A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

El_Arrow

  • one year ago

test the series for convergence or divergence

  • This Question is Closed
  1. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sum_{n=1}^{\infty} 1/(6n+1)\]

  2. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i used the comparison test but im not sure if its right

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    eyeball test the degree of the denominator is ?

  4. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and the numerator?

  6. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it doesnt have one

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    of course it has one

  8. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it 1

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no

  10. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    0

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    less

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now we compare degrees \[1-0=1\]

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    since the difference is one, the sum does NOT converge in order for a rational function to converge, the degree of the denominator has to be larger than the degree of the numerator by MORE than one anything more than one will do but one does not work

  15. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so we dont use any tests for this one

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can use a test if you like you could use the limit comparison test for example compare to \(\sum\frac{1}{n}\) which is known to diverge

  17. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but doesnt it converge when an is less than bn when using the limit comparison test

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you cannot use the direct comparison test however since \[\frac{1}{6n+1}<\frac{1}{n}\] but the limit comparison test will work in general however what i said above is true

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, it does NOT converge again, you can see it with your eyeballs

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    degree of the denominator has to be MORE than one larger then the degree of the numerator \[\sum\frac{1}{n}\] does not converge, neither does \[\sum\frac{a}{bx+c}\]

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops i mean neither does \[\sum_{n=1}^{\infty}\frac{a}{bn+c}\]

  22. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so when i used the limit comparison test i got 1/6

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can also use the integral test for this if you like you will get a log, and the log does not have a limit as \(n\to \infty\)

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right and you compared it to a series that does NOT converge right?

  25. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes 1/n

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok then since the limit is a finite non zero number, your series does not converge either

  27. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alright and the same thing goes for direct comparison test right

  28. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when its a finite non zero number it does not converge?

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    direct comparison you just compare if the terms are greater than a series that does not converge, NO if the terms are less than a series that does converge, YES but you can't really use it here

  30. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alright thanks

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yw

  32. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh one more question when does the integral test diverge or converge

  33. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @satellite73 @freckles

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if the integral diverges, then so does the series and if the integral converges so does the series

  35. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but like does it diverge when its a finite non zero number or?

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for example you have \[\sum\frac{1}{6n+1}\] so \[a_n=\frac{1}{6n+1}\] put \[f(x)=\frac{1}{6x+1}\] and compute \[\int_1^{\infty}\frac{1}{6x+1}=\lim_{t\to \infty}\frac{1}{6}\log(6t+1)\]

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the integral diverges, so does the sum

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or perhaps i should say "the integral is infinite (does not exist) therefore neither does the sum"

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    to answer you question, NO if the integral is finite, then it exists (is a number) therefore the corresponding sum will be finite (not the same number as the integral, just finite)

  40. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay thanks

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yw

  42. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can still use a direct comparison: $$\sum_{n=1}^\infty \frac1{6n+1}=\sum_{n=2}^\infty\frac1{6(n-1)+1}=\sum_{n=2}^\infty\frac1{6n-5}$$now consider \(6n-5\le 6n\implies\frac1{6n-5}\ge\frac1{6n}\) and we know \(\sum_{n=2}^\infty\frac1{6n}=\frac16\sum_{n=1}^\infty\frac1n-\frac16\); clearly this diverges so it follows \(\sum_{n=2}^\infty\frac1{6n-5}\) diverges too

  43. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @oldrin.bataku what about for this one \[\sum_{n=1}^{\infty} \frac{ n^6+1 }{ n^7+1 }\]

  44. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you use a comparison test like 1/n?

  45. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or the divergence test maybe?

  46. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    $$\frac{n^6+1}{n^7+1}=\frac{1+1/n^6}{n+1/n^6}$$ consider that for \(n\ge1\) we have \(n+1\ge n+1/n^6\implies \frac1{n+1}\le\frac1{n+1/n^6}\le\frac{1+1/n^6}{n+1/n^6}\); since we know \(\frac1{n+1}\) diverges by an argument like the previous it follows that this series diverges too

  47. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.