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    Would I be solving this the correct way if I first divided everything by n^2 then since everything would be fractions on the denominator which ----> 0
    3/0 ----> inf and so the sequence is divergent? Or would I need to consider the n+1 term of the sequence?
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    anyone?
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    (Original post by Black_Materia_)
    anyone?
    In such cases, we generally divide by the highest power of n and use

    \lim_{n \to \infty} \frac{1}{n} = 0
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    As someone has already pointed out, you want to get the fraction into a form where the denominator tends to 0 with the numerator tending to something finite, other than 0, which will be sufficient in showing it is divergent. So your method is correct.

    It's also worth remembering the 'power' of certain functions when it comes to convergence and divergence.

    Clearly n^2 grows a lot more quickly than n, which is the reason why your function is divergent.

    Another way to come to the conclusion it's divergent is by noticing that

    \frac{3n^2}{2n+1} = \frac{3n}{2 + \frac{1}{n}} > \frac{3n}{3} = n

    and obviously the series of n is divergent, and so by comparison, so is your series.
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    (Original post by Noble.)
    As someone has already pointed out, you want to get the fraction into a form where the denominator tends to 0 with the numerator tending to something finite, other than 0, which will be sufficient in showing it is divergent. So your method is correct.

    It's also worth remembering the 'power' of certain functions when it comes to convergence and divergence.

    Clearly n^2 grows a lot more quickly than n, which is the reason why your function is divergent.

    Another way to come to the conclusion it's divergent is by noticing that

    \frac{3n^2}{2n+1} = \frac{3n}{2 + \frac{1}{n}} > \frac{3n}{3} = n

    and obviously the series of n is divergent, and so by comparison, so is your series.
    Can you possibly use the Ratio Test on this? Or would that make things more complicated? Just when I mentioned maybe considering the (n+1)th term, then taking the limit as n --> infinity of b(n+1) / b(n)
    to see if it is less than, greater than or equal to 1? In which greater than 1 would suggest divergence.

    Thanks a lot for your help by the way! (+rep to both of you)
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    (Original post by Black_Materia_)
    Can you possibly use the Ratio Test on this? Or would that make things more complicated? Just when I mentioned maybe considering the (n+1)th term, then taking the limit as n --> infinity of b(n+1) / b(n)
    to see if it is less than, greater than or equal to 1? In which greater than 1 would suggest divergence.

    Thanks a lot for your help by the way! (+rep to both of you)
    The ratio test would be used if the question was to work out if \displaystyle\sum_{n=1}^\infty b_n is convergent or not, but it's still pointless using the ratio test for this because b_n \nrightarrow 0 as n \rightarrow \infty so it's impossible for \displaystyle\sum_{n=1}^\infty b_n to be finite.
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    Also \displaystyle\sum_{n=1}^\infty b_n not being finite does not imply b_n is divergent. For example, if b_n = \frac{1}{n} then the sum is not finite, yet b_n \rightarrow 0 as n \rightarrow \infty and so is convergent.
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    (Original post by Black_Materia_)
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    Would I be solving this the correct way if I first divided everything by n^2 then since everything would be fractions on the denominator which ----> 0
    3/0 ----> inf and so the sequence is divergent? Or would I need to consider the n+1 term of the sequence?
    The key to solving most A/B limits is to use over/under estimates of the numerator/denominator to get a simpler sequence.

    Here 3n^2 / (2n+1) >= 3n^2 / 3n = n.
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    (Original post by Noble.)
    Also \displaystyle\sum_{n=1}^\infty b_n not being finite does not imply b_n is divergent. For example, if b_n = \frac{1}{n} then the sum is not finite, yet b_n \rightarrow 0 as n \rightarrow \infty and so is convergent.
    ...But it is about the sum, not the terms.

    "Formally, the infinite series is convergent if the sequence of partial sums is convergent. Conversely, a series is divergent if the sequence of partial sums is divergent."

    --Wolfram MathWorld
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    (Original post by aznkid66)
    ...But it is about the sum, not the terms.

    "Formally, the infinite series is convergent if the sequence of partial sums is convergent. Conversely, a series is divergent if the sequence of partial sums is divergent."

    --Wolfram MathWorld
    Actually, the sequence (a_n) converging and the infinite series \displaystyle\sum_{n=1}^{\infty} a_n converging are two completely different things. The definition of both are as follows:

    (a_n) converges to L if

    \forall \epsilon > 0 \ \exists N \in \mathbb{N} \ \forall n \geq N \ |a_n - L| < \epsilon

    Alternatively, you can take the definition of a Cauchy sequence (and a Cauchy sequence is convergent)

    \forall \epsilon > 0 \ \exists N \in \mathbb{N} \ \forall n,m \geq N \ |a_m - a_n| < \epsilon

    In both definition, the sum doesn't even come into it and isn't even relevant. If we're talking about \displaystyle\sum_{n=1}^{\infty} a_n being convergent then the (Cauchy) definition is:

    \forall \epsilon > 0 \ \exists N \in \mathbb{N} \ \forall m,n \geq N \ |s_n - s_m| = \left| \displaystyle\sum_{m+1}^n a_k \right| < \epsilon

    Where s_n is the nth partial sum.

    Both definition are vastly different. If you know \displaystyle\sum_{n=1}^{\infty} a_n is convergent, then you an infer that a_n \rightarrow 0 as n \rightarrow \infty

    But this is pretty much the only deduction you can make (other than the obvious one that if a_n \nrightarrow 0 then the sum is not finite)

    The classic one to look at is when a_n = \frac{1}{n} clearly (a_n) is convergence/Cauchy as it clearly satisfies the definition. However, the \displaystyle\sum_{n=1}^{\infty} a_n is not finite.
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    (Original post by aznkid66)
    ...But it is about the sum, not the terms.

    "Formally, the infinite series is convergent if the sequence of partial sums is convergent. Conversely, a series is divergent if the sequence of partial sums is divergent."

    --Wolfram MathWorld
    No. The convergence of a sequence is completely different from the convergence of the sum of a sequence.
 
 
 
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