# Thread: calc 3..3 questions left NEED HELP URGENTLY

1. ## 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.

2. 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 $\displaystyle \lim_{n\rightarrow\infty}\frac{3^{2n}}{2^{3n+1}}=\ frac{9}{8}.$
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, $\displaystyle \frac{1}{\sqrt{n}+\sqrt{n-1}}\rightarrow\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1} {2\sqrt{n}}$. This is known to diverge by the integral test

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

Hope that helps!

3. 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 $\displaystyle \lim_{n\rightarrow\infty}\frac{3^{2n}}{2^{3n+1}}=\ frac{9}{8}.$
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, $\displaystyle \frac{1}{\sqrt{n}+\sqrt{n-1}}\rightarrow\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1} {2\sqrt{n}}$. This is known to diverge by the integral test

l) Again, this sequence tends to the limit $\displaystyle 1$ 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 ....... $\displaystyle \frac{1}{\sqrt{n}+\sqrt{n-1}}> \frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}}$ therefore ......

4. 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 ....... $\displaystyle \frac{1}{\sqrt{n}+\sqrt{n-1}}> \frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}}$ 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

5. He4llo, aloufy!

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

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

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

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

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

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

6. For i:

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

Rewrite as:

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

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

$\displaystyle 2\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)+\frac{1}{3}\left(\frac{\frac{3 }{4}}{1-\frac{3}{4}}\right)=3$

7. 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 $\displaystyle \frac{1}{3\cdot{2^0}}+\frac{1}{3\cdot{2^1}}+\frac{ 1}{3\cdot{2^2}}+\frac{1}{3\cdot{2^3}}...+\frac{1}{ 3\cdot{2^n}}$

So the series is really

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

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

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

$\displaystyle \therefore{a_{n+1}<a_n}\forall{n}\in\mathbb{N}$

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

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

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

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 $\displaystyle \frac{2^n}{3^n}$

e) Utilizing the root test we see that

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

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

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

therefore obviously divergent

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

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

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

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

therefore divergent

EDIT: I didnt see those last four

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

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

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

therefore this series is divergent

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

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

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

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

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

therefore divergent