Question

test the series for convergence or divergence. Use the comparison or limit comparison test

Answer 1

\[\sum_{k=1}^{\infty} \frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}}\]

Answer 2

help

Answer 3

you can try limit comparison

Answer 4

would it be an = the equation above and bn= \[\frac{ 3k^\frac{ 1 }{ 2 }}{ k^\frac{ 3 }{ 2 } }\]

Answer 5

?

Answer 6

yes I would compare it to sum 1/ k ^(3/2)

Answer 7

a direct comparison also works

Answer 8

Can you also simplify that last expression to \(\frac{3}{k}\) and compare to the series \(\frac{1}{k}\)?

Answer 9

$$
\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} \leq \frac{3\sqrt k + 2}{\sqrt{k^3}}

$$

Answer 10

and the right side can be split apart

Answer 11

i like the eyeball test

Answer 12

whats the eyeball test

Answer 13

the degree of the denominator needs to be larger than the degree of the numerator by more than one

Answer 14

is that why he has the +2 there?

Answer 15

for the bn not the an

Answer 16

the degree of the numerator is \[\frac{1}{2}\] and the degree of the denominator is \[\frac{3}{2}\] so no way

Answer 17

so what happens if the degree of the numerator is less than the denominator

Answer 18

$$ \large {
\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} \leq \frac{3\sqrt k + 2}{\sqrt{k^3}} = \frac{3}{\sqrt {k^2}} + \frac{2}{\sqrt{k^3}}
\\ = \frac{3}{k} + \frac{2}{k^{3/2}}
}

$$

Now we see that the direct comparison test fails , because the right side diverges as a series. (harmonic series).
We can use limit comparison test. and by limit comparison test it diverges.

Answer 19

it is easier if we simplify the original series

Answer 20

why are diving by \[\sqrt{k}\]

Answer 21

or we can try satellites solution. subtract the degree of the denominator minus the degree of the numerator.

Answer 22

cant it says to do it using comparison or limit comparison

Answer 23

if you want to use the limit comparison test.
this is it:

$$ \Large {
\lim_{k \to \infty} \frac{\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} }{\frac1k}


}
$$

Answer 24

okay and i cancel out the 1/k making it \[\frac{ 3+2 }{ \sqrt{k^2+3k+1} }\]

Answer 25

right?

Answer 26

if you want to use the limit comparison test.
this is it:

$$ \Large {
\lim_{k \to \infty} \frac{\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} }{\frac1k}
\\ \therefore \\
\lim_{k \to \infty} {\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac k 1

}
$$

Answer 27

so the numerator is 3k+2k = 5k right

Answer 28

i dont see how how you got that, the step before either

Answer 29

multiply the k times the sqrt(k) equals k and 3*k is 3k

Answer 30

no my bad its not that

Answer 31

$$


\Large {
\lim_{k \to \infty} \frac{\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} }{\frac1k}
\\ \therefore \\
\lim_{k \to \infty} {\frac{( 3\sqrt{k}+2) }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac k 1 \\
\lim_{k \to \infty} {\frac{ 3\sqrt{k}\cdot k +2k }{ \sqrt{k^3 + 3k^2 +1}} }



}





$$

Answer 32

its \[3k \sqrt{k}+2k\]

Answer 33

right :)

Answer 34

so after that do i just plug in infinity for k

Answer 35

that will give you infinity / infinity

Answer 36

but we can divide top and bottom by sqrt(k^3)

Answer 37

really?

Answer 38

$$
\Large {
\lim_{k \to \infty} \frac{\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} }{\frac1k}
\\ \therefore \\
\lim_{k \to \infty} {\frac{( 3\sqrt{k}+2) }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac k 1 \\
\lim_{k \to \infty} {\frac{ 3\sqrt{k}\cdot k +2k }{ \sqrt{k^3 + 3k^2 +1}} }
\\ \therefore \\
\lim_{k \to \infty} {\frac{ 3\sqrt{k}\cdot k +2k }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac{\frac{1}{\sqrt{k^3}}}{\frac{1}{\sqrt{k^3}}}
}

$$

Answer 39

also note that :
\( \Large 3\sqrt{k}\cdot k = 3~ k^{1/2}\cdot k = 3 k^{3/2} = 3 \sqrt{k^3} \)

Answer 40

oh yes i see

Answer 41

your dividing everything by the largest degree right?

