Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

agentx5

  • 2 years ago

Ok honestly the notes make no sense on this one... How do I do this? \[\huge \sum_{n=1}^{\infty} \frac{\sin(\frac{\pi n}{2})}{n!}\]

  • This Question is Closed
  1. agentx5
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got to go for now but I'll be back tomorrow to see if anybody has figured this one out :-)

  2. agentx5
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Convergence/Divergence and if it converges to what are the typical questions for these...

  3. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    hi agentx5 1.Convergence/Divergence u can write \( \large 0 \le |\frac{\sin \frac{n \pi}{2}}{n!}| \le \frac{1}{n!} \) and use this \(\large \sum_{n=1}^{\infty} \frac{1}{n!}=e-1\) ------------------------------------------------------------------------- 2.Convergence to What see \(\large \sin \frac{n \pi}{2}=0 \ \ for \ \ n=2,4,6,... \) and \(\large \sin \frac{n \pi}{2}=1 \ or \ -1 \ \ for \ \ n=1,3,5,... \) so u can rearrange the sigma like this \[\large \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\] now think of Maclaurin series for sin to Evaluate the sum

  4. agentx5
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So kind of like the Squeeze Theorem for the first one, gotcha. Does this test have a name though? Typically we have to say something like "convergent by the alternating series test" or something like that. I don't think we've gotten to MacLaurin series yet, but I'll try to understand :-) Would #46 from this be what I'm looking for? http://mathworld.wolfram.com/MaclaurinSeries.html

  5. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1342891305015:dw|

  6. FoolAroundMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Using the power series expansion for e^(x) around x=0 (which is equivalent to Maclaurin's series) \[e^{i} = 1 + i - 1/2! -i/3! + 1/4! + i/5! - 1/6! - i/7! + ....\]\[e^{-i} = 1 - i - 1/2! + i/3! + 1/4! - i/5! - 1/6! + i/7! + ...\] what we need is 1 - 1/3! + 1/5! - 1/7! + ... try subtracting second series from first to get \[e^{i} - e^{-i} = 2i(1 - 1/3! + 1/5! - 1/7! + ... )\]So, 1 - 1/3! + 1/5! - 1/7! + ... = \((e^{i} - e^{-i})/2i = sin(1)\)

  7. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    @agentx5 http://www.sosmath.com/calculus/series/absolute/absolute.html

  8. agentx5
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Awesome thanks for your help! :-)

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

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.