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    I've been attempting these questions.. like, among others.. think I have the first one ok but want to check it, and I'd like some hints for the other.. if anyone would be so kind
    They're both for finding the lim as n tends to infinity.

    1. xn = (n2 + en)1/n

    2. xn = [ (3n + 2)/(2n+1) ] n


    For 1 I took logs, then got 1/n log n2 + 1/n log(1 + en/n2).. can I say that the log(1 + en/n2) tends to log1, and then limit of log1/n is 0?
    I know the first term tends to 0, btw.

    Ermmm and for 2 I'm totally lost. I tried getting it into the (1 + m/n)^n form but it totally failed.. and I've tried using logs and that failed.. HELP PLEASE

    Also fyi, in these questions there's also the option of that the limit does not exist.

    Thanks
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    why cant u take logs of the second thing?

    Hmm

    [ (3n + 2)/(2n+1) ]^n

    = (3n + 2)^n / (2n+1)^n
    = nlog(3n+2) - nlog(2n+1)
    ?

    No thats correct... lol
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    (Original post by TheRandomer)
    I've been attempting these questions.. like, among others.. think I have the first one ok but want to check it, and I'd like some hints for the other.. if anyone would be so kind
    They're both for finding the lim as n tends to infinity.

    1. xn = (n2 + en)1/n

    2. xn = [ (3n + 2)/(2n+1) ] n


    For 1 I took logs, then got 1/n log n2 + 1/n log(1 + en/n2).. can I say that the log(1 + en/n2) tends to log1, and then limit of log1/n is 0?
    no. 1+ e^n/n^2 does not tend to 1.
    I know the first term tends to 0, btw.

    Ermmm and for 2 I'm totally lost. I tried getting it into the (1 + m/n)^n form but it totally failed.. and I've tried using logs and that failed.. HELP PLEASE

    Also fyi, in these questions there's also the option of that the limit does not exist.

    Thanks
    for 1) remember that exponentials (a^x) increase much faster than x^b as x increases.

    For large n, (3n+1)/(2n+1) is close to to 3n/2n = 3/2
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    (Original post by Ewan)
    why cant u take logs of the second thing?

    Hmm

    [ (3n + 2)/(2n+1) ]^n

    = (3n + 2)^n / (2n+1)^n
    = nlog(3n+2) - nlog(2n+1)
    ?

    No thats correct... lol

    But then how we know which term 'wins' kind of thing?
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    (Original post by yusufu)
    no. 1+ e^n/n^2 does not tend to 1.

    for 1) remember that exponentials (a^x) increase much faster than x^b as x increases.

    Ooh yeah.. I meant the other way around.. I looked at the wrong solution on my working. So I get that the overall limit is 1. Is this right? n^2/e^n tends to 0?

    The second one.. I'm still stuck.
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    (Original post by TheRandomer)
    But then how we know which term 'wins' kind of thing?
    I dunno, I don't like that method, doesn't come out with what I'd intuitively think at all... think about what happens as all n's are large... with large n it will tend to 3/2 on the inside.. then with that large n as a power, what will 3/2 become.

    Say you have 10000

    you have [1.500024999]^10000 ....
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    (Original post by TheRandomer)
    Ooh yeah.. I meant the other way around.. I looked at the wrong solution on my working. So I get that the overall limit is 1. Is this right? n^2/e^n tends to 0?

    The second one.. I'm still stuck.
    n^2 + e^n tends to e^n.

    See edit of previous post for 2.
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    (Original post by TheRandomer)

    2. xn = [ (3n + 2)/(2n+1) ] n

    Ermmm and for 2 I'm totally lost. I tried getting it into the (1 + m/n)^n form but it totally failed.. and I've tried using logs and that failed.. HELP PLEASE

    Also fyi, in these questions there's also the option of that the limit does not exist.

    Thanks
    (3n+2)/(2n+1) = (2n+1+n+1)/(2n+1)

    =1 + (n+1)/(2n+1)

    > 1 + (n+1)/(2n+2)

    =1+ 1/2

    Does that help?
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    (Original post by yusufu)
    n^2 + e^n tends to e^n.

    See edit of previous post for 2.

    I meant n^2 OVER e^n though
    Does that not tend to zero?

    If I use COLT on the bracket in q2 I get 3/2 right.. but how do I find what happens when it's raised to the power n?

    Thanks for helping guys =]
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    (Original post by TheRandomer)
    I meant n^2 OVER e^n though
    Does that not tend to zero?
    ?
    x_n = (n^2 +e^n)1/n. Is this the correct expression?

    If I use COLT on the bracket in q2 I get 3/2 right.. but how do I find what happens when it's raised to the power n?

    Thanks for helping guys =]
    COLT?

    for x>1, x^n gets very large as n tends to infinity.
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    (Original post by TheRandomer)
    If I use COLT on the bracket in q2 I get 3/2 right.. but how do I find what happens when it's raised to the power n?

    Thanks for helping guys =]
    Think about it :p: a SMALL number to the power of a BIG number
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    (Original post by yusufu)
    ?
    x_n = (n^2 +e^n)1/n. Is this the correct expression?


    COLT?

    for x>1, x^n gets very large as n tends to infinity.
    Yeah.. I took logs. so then
    1/n log en + 1/n log(1 + n2/en)

    Is the limit of the RHS term 0? Because we have 1/n log(1 + something with limit of 0)??


    Calculus of limits theorem? :p:
    How should I show that in a proofy kind of way?
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    Do they want a formal proof, or is it just a "show that" kind of question? I would think "show that", so just show all your steps as to how you come to the outcome. Ie explain that if r>1 r^n tends to infinity as n tends to infinity [no limit]
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    (Original post by TheRandomer)
    Yeah.. I took logs. so then
    1/n log en + 1/n log(1 + n2/en)

    Is the limit of the RHS term 0? Because we have 1/n log(1 + something with limit of 0)??
    Yes/


    Calculus of limits theorem? :p:
    How should I show that in a proofy kind of way?
    \displaystyle (\frac{3n+2}{2n+1})^n \to (\frac{3n}{2n})^n = (3/2)^n \to \infty as n \to \infty
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    (Original post by Ewan)
    Do they want a formal proof, or is it just a "show that" kind of question? I would think "show that", so just show all your steps as to how you come to the outcome.
    I just says 'show that no limit exists' .. so can I just write that as n-> 3/2^n has no limit?
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    Thankyouuu! =]
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    (Original post by TheRandomer)
    I just says 'show that no limit exists' .. so can I just write that as n-> 3/2^n has no limit?
    Write what Yusufu wrote, and since it tends to infinity there is obviously no limit.
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    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?
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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...?
    |nsinn| \le n
    [latex]\frac{1}{4^n} decreases at a much faster rate than n as n increases.

    so you're adding/subtracting smaller and smaller terms that are tending to 0 in magnitude.
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    (Original post by yusufu)
    |nsinn| \le n
    [latex]\frac{1}{4^n} decreases at a much faster rate than n as n increases.

    so you're adding/subtracting smaller and smaller terms that are tending to 0 in magnitude.
    So how do I prove whether it's convergent/divergent? The fact that the terms are a null sequence wont prove the sum is convergent.
 
 
 
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