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Math Help - Sequences convergence or not

  1. #1
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    Question Sequences convergence or not

    For the sequence below, determine whether they converge or not, and find the limit when the sequence converges

    a)
    an=4^n/n!
    a=n+1/an=4/n+1 a=n+1 <a=n convergent


    b)an=nsin(1/n)
    an=sin(1/n)/(1/n) =1 convergent

    an=(3/2)n
    3/2 >1 divergent
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  2. #2
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    Re: Sequences convergence or not

    Quote Originally Posted by LAPOSH42 View Post
    For the sequence below, determine whether they converge or not, and find the limit when the sequence converges

    a)
    an=4^n/n!
    a=n+1/an=4/n+1 a=n+1 <a=n convergent
    This is really very confusingly written. I think you mean an+1 where you wrote "a= n+1". That is, you are using the ratio test: \frac{a_{n+1}}{a_n}= \frac{4^{n+1}}{(n+1)!}\frac{n!}{4^n}= \frac{4}{n+1} which goes to 0< 1 as n goes to infinity so the series converges.
    (But "an+1< an alone does NOT imply \sum a_n converges.)

    As for what it converges to, [itex]\sum_{n=0}^\infty \frac{x^n}{n!}= e^x[/tex].


    b)an=nsin(1/n)
    an=sin(1/n)/(1/n) =1 convergent
    What??? sin(1/n)/(1/n) is certainly NOT equal to 1! And even if it were, that would NOT prove it was convergent. What are you trying to say here?

    an=(3/2)n
    3/2 >1 divergent
    Yes, this is a geometric series \sum r^n with r> 1 and so diverges.
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    Re: Sequences convergence or not

    Ivy, I think the original poster has to determine whether the sequences converge, not the corresponding series. I think he (or she) has got the right idea for the first two problems - showing that the sum of a sequence converges is sufficient to show that the sequence converges (to zero), and the second problem involves evaluating \textstyle\lim_{n\to\infty}\frac{\sin 1/n}{1/n}. I do agree, however, that this is not at all clear with the confusing notation that the poster is using.
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