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# Proving a sequence converges using the formal definition

Applying the formal definition of the limit of a sequence to prove that a sequence converges. Created by Sal Khan.

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• what is the proof for showing sin(x) is divergent
• A function is divergent if it fails to converge to a single number. It doesn't have to veer off to some large value to be considered divergent. The function sin(x) oscillates between 1 and -1 forever, so it never converges to a single number.
• I understand everything, but how to determine the value of epsilon?
• Well, epsilon is an arbitrary value that is chosen for the purpose of proving a limit. In this case, Sal started drawing a line for his arbitrary value, and then said it looked like he had chosen a value of 1/2 for his epsilon. He could have chosen another value, but it is a good idea not to go too far along in the sequence, or you might wonder whether the proof holds only in the tail end of the sequence or perhaps something interesting happens in the first values.

Once he "chose" this value, he then used it as he talked through the proof. When he chose 1/2, the reciprocal was 2, so that was where he drew his M.

If he had chosen an epsilon of 1/3, his value of M would be 3, and it is still true that for values of n greater than M, the values of the sequence would be within the range defined by that epsilon. By looking at his graph, you can see that the first value was outside his criteria according to the epsilon he chose, but for convergence, we really care about what happens further along in the sequence.

Having said that, if you have to do this for another problem, you can solve the inequality to find suitable values of epsilon for the particular sequence.
• why does the exponent n+1 disappear after you introduce the insertion of the obsolete value signs?
• The exponent n+1 is only making the sequence oscillate between positive and negative (i.e. the first term is positive, the second is negative, the third is positive, the fourth is negative, etc.). However, the absolute value of any number is (by definition) positive, so the exponent has no purpose anymore (because the "new sequence" is now entirely positive and does not oscillate between positive and negative).
• If a sequence is undefined at a certain integer value e.g ((-1)^n)/4-n^2 is clearly undefined at n=2. What are the implications of this in regard to the sequence and its convergence?
• If you would draw it, a clear asymptote would be seen at 2, but after that it normally converges to +/- 0. Since you care about the convergence when n goes to infinity, a tinsy number like 2 has no significance there. To give a full proper answer though, you'd have to split it into three questions and solve them separately, |f(n)| < 2, |f(n)| = 2 and |f(n)| > 2 . The middle case has another twist, since the limit is different if you approach it from the left or right.
• Sal begins the video by "claiming" that the limit as the alternating harmonic sequence approaches infinite terms is zero, and then proceeds to plug in L=0. But suppose, for some reason, you thought that L=1. If you then plug that in, what would break down?
• To show that 1 is not the limit, we would need to find an ε>0 and an M>0 such that, for all n>M, |a_n-1|>ε. This is the negation of the limit definition.

If we take ε=1/2, M=3, we just need to show that |(-1)ⁿ/n -1|>1/2 for all n>3. We can prove this by induction or just observe that the numbers within a distance 1/2 of 1 are those in the interval (1/2, 3/2), which the remainder of this sequence stays outside of.
• How would you prove a sequence that's limit as n-> infinity = infinity
• Let a_n be a sequence. To prove that limit as n-> infinity a_n = infinity, show that for every positive number M, there exists a positive integer N such that a_n > M for all n > N. In more concrete language, this means showing that for every threshold level (M), there is a position in the sequence (N) beyond which the terms of the sequence are above that threshold level forever.

Have a blessed, wonderful day!
• So it's true is M is positive and not true when M is negative?
(1 vote)
• Sal's statement that M > 0 is superfluous. This sequence happens to start at 1, so our formula for M will produce a positive number, but we could just as easily have a sequence that begins with a negative number and then it would be acceptable for M to be negative. For example, if this sequence were shifted 10 units to the left, then our formula for M would be 1/ε - 10, and there would be no problem in the fact that some values of ε, such as 1/4, would produce a negative M.
• So all the proof is saying is that if you set up some boundaries for a function and the values of the function stay in that range the function will converge at some point ?
• In a sense, yes, in that L + ε and L - ε are your boundaries, and as the boundaries become increasingly stringent, in other words as ε approaches 0, there must be a value of n that is greater than M for which | a sub n - L | < ε. As a result, you must be able to find a function, M(ε), that defines M for any value of ε. In the video, this function was 1/ε. If there is a value of ε where a value of M cannot be found, then the sequence is not convergent.
(1 vote)
• Wouldn't the series (-1)^n converge according to the definition if I pick epsilon=2, m=1?
The sequence (1,-1,1,-1,1,-,1...) is always epsilon within 0, but it obviously doesn't converge to 0
• No, the series does not converge. Neither does the sequence.
But, remember this video was about whether the sequence converges, not the series.
Remember this test must work for ALL ε > 0, not just a convenient value. So, you can pick 0.01 for epsilon, which obviously won't work.
So, what is L? 0? 1? A simple check will show it is none of these:
Let us pick ε = 0.01
Try L = 1
[an - 1| < 0.01
an will oscillate between -1 and 1, when an is -1, the test fails.
Try L = 0, when an is 1 or -1 the test fails.

In fact, try any value of M whatsoever and the test will fail with a sufficiently low ε no matter what we pick for a potential L.
Thus, this sequence diverges.