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# Proving a sequence is divergent Watch

1. 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?
2. anyone?
3. (Original post by Black_Materia_)
anyone?
In such cases, we generally divide by the highest power of n and use

4. 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 grows a lot more quickly than , which is the reason why your function is divergent.

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

and obviously the series of is divergent, and so by comparison, so is your series.
5. (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 grows a lot more quickly than , which is the reason why your function is divergent.

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

and obviously the series of 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)
6. (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 is convergent or not, but it's still pointless using the ratio test for this because as so it's impossible for to be finite.
7. Also not being finite does not imply is divergent. For example, if then the sum is not finite, yet as and so is convergent.
8. (Original post by Black_Materia_)

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.
9. (Original post by Noble.)
Also not being finite does not imply is divergent. For example, if then the sum is not finite, yet as 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
10. (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 converging and the infinite series converging are two completely different things. The definition of both are as follows:

converges to L if

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

In both definition, the sum doesn't even come into it and isn't even relevant. If we're talking about being convergent then the (Cauchy) definition is:

Where is the nth partial sum.

Both definition are vastly different. If you know is convergent, then you an infer that as

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

The classic one to look at is when clearly is convergence/Cauchy as it clearly satisfies the definition. However, the is not finite.
11. (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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Updated: April 11, 2013
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