Question 1

I am confused how at 2:00 Sal is sure that is converges. I understand that the N^2 is going to grow at the faster rate, but wont the numerator become larger and larger than the denominator? In the denominator as "n" gets larger and larger the 10n is being subtracted from n^2. Whereas on the top the "n" is recieving addition by 9n. At what point is the "10n" or the "8n" enough to change the limiting value.

Accepted Answer

The key is that the absolute size of 10n doesn't matter; what matters is its size relative to n^2. When n=100, n^2 is 10,000 and 10n is 1,000, which is 1/10 as large. When n=1,000, n^2 is 1,000,000 and 10n is 10,000. We increased 10n by a factor of 10, but its significance in computing the value of the fraction dwindled because it's now only 1/100 as large as n^2. Each time we add a zero to n, we multiply 10n by another 10 but multiply n^2 by another 100. Eventually 10n becomes a microscopic fraction of n^2, contributing almost nothing to the value of the fraction.

Question 2

if i had a non convergent seq.  that's mean it's divergent ?

Accepted Answer

Yes.  This can be confusing as some students think "diverge" means the sequence goes to plus of minus infinity.  This is NOT the case.  For example, a sequence that oscillates like -1, 1, -1, 1, -1, 1, -1, 1, ...  is a divergent sequence.

Question 3

So if  a series doesnt diverge it converges and vice versa? Is there no in between? And why does the C example diverge? It converges to n i think because if the number is huge you basically get n^2/n which is closer and closer to n.

Accepted Answer

There is no in-between.  All series either converge or do not converge.  By definition, a series that does not converge is said to diverge.
 
However, not all divergent series tend toward positive or negative infinity.  Some series oscillate without ever approaching a single value.
 
Now, there is a special kind of convergent series called a "conditionally convergent series".  In this type of series half of its terms diverge to positive infinity and half of them diverge to negative infinity; however, the overall sum actually converges to some number.
 
An example of a conditionally convergent series is:
∑ n=1 to infinity of { (-1)^(n+1)/(ln(8)*n)}
This converges to ⅓.  However, its negative terms diverge to negative infinity and its positive terms diverge to positive infinity.

Question 4

Why does the first equation converge?  I thought that the limit had to approach 0, not 1 to converge?

Accepted Answer

I think you are confusing sequences with series. Remember that a sequence is like a list of numbers, while a series is a sum of that list. Notice that a sequence converges if the limit as n approaches infinity of An equals a constant number, like 0, 1, pi, or -33. However, if that limit goes to +-infinity, then the sequence is divergent. If the first equation were put into a summation, from 11 to infinity (note that n is starting at 11 to avoid a 0 in the denominator), then yes it would diverge, by the test for divergence, as that limit goes to 1. However, since it is only a sequence, it converges, because the terms in the sequence converge on the number 1, rather than a sum, in which you would eventually just be saying 1+1+1+1+1+1+1...

Question 5

In the option D) Sal says that it is a divergent sequence......
Lets assume S = 1-1+1-1+1-1.... then
 1 - S = 1- (1-1+1-1+1-1+1...) which is 1-S = 1-1+1-1+1-1... which is same as S
1-S =S  ..... S= 1/2 ..... doesn't that means that it is convergent if we get a specific  value......

Accepted Answer

You cannot assume the associative property applies to an infinite series, because it may or may not hold. And, in this case it does not hold.
In short, S fails to exist for a divergent series, thus computations with S are meaningless. They have no more meaning than the "proofs" 1=2 which contain a hidden division by zero.
 
Let me add a counter-example.
S =  1-1+1-1+1-1....
Notice that if I add another copy of the sum 1-1+1-1+1-1.... at the beginning of the 
sum, I get exactly the same sum. It is still  1-1+1-1+1-1....
Thus,
S + S = S
2S = S
And since you "proved" S = ½
It must be the case that 
2S = ½
Thus, S = ¼
And, of course, I can add as many sets of S to each other as a like and they will still be the same sum, they will still be  1-1+1-1+1-1.... So let us say that I added S to itself 999 times, giving me 1000S = S.
Since S = ½ and S = ¼
And since S = 1000S
Then,
1000S = ½. Thus, S = 1/2000
And 1000S = ¼. Thus S =1/4000
This is obviously absurd and self-contradictory. Thus, we know that the math is wrong. It Is NOT the case that S=½ nor any of the other values we could come up with. It is not the case that the associative property holds for this particular series. And, for that matter, it does not hold that S + S = 2S.
 Since it is possible to have multiple contradictory sums for S, it must be the case that S fails to exist.
Thus when S fails to exist, it is possible to get various nonsensical and contradictory solutions.
 
Basic mathematical operations all require that S exists, if S does not exist the operations can still produce "answers" but they will be nonsense.
 
BTW, the Numberphile video where they "proved" that S of the positive integers was -1/12 made use of such nonsense with divergent series. Their "proof" was utter nonsense.

Question 6

what is exactly meant by a conditionally convergent sequence ?

Accepted Answer

It is a series, not a sequence.
A series is defined to be conditionally convergent if and only if it meets ALL of these requirements:
1.  It is an infinite series.
2. The series is convergent, that is it approaches a finite sum.
3. It has both positive and negative terms.
4. The sum of its positive terms diverges to positive infinity.
5. The sum of its negative terms diverges to negative infinity.
 
The commutative and associative properties do not hold for conditionally convergent series. Thus, it is possible (by using the associative property and/or the commutative property) to group the terms of a conditionally convergent series to make it look like the series converges to any arbitrarily chosen number or to make it look like the the series diverges. Thus, conditionally convergent series are quite difficult to work with and it is very easy to get nonsense answers that look like they are correct.
 
In other words, a conditionally convergent series has the property that you get different answers if you rearrange or regroup the terms.

Question 7

At 2:07 Sal says that the exponential grows much faster than the polynomial, and I kinda see that, but is there a video somewhere that proves this, or for example if i had 10^100*en in the denominator it would still diverge right?

Accepted Answer

Assuming you meant to write "it would still diverge," then the answer is yes. As x goes to infinity, the exponential function grows faster than any polynomial. Not sure where Sal covers this, but one fairly simple proof uses l'Hospital's rule to evaluate a fraction e^x/polynomial, (it can be any polynomial whatever in the denominator) which is infinity/infinity as x goes to infinity. Repeated application of l'Hospital's rule will eventually reduce the polynomial to a constant, while the numerator remains e^x, so you end up with infinity/constant which shows the expression diverges no matter what the polynomial is.

Question 8

Is there any videos of this topic but with factorials? I need to understand that. I found a few in the pre-calculus area but I don't think it was that deep. Any suggestions?

Accepted Answer

Here they are:
https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc

Question 9

can I multiply a divergent seq to a divergent seq and get a convergent seq?

Accepted Answer

Sure. If you multiply each term of
{1, 0, 2, 0, 3, 0, 4, 0,...} by
{0, 1, 0, 2, 0, 3, 0, 4,...} you get
{0, 0, 0, 0, 0, 0, 0, 0,...}. Two divergent sequences whose product is convergent.

Course: AP®︎ Calculus BC (2017 edition) > Unit 12

Worked example: sequence convergence/divergence

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