate inequality in order to run the argument ....... $\frac{1}{\sqrt{n}+\sqrt{n-1}}> \frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}}$ therefore ......

thanks for the help. I used the alternating series test and the rati test to prove the problems. I came up with the same answer. Thanks alot
• May 29th 2008, 11:54 AM
Soroban
He4llo, aloufy!

Quote:

$(f)\;\;\sum^{\infty}_{n=1}(-1)^{n-1}\frac{3^{2n}}{2^{3n+1}}$
Ratio test: . $R \:=\:\left|\frac{a_{n+1}}{a_n}\right| \:=\:\frac{3^{2n+2}}{2^{3n+4}}\cdot\frac{2^{3n+1}} {3^{2n}} \;=\;\frac{3^2}{2^3} \:=\:\frac{9}{8}$

$\text{Since }\lim_{n\to\infty}|R| \:\geq \:1\text{, the series diverges.}$

Quote:

$(l)\;\;\sum^{\infty}_{n=1}\frac{4^n}{4^n + 5}$
The $n^{th}$ term is: . $a_n \:=\:\frac{4^n}{4^n + 5}$

Divide top and bottom by $4^n\!:\;\;a_n \:=\:\frac{1}{1 + \frac{5}{4^n}}$

Then: . $\lim_{n\to\infty} a_n \;=\;\lim_{n\to\infty}\frac{1}{1 + \frac{5}{4^n}} \;=\;1$

$\text{Since }\lim_{n\to\infty} a_n \:\neq \:0\text{, the series diverges.}$

• May 29th 2008, 12:04 PM
galactus
For i:

We could use the ratio test, but we can do one better by finding what it actually converges to.

Rewrite as:

$2\sum_{n=1}^{\infty}\frac{1}{2^{n}}+\frac{1}{3}\su m_{n=1}^{\infty}\left(\frac{3}{4}\right)^{n}$

We can see both of these are geometric series. We can use the formula.

$2\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)+\frac{1}{3}\left(\frac{\frac{3 }{4}}{1-\frac{3}{4}}\right)=3$
• May 29th 2008, 08:33 PM
Mathstud28
Quote:

Originally Posted by aloufy
Helllo people. I have been doing this assignment and got stuck on 3 problems that I cant seem to figure out. I know its late since the assignment is due well later on today but any help will be greatly appreciated. I uploaded the section I need help with. The questions i couldnt figure out were f,k and l. For some reason my mind just hit a blank.

I will, show a couple of ways of doing some of these

a) this series can be rewritten as $\frac{1}{3\cdot{2^0}}+\frac{1}{3\cdot{2^1}}+\frac{ 1}{3\cdot{2^2}}+\frac{1}{3\cdot{2^3}}...+\frac{1}{ 3\cdot{2^n}}$

So the series is really

$\sum_{n=0}^{\infty}\frac{1}{3\cdot{2^n}}$

This one can really be done with almost any of the tests in specific, root,ratio, limit comparison test, direct comparison test, I will opt for the LCT(limit comparison test)

Since $\lim_{n\to\infty}\frac{\frac{1}{3\cdot{2^n}}}{\fra c{1}{2^n}}=\frac{1}{3}$ the series share convergence/divergence, and since $\frac{1}{2^n}$ is a convergent geometric series both are convergent

c)Since we can see that $2^{n+1}>2^n, \forall{n}\in\mathbb{N}$
we can see that
$\frac{1}{2^{n+1}}<\frac{1}{2^n}$

$\therefore{a_{n+1}

and since $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{2^{ n+1}}+0$ this series is conditionally convergent at least,

Alternately seeing that $\bigg|\frac{(-1)^n}{2^{n+1}}\bigg|=\frac{1}{2^{n+1}}$ which is a convergent geo series that this series is absolutely convergent, therefore conditionally convergent

d) Rewriting this $\sum_{n=0}^{\infty}\frac{2^{n+1}}{3^{n-3}}$ as $\sum_{n=0}^{\infty}\frac{2\cdot{2^n}}{\frac{1}{27} \cdot{3^n}}=54\sum_{n=0}^{\infty}\frac{2^n}{3^n}$

we can easily see this is a convergent geo series, you could have also used the ratio or root test, as well as the limit compariosn test with $\frac{2^n}{3^n}$

e) Utilizing the root test we see that

$\lim_{n\to\infty}\bigg|\frac{1}{3^{\frac{n}{2}}}\b igg|^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{3^{\f rac{1}{2}}}=\frac{1}{\sqrt{3}}<1$

therefore convergent, also you could use the ratio test

f) This should be an automatic recognition of a series divergent by the n-th term test

$\lim_{n\to\infty}\frac{3^{2n}}{2^{3n+1}}=\lim_{n\t o\infty}\frac{1}{2}\frac{9^n}{8^n}=\infty$

therefore obviously divergent

g) This is probably most easily done by either the Direct Comparion test or LCT with $\frac{1}{4n^2}$

$\lim_{n\to\infty}\frac{\frac{4}{4n^2-1}}{\frac{1}{4n^2}}=4$

therefore convergent

but you can also use the integral test, ratio test, root test

h)THis once again should be a recogintion of nth term test

$\lim_{n\to\infty}\frac{3n^2}{(3n-1)^2}=\lim_{n\to\infty}\frac{3n^2}{9n^2-6n+1}=\frac{1}{3}\ne{0}$

therefore divergent

EDIT: I didnt see those last four

i) NOting that
$\sum_{n=0}^{\infty}\frac{2^{n+1}+3^{n-1}}{4^n}=2\sum_{n=0}^{\infty}\frac{2^n}{4^n}+\frac {1}{3}\sum_{n=0}^{\infty}\frac{3^n}{4^n}$
so since this series is the sum of two convergent geo series it itself is convergent

j) Seeing that
$\sum_{n=0}^{\infty}\ln(4n+1)-\ln(2n+1)=\sum_{n=0}^{\infty}\ln\bigg(\frac{4n+1}{ 2n+1}\bigg)$
now n-th term test should be screaming

let
$L=\lim_{n\to\infty}\ln\bigg(\frac{4n+1}{2n+1}\bigg )\Rightarrow{e^L=\lim_{n\to\infty}\frac{4n+1}{2n+1 }=2}$
so
$e^L=2\Rightarrow{L=\ln(2)\ne{0}}$

therefore this series is divergent

k) YOu can either use LCT with $\frac{1}{2\sqrt{n}}$ or see that

$\exists{N}\backepsilon\forall{n>N}\frac{1}{\sqrt{n }+\sqrt{n+1}}>\frac{1}{2\sqrt{n}}$

and sicne the right side is a divergent p-series this series diverges by the Direct comparison tset

l) This should scream n-th term test as well

$\lim_{n\to\infty}\frac{4^n}{4^n+5}=1\ne{0}$

and just in case you are rusty on limits
$\lim_{n\to\infty}\frac{4^n}{4^n+5}=\lim_{n\to\inft y}\frac{4^n}{4^n+5}\cdot\frac{\frac{1}{4^n}}{\frac {1}{4^n}}=\lim_{n\to\infty}\frac{1}{1+\frac{5}{4^n }}=\frac{1}{1+0}\ne{0}$

therefore divergent