Log in Robert Checco 10 years agoPosted 10 years ago. Direct link to Robert Checco's post “I am confused how at 2:00...” I am confused how at • (16 votes) Creeksider 10 years agoPosted 10 years ago. Direct link to Creeksider's post “The key is that the absol...” 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. (73 votes) Oya Afify 10 years agoPosted 10 years ago. Direct link to Oya Afify's post “if i had a non convergent...” if i had a non convergent seq. that's mean it's divergent ? • (11 votes) Mr. Jones 10 years agoPosted 10 years ago. Direct link to Mr. Jones's post “Yes. This can be confusi...” 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. (28 votes) Oskars Sjomkāns 10 years agoPosted 10 years ago. Direct link to Oskars Sjomkāns's post “So if a series doesnt di...” 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. • (3 votes) Just Keith 10 years agoPosted 10 years ago. Direct link to Just Keith's post “There is no in-between. ...” 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: (21 votes) idkwhat 9 years agoPosted 9 years ago. Direct link to idkwhat's post “Why does the first equati...” Why does the first equation converge? I thought that the limit had to approach 0, not 1 to converge? • (2 votes) Derek M. 9 years agoPosted 9 years ago. Direct link to Derek M.'s post “I think you are confusing...” 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... (8 votes) Jayesh Swami 9 years agoPosted 9 years ago. Direct link to Jayesh Swami's post “In the option D) Sal says...” In the option D) Sal says that it is a divergent sequence...... • (2 votes) Just Keith 9 years agoPosted 9 years ago. Direct link to Just Keith's post “You cannot assume the ass...” 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. Let me add a counter-example. 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. (5 votes) Ahmed Rateb 9 years agoPosted 9 years ago. Direct link to Ahmed Rateb's post “what is exactly meant by ...” what is exactly meant by a conditionally convergent sequence ? • (2 votes) Just Keith 9 years agoPosted 9 years ago. Direct link to Just Keith's post “It is a series, not a seq...” It is a series, not a sequence. 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. (4 votes) David Procházka 10 years agoPosted 10 years ago. Direct link to David Procházka's post “At 2:07 Sal says that the...” At • (2 votes) Creeksider 10 years agoPosted 10 years ago. Direct link to Creeksider's post “Assuming you meant to wri...” 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. (3 votes) Daniel Santos 9 years agoPosted 9 years ago. Direct link to Daniel Santos's post “Is there any videos of th...” 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? • (2 votes) Stefen 9 years agoPosted 9 years ago. Direct link to Stefen's post “Here they are:https://ww...” Here they are: (2 votes) Akshaj Jumde 8 years agoPosted 8 years ago. Direct link to Akshaj Jumde's post “The crux of this video is...” The crux of this video is that if lim(x tends to infinity) exists then the series is convergent and if it does not exist the series is divergent. Am I right or wrong ? • (2 votes) Stefen 8 years agoPosted 8 years ago. Direct link to Stefen's post “That is the crux of the b...” That is the crux of the biscuit, yes. (1 vote) Sophie Nikol 3 years agoPosted 3 years ago. Direct link to Sophie Nikol's post “can I multiply a divergen...” can I multiply a divergent seq to a divergent seq and get a convergent seq? • (1 vote) kubleeka 3 years agoPosted 3 years ago. Direct link to kubleeka's post “Sure. If you multiply eac...” Sure. If you multiply each term of (3 votes)Want to join the conversation?
2:00
∑ 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.
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......
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.
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.
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. 2:07
https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc
A more rigorous set of rules is upcoming.
{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.
Video transcript
So we've explicitly definedfour different sequences here. And what I wantyou to think about is whether these sequencesconverge or diverge. And remember,converge just means, as n gets larger andlarger and larger, that the value of our sequenceis approaching some value. And diverge means that it'snot approaching some value. So let's look at this. And I encourage youto pause this video and try this on your ownbefore I'm about to explain it. So let's look at this firstsequence right over here. So the numerator n plus 8 timesn plus 1, the denominator n times n minus 10. So one way to think aboutwhat's happening as n gets larger and larger is lookat the degree of the numerator and the degree ofthe denominator. And we care about the degreebecause we want to see, look, is the numerator growingfaster than the denominator? In which case this thingis going to go to infinity and this thing'sgoing to diverge. Or is maybe the denominatorgrowing faster, in which case this might converge to 0? Or maybe they're growingat the same level, and maybe it'll convergeto a different number. So let's multiply out thenumerator and the denominator and figure that out. So n times n is n squared. n times 1 is 1n, plus 8n is 9n. And then 8 times 1 is 8. So the numerator is nsquared plus 9n plus 8. The denominator isn squared minus 10n. And one way tothink about it is n gets really, really, really,really, really large, what dominates in thenumerator-- this term is going to represent most of the value. And this term is going torepresent most of the value, as well. These other termsaren't going to grow. Obviously, this 8doesn't grow at all. But the n terms aren't goingto grow anywhere near as fast as the n squared terms,especially for large n's. So for very, verylarge n's, this is really goingto be approaching n squared over n squared, or 1. So it's reasonable tosay that this converges. So this one converges. And once again, I'm notvigorously proving it here. Or I should sayI'm not rigorously proving it over here. But the giveaway is thatwe have the same degree in the numeratorand the denominator. So now let's look atthis one right over here. So here in the numeratorI have e to the n power. And here I have e times n. So this grows much faster. I mean, this ise to the n power. Imagine if when youhave this as 100, e to the 100th power is aginormous number. e times 100-- that's just 100e. Grows much faster thanthis right over here. So this thing is justgoing to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at thisone right over here. Well, we have ahigher degree term. We have a higherdegree in the numerator than we have in the denominator. n squared, obviously, is goingto grow much faster than n. So for the same reasonas the b sub n sequence, this thing is going to diverge. The numerator is goingto grow much faster than the denominator. Or another way to thinkabout it, the limit as n approaches infinityis going to be infinity. This thing's goingto go to infinity. Now let's think aboutthis right over here. So as we increasen-- so we could even think about what thesequence looks like. When n is 0, negative1 to the 0 is 1. When n is 1, it'sgoing to be negative 1. When n is 2, it's going to be 1. And so this thing isjust going to keep oscillating betweennegative 1 and 1. So it's not unbounded. It's not going to go toinfinity or negative infinity or something like that. But it just oscillatesbetween these two values. So it doesn't convergeto one particular value. So even though this oneisn't unbounded-- it doesn't go to infinity-- thisone still diverges. It doesn't go to one value. So let me write that down. This one diverges.
