Question

determine whether the series converges absolutely, converges conditionally or diverges

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^n *(2n)! }{ 5n!}\]

Answer 2

@perl need help with this problem

Answer 3

the answer is diverge but how do i proof or show it diverges?

Answer 4

try ratio test

Answer 5

by the way is the bottom 5*n!
or (5n)!

Answer 6

it doesn't matter
we would still use ratio test

Answer 7

but anyways let me know if you need anymore help

Answer 8

useful hints:
(n+1)!=(n+1)*n!


or you know
like
(2n+1)!=(2n+1)*(2n)!

Answer 9

HI!!

Answer 10

ratio if it is what you wrote, the terms don't go to zero, do they?

Answer 11

@freckles yes its 5*n!

Answer 12

@freckles sorry i was away from keyboard

Answer 13

its k still here

Answer 14

so i have the multiply the numerator and denominator

Answer 15

\[a_n=\frac{(-1)^n(2n)!}{5 \cdot n!} \\ a_{n+1}=\frac{(-1)^{n+1}(2[n+1])!}{5 (n+1)!} =\frac{(-1)^n(-1)(2n+2)!}{5(n+1)!}\\ =\frac{(-1)(-1)^n(2n+2) \cdot (2n+1) \cdot (2n)! }{5(n+1) \cdot n!}\]

Answer 16

\[\lim_{n \rightarrow \infty}|\frac{a_{n+1}}{a_n}|=?\]

Answer 17

\[\frac{ (-1)^n (-1)(2n+2) }{ 5(n+1)! }/\frac{ (-1)^n * (2n)! }{ 5*n! } = \frac{ 10n+10 * n! }{ 10n(n+1)! }\]

Answer 18

sort of right?

Answer 19

I'm not quite sure how you got that

Answer 20

its indeterminate so we got to use l'hopital's rule 10+10n!/10n^2+10n!

Answer 21

\[\frac{(-1)\cancel{(-1)^n}(2n+2)(2n+1) \cancel{(2n)!}}{\cancel{5}(n+1) \cancel{n!}} \cdot \frac{\cancel{5} \cancel{n!}}{\cancel{(-1)^n} \cancel{(2n)!}}\]

Answer 22

oh

Answer 23

why 2n+1?

Answer 24

i mean where did you get it from?

Answer 25

remember
n!=n(n-1)(n-2)(n-3)!

so if you have
(2n)!=(2n)*(2n-1)*(2n-2)*(2n-3)*(2n-4)!
or if you don't won't to write that many factors
(2n)!=(2n)*(2n-1)!



so if you have
(2n+2)!=(2n+2)*(2n+1)*(2n)*(2n-1)*(2n-2)!

or
as I wrote it above
(2n+2)!=(2n+2)*(2n+1)*(2n)!

Answer 26

like you know
15!=15*14*13*12!

16!=16*15!

Like 2n+2=16 when n=7
so
16!=(2*7+2)!=(2*7+2)*(2*7+1)*(2*7)!

you are just taking one away from the previous and tacking on a ! if you don't keep writing

Answer 27

oh okay so i got infinity as the result does that mean we have to use l'hopital's rule?

Answer 28

which part gave you infinity

Answer 29

\[\lim_{n \rightarrow \infty} (2n+2)(2n+1)\]

Answer 30

and that is over (n+1) right

Answer 31

oh yes

Answer 32

but its like you basically have 4n^2/n=4n
and yes 4n goes to infinity as n goes to infinity

Answer 33

and thats diverge

Answer 34

yep

Answer 35

no its indeterminate

Answer 36

L is definitely bigger than 1 here

Answer 37

whats the L? the limit?

Answer 38

I'm calling L=|-infty|=infty since that was the limit we got here
L>1 which means it diverges

Answer 39

but its an infinity over infinity

Answer 40

yes but the top goes to infinity faster than the bottom

Answer 41

because it has a much larger value right?

