Sequence diverges or converges

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  • #1
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Question 1

write the first 4 terms in the sequence defined by a 1 = -2, a n+1 = an/n!

this what I tried

a2 = -2/1
a3 = -2/2
a4 = -1/6

Question 2

determine the following sequences converges or diverges

1- an = sqrt(n^2 -3 )/ 5th rt(n^2)

what I would try is to divide both top and bottom by n^2 but here doesn't work that
way because the denominator is 5th rt

2- a1 = 3, an+1 = (-1)^n+1 an + 3

a2 = 6
a3 = -3

but the ans would be ( 3,0,3,0,3,0)

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Answers and Replies

  • #2
jbunniii
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Question 1

write the first 4 terms in the sequence defined by a 1 = -2, a n+1 = an/n!

this what I tried

a2 = -2/1
a3 = -2/2
a4 = -1/6
Looks fine to me.
Question 2

determine the following sequences converges or diverges

1- an = sqrt(n^2 -3 )/ 5th rt(n^2)

what I would try is to divide both top and bottom by n^2 but here doesn't work that
way because the denominator is 5th rt
When ##n## is large, ##\sqrt{n^2 - 3}## is nearly the same as ##\sqrt{n^2}##. So you can use ##\sqrt{n^2}## instead of ##\sqrt{n^2 - 3}## to determine whether the sequence converges or not. Then you can make the argument more precise by comparing ##\sqrt{n^2 - 3}## with an appropriate sequence. (Hint: try comparing it with ##\sqrt{n^2}/2##.)
2- a1 = 3, an+1 = (-1)^n+1 an + 3

a2 = 6
a3 = -3

but the ans would be ( 3,0,3,0,3,0)
OK, so what can you conclude? Does the sequence converge?
 
  • #3
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question 2

A-
actually this section was before the comparison theorem so let's assume we can't compare it for now

what is the other way


B-it diverges but I don't know how to come up with these values as my values differ ( 3,0,3,0,3,0)
 
  • #4
jbunniii
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question 2

A-
actually this section was before the comparison theorem so let's assume we can't compare it for now

what is the other way
OK, try simplifying: ##\sqrt{n^2 - 3}/\sqrt[5]{n^2} = (n^2 - 3)^{1/2}n^{-2/5} = (n^2 - 3)^{1/2}(n^{-4/5})^{1/2} = ?##

B-it diverges but I don't know how to come up with these values as my values differ ( 3,0,3,0,3,0)
Check your values of ##(-1)^{n+1}##. This is ##-1## when ##n## is even, and ##+1## when ##n## is odd. I suspect you have it reversed.
 
  • #5
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OK, try simplifying: ##\sqrt{n^2 - 3}/\sqrt[5]{n^2} = (n^2 - 3)^{1/2}n^{-2/5} = (n^2 - 3)^{1/2}(n^{-4/5})^{1/2} = ?##


Check your values of ##(-1)^{n+1}##. This is ##-1## when ##n## is even, and ##+1## when ##n## is odd. I suspect you have it reversed.


for part A, now it looks complicated to me than before what my prof did is

moving the bottom to the numerator and you get infinity

for part B I still unable to get ( 3,0,3,0,3,0)
 
  • #6
jbunniii
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for part A, now it looks complicated to me than before what my prof did is

moving the bottom to the numerator and you get infinity
Can you show what you mean by this?
for part B I still unable to get ( 3,0,3,0,3,0)
Let's go back to the definition:
##a_1 = 3, a_{n+1} = (-1)^{n+1} a_n + 3##
To get ##a_2## we plug in ##n=1##: this gives us ##a_2 = (-1)^2 a_1 + 3 = (1)(3) + 3 = 6##. So already this is different from your ##3,0,3,0,...##.
 
  • #7
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Can you show what you mean by this?

Let's go back to the definition:

To get ##a_2## we plug in ##n=1##: this gives us ##a_2 = (-1)^2 a_1 + 3 = (1)(3) + 3 = 6##. So already this is different from your ##3,0,3,0,...##.

( 3,0,3,0,3,0) is my professor answer to this part

but as you see we are not getting close to that
 
  • #8
LCKurtz
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Can you show what you mean by this?

Let's go back to the definition:$$
a_1 = 3, a_{n+1} = (-1)^{n+1} a_n + 3$$

To get ##a_2## we plug in ##n=1##: this gives us ##a_2 = (-1)^2 a_1 + 3 = (1)(3) + 3 = 6##. So already this is different from your ##3,0,3,0,...##.

( 3,0,3,0,3,0) is my professor answer to this part

but as you see we are not getting close to that

I'm guessing there is a typo somewhere and the recursion is supposed to be$$
a_1 = 3, a_{n+1} = (-1)^{n} a_n + 3$$
 

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