Answer 42

correct

Answer 43

$$
\Large {
\lim_{k \to \infty} \frac{\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} }{\frac1k}
\\ \therefore \\
\lim_{k \to \infty} {\frac{( 3\sqrt{k}+2) }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac k 1 \\
\lim_{k \to \infty} {\frac{ 3\sqrt{k}\cdot k +2k }{ \sqrt{k^3 + 3k^2 +1}} }
\\ \therefore \\
\lim_{k \to \infty} {\frac{ 3\sqrt{\color{red}{k^3}} +2k }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac{\frac{1}{\sqrt{k^3}}}{\frac{1}{\sqrt{k^3}}}
\\ \therefore \\
\lim_{k \to \infty} {\frac{ 3\sqrt{ \frac{k^3}{k^3}} +2\cdot \frac{k}{\sqrt{k^3}} }{ \sqrt{\frac{k^3}{k^3} + 3\frac{k^2}{k^3}+\frac{1}{k^3}}} }
\\ \therefore \\
\lim_{k \to \infty} {\frac{ 3\cdot 1 +2\cdot \frac{1}{\sqrt{k}} }{ \sqrt{1 + \frac{3}{k}+\frac{1}{k^3}}} }

}

$$

Answer 44

now you can plug in infinity

Answer 45

it will equal to 1 right

Answer 46

not quite

Answer 47

$$
\Large {
\lim_{k \to \infty} \frac{\frac{ 3\sqrt{k}+2 }{ \sqrt{k^3 + 3k^2 +1}} }{\frac1k}
\\ \iff \\
\lim_{k \to \infty} {\frac{( 3\sqrt{k}+2) }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac k 1 \\
\\ \iff \\
\lim_{k \to \infty} {\frac{ 3\sqrt{k}\cdot k +2k }{ \sqrt{k^3 + 3k^2 +1}} }
\\ \iff \\
\lim_{k \to \infty} {\frac{ 3\sqrt{\color{red}{k^3}} +2k }{ \sqrt{k^3 + 3k^2 +1}} }\cdot \frac{\frac{1}{\sqrt{k^3}}}{\frac{1}{\sqrt{k^3}}}
\\ \iff \\
\lim_{k \to \infty} {\frac{ 3\sqrt{ \frac{k^3}{k^3}} +2\cdot \frac{k}{\sqrt{k^3}} }{ \sqrt{\frac{k^3}{k^3} + 3\frac{k^2}{k^3}+\frac{1}{k^3}}} }
\\ \iff\\
\lim_{k \to \infty} {\frac{ 3\cdot 1 +2\cdot \frac{1}{\sqrt{k}} }{ \sqrt{1 + \frac{3}{k}+\frac{1}{k^3}}} } =
{\frac{ 3\cdot 1 +2\cdot \frac{1}{\sqrt{\infty }} }{ \sqrt{1 + \frac{3}{\infty}+\frac{1}{\infty^3}}} }
\\ = {\frac{ 3\cdot 1 +2\cdot \color{red}0 }{ \sqrt{1 + \color{red}0+ \color{red}0}} }
= \color{blue}3
}

$$

Answer 48

so it converges since its greater than 1 right?

Answer 49

by the limit comparison test, both of the series have the same behavior. if one diverges, the other diverges. if one converges, the other converges.
and we know that $$ \Large \sum_{k=1}^{\infty} \frac{1}{k} $$ diverges

Answer 50

therefore the other series must diverge as well .

Answer 51

how did we get 1/k again?

Answer 52

that part you had to use reasoning or 'eyeballing'

Answer 53

because overall the degree of the numerator was 1/2. the degree of the denominator was 3/2
and 3/2 - 1/2 = 2/2 = 1 . so you will have 1\k

Answer 54

oh okay

Answer 55

$$ \Large
\frac {k^{1/2}} {k^{3/2}} = \frac{1}{k^{3/2 - 1/2}} = \frac{1}{k^{2/2}} = \frac1k

$$

Answer 56

is it the same when we have \[\frac{ 1+3^n }{ 1+4^n} = \frac{ 3^n }{ 4^n }\]

Answer 57

$$\Large \frac{ 1+3^n }{ 1+4^n} \sim \frac{ 3^n }{ 4^n } = \left( \frac{3}{4}\right)^n \text{, for large n} $$

Answer 58

or \[\frac{ 1+n^2 }{ 2n+1 } = \frac{ n^2 }{ 2n } = \frac{ n }{ 2}\]

Answer 59

yes , they are equivalent for large n . and that clearly diverges

Answer 60

$$
\Large{

\frac{ 1+n^2 }{ 2n+1 } \sim \frac{ n^2 }{ 2n } = \frac{ n }{ 2}


}

$$

Answer 61

n/2 would diverge as well correct?

Answer 62

you can show it rigorously using limits and dividing as we did above by the highest degree

Answer 63

correct,
$$\Large{
\lim_{n \to \infty} \frac n2 = \frac{\infty}{2}= \infty
}
$$

Answer 64

you mean i can divide the equation by 1/n like we did last problem

Answer 65

so that series has no chance of converging, since the limit of the a_n diverges.

Answer 66

correct

Answer 67

okay thanks a lot perl

Answer 68

yw :)

Answer 69

how long are you going to be on for cause i might have questions later

Answer 70

I will be here for another hour or so.
Also you can post and tag me and i will come to it later , if i'm not here.

Answer 71

okay thanks