Answer 42

x^2>x for large x

Answer 43

\[\frac{x^2}{x}-> \infty \text{ as } x-> \infty \]

Answer 44

the degree right

Answer 45

yes 2>1

Answer 46

what if we had \[\sum_{n=1}^{\infty} \frac{ n ^{10} }{ 10^{-n} }\]

Answer 47

do the n and the 10 cancel out

Answer 48

I would try ratio test again

Answer 49

like this \[\frac{ n ^{10}}{ 10^{-n} } * \frac{ 10^{-n} }{ n }\]

Answer 50

now you have to replace the n's in the first quotient with (n+1)
like we did before

Answer 51

\[\frac{ n ^{10} }{ 10^{-n}} * \frac{ 10^{-(n+1)} }{ (n+1)^{10} }\]

Answer 52

ok well I was hoping you would replace the n's in the first quotient
maybe fine we will find the recipocral of our limit I guess later

Answer 53

i didnt know you were suppose to replace the n's in both of them

Answer 54

my mistake

Answer 55

no you aren't

Answer 56

just the first quotient

Answer 57

let me redo it real quick

Answer 58

\[\frac{ (n+1)^{10} }{ 10^{-(n+1)}}*\frac{ 10^{-n} }{ n ^{10} }\]

Answer 59

\[L=\lim_{n \rightarrow \infty}|\frac{a_{n+1}}{a_n}|=\lim_{n \rightarrow \infty}|\frac{(n+1)^{10}}{10^{-(n+1)}} \cdot \frac{10^{-n}}{n^{10}}|\]
great

Answer 60

you can cancel the 10^(-n) but you would be left with a 10^(-1) on bottom
and you also have the ((n+1)/n)^10 to worry about that but that isn't too bad

Answer 61

so its always the first one that gets the n's replaced by n+1

Answer 62

yes that is what the a_(n+1) divided by a_n tells us to do

Answer 63

so i'll be left with (n+1)^10 / 10^1 * 1 / n^10

Answer 64

almost you have 10^(-1) on bottom

Answer 65

\[10^{-(n+1)}=10^{-n-1}=10^{-n}10^{-1}\]

Answer 66

the 10^(-n)'s will cancel

Answer 67

you will have 10^(-1) on bottom

Answer 68

\[L=\lim_{n \rightarrow \infty}|\frac{a_{n+1}}{a_n}|=\lim_{n \rightarrow \infty}|\frac{(n+1)^{10}}{10^{-(n+1)}} \cdot \frac{10^{-n}}{n^{10}}| \\= \lim_{n \rightarrow \infty}| \frac{1}{10^{-1}} (\frac{n+1}{n})^{10}| \\ =\lim_{n \rightarrow \infty} 10 (\frac{n+1}{n})^{10}\]

Answer 69

now find the limit as n->infty inside the ( ) part

Answer 70

you basically have n/n since n is really really large
n/n=?

Answer 71

infinity over infinity

Answer 72

do you know what 5/5=?
or 5252352/5252352=?

Answer 73

n/n=1

Answer 74

\[\lim_{n \rightarrow \infty}10(1)^{10}=10(1)=10\]

Answer 75

L=10>1

Answer 76

but dont you add the 1+1?

Answer 77

I'm not sure why you are asking that

Answer 78

where does 1+1 come from

Answer 79

pretend n is large
n+1 is approximately n

you have n/n

and n/n=1

Answer 80

this is the same method I used before
you basically had 4n^2/n for the previous one
and this simplified to 4n

Answer 81

n+1/n = n/n + 1/n = 1

Answer 82

well that is actually 1+0=1

Answer 83

now 1+1

Answer 84

not*

Answer 85

okay and since its greater than 1it converges?

Answer 86

no if L>1 then it diverges

Answer 87

you have 10(1)=10>1

Answer 88

it says converges on the back of the book

Answer 89

what did we wrong?

Answer 90

\[\sum_{n=1}^{\infty} \frac{n^{10}}{10^{-n}} \text{ diverges }\]

Answer 91

yeah i think it diverges too i dont know why it says converge

Answer 92

can we do one more?

Answer 93

did you give the correct problem above

Answer 94

\[\sum_{n =1}^{\infty} (-1)^{n-1}\frac{ \frac{ 1 }{ 2 }^{n} }{ n^2}\]

Answer 95

yeah everything is typed in right

Answer 96

ok I'm going to give you a chance to play with it
i be right back

Answer 97

alright for this next one \[-1^{n+1-1}\frac{ \frac{ 1 }{ 2 }^{n+1} }{ (n+10)^2 } * -1^{n-1}\frac{ n^2 }{ \frac{ 1 }{ 2 }^n }\]