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    (Original post by silent ninja)
    Since we're discussing convergence how about  \sum_{n=1}^{\infty}\frac{nsinn}{  4^n} ?
    I figure it's divergent because 1/4^n is just a geometric sequence and we're multiplying by increasing natural numbers so that's going to tend to infinity. But sin(n) sign keeps changing so it must be divergent...?
    The question is to do this via comparison test so can I compare it to say  \sum\frac{-n}{4^n} and  \sum\frac{n}{4^n} on either side? One tend to positive infinity, the other negative... so the test wont work?
    Well they tend to positive and minus infinity, but as Yusu inferred, when you put sin back in and think about it... its a bit like a sin graph getting closer and closer to zero I think. Your dealing with a series rather than a sequence though so.. Doesn't converge I think, yea your right. Like 1 -1 + 1 - 1 will never converge
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    (Original post by yusufu)
    \displaystyle (\frac{3n+2}{2n+1})^n \to (\frac{3n}{2n})^n \to (3/2)^n \to \infty as n \to \infty
    You can't write that, surely?? (That is, you can't write f(n) --> g(n) as n tends to anything. I'm fairly certain you'll need to set up the comparison test here to be rigorous about it.)
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    (Original post by Ewan)
    Well they tend to positive and minus infinity, but as Yusu inferred, when you put sin back in and think about it... its a bit like a sin graph getting closer and closer to zero I think.
    I see that but the problem is the sum of a null sequence is not necessarily convergent (harmonic series is the famous example!). Hence my confusion-- do the terms tend to zero fast enough? What can I compare it to to do a comparison test?
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    (Original post by generalebriety)
    You can't write that, surely?? (That is, you can't write f(n) --> g(n) as n tends to anything. I'm fairly certain you'll need to set up the comparison test here to be rigorous about it.)
    Probably not. I actually wanted to use the squiggly equals sign in place of the first arrow, but didn't know the latex for it! (and = sign in place of the 2nd arrow)
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    (Original post by silent ninja)
    I see that but the problem is the sum of a null sequence is not necessarily convergent (harmonic series is the famous example!). Hence my confusion-- do the terms tend to zero fast enough? What can I compare it to to do a comparison test?
    compare it to n/n^3.
    and note that 4^n> n^3 for n > 2
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    (Original post by silent ninja)
    I see that but the problem is the sum of a null sequence is not necessarily convergent (harmonic series is the famous example!). Hence my confusion-- do the terms tend to zero fast enough? What can I compare it to to do a comparison test?
    To be honest, my intuition says that it does converge fast enough. After all, we know that |n sin n| <= n, and we know that |n / 2^n| < 1 for large enough n - use those two inequalities to compare it to sum 1/2^n.

    (Original post by yusufu)
    Probably not. I actually wanted to use the squiggly equals sign in place of the first arrow, but didn't know the latex for it! (and = sign in place of the 2nd arrow)
    Fair enough. :p:

    Incidentally, this drops out nicely enough with the triangle inequality: |(3/2)^n| \leq |(3/2)^n - (\frac{3n+2}{2n+1})^n| + |(\frac{3n+2}{2n+1})^n|. (Edit: oh, wait. That may need a bit more work than I thought, but I think it still works.)
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    (Original post by yusufu)
    compare it to n/n^3.
    and note that 4^n> n^3 for n > 2
    (Original post by generalebriety)
    To be honest, my intuition says that it does converge fast enough. After all, we know that |n sin n| <= n, and we know that |n / 2^n| < 1 for large enough n - use those two inequalities to compare it to sum 1/2^n.
    Got it. Thanks for the help!


    I'll now stop overtaking this thread. For this cant we just do

      [ \frac{3n+2}{2n+1}]^n = [\frac{3+ \frac{2}{n}}{2+\frac{1}{n}}]^n as  n \to \infty get  [\frac{3+0}{2+0}]^n
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    (Original post by silent ninja)
    Got it. Thanks for the help!


    I'll now stop overtaking this thread. For this cant we just do

      [ \frac{3n+2}{2n+1}]^n = [\frac{3+ \frac{2}{n}}{2+\frac{1}{n}}]^n as  n \to \infty get  [\frac{3+0}{2+0}]^n
    Nope. After all, (1 + 1/n)^n --> e. You're applying the limit to the "1/n" bits before applying it to the "^n" bit, whereas you really have to apply them both simultaneously.

    Edit: typo.
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    (Original post by generalebriety)
    Nope. After all, (1 + 1/n)^n --> e. You're applying the limit to the "1/n" bits before applying it to the "^n" bit, whereas you realy have to apply them both simultaneously.
    Oops, power series, silly mistake. Of course! Good thing you pointed that out; we did it weeks ago and so many series/sequences things sound similar but are not. Hopefully it will all be a lot clearer around revision time.
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    (Original post by generalebriety)
    Nope. After all, (1 + 1/n)^n --> e. You're applying the limit to the "1/n" bits before applying it to the "^n" bit, whereas you really have to apply them both simultaneously.

    Edit: typo.
    Ok.. I think I see what's happening now.. why we can't do that.
    What's the *right* method for me to find the limit?
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    Comparison. Show (\frac{3n+2}{2n+1})^{n} is greater than or equal to something which tends to infinity. Make the numerator smaller and denominator bigger to do the inequality. You want to get something like x^{n} with x>1.
 
 
 
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