Theorem 1 and the Test for the Divergence of Infinite Series by @mes on 3Speak
In this video, I go over Theorem 1, which states that the terms of a convergent series approach zero. Conversely, if the terms don't approach zero, the series diverges, hence providing a useful tes...
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theorem 1. So in the following series, yeah, if the following series is convergent, I'll just write this, infinity n equals to 1 of A.N, there's a general series, so if it's convergent, then what we have is the limit as N approaches infinity of the sequence A.N, so if that series is convergent, then the sequence is approaching zero, or the terms in the series, which is a sequence, yeah, is approaching zero, so that's a bit. very interesting theorem and let's look at the proof for this theorem so what I'm going to do now is I'm going to let let s n equals to just partial sum a a 1 plus a 2 plus a 3 plus dot dot dot all the way until we have a n minus 1 that's term just before it and then a n like that at the end so what I want to make a note is well note that before we get to the AN, this could be written as, yeah, this right here, can be written as AN minus 1. I mean, SN minus 1, so the partial sum up to SN minus 1. So then what we have is, it's a very interesting algebra here, so we have, yeah, here, SN equals to SN minus 1 plus AN. like that so we could write the sequence a n equals 2 s n minus s n minus 1 like that so just write it like that all right it like that is because well since we're given that the series from n equals 1 to infinity of a n is convergent so if we're given that and then over here, A.N is equal to S-N minus S-N minus 1. So if you take the summation of this, if that's convergent, this means this has to be convergent. So in other words, S-N and S -N-min-1 have to be convergent, because you can't be subtracting infinities there. So since the summation or the series A -N is convergent, the sequence of partial sums SN is also convergent, because this A-N is equal to formal with the partial sums like that. So, going further, let limit as n approaches infinity of s n equals to s. In other words, let the limit of the partial sum as an approach infinity equals a total sum of the series, s. So since n minus 1 approaches infinity as n approaches infinity, we also have Yeah, we also have limit as an approach infinity of s n minus 1 is equal to s. So it doesn't matter if you have s n minus 1 or s n like that. If they're both approaching if you're approaching infinity, they're both going to be approaching s the partial sum. Yeah, I mean the total sum here. So it doesn't matter if you're saying as n as you're taking an infinite number of terms is going to go equals to s. So therefore, limit. as n approaches infinity of a n. Yeah, so a n like this. So we're going to take the limit over there, a n, equals to the limit as n approaches infinity of s n minus s n minus 1 like that. And you could break these up, but this just equals to the limit as n approaches infinity of s n and minus limit. as n approaches infinity of s n minus 1, and these are both equal to the same thing, so equals to s minus s equals to 0. So hence, if it's convergent, the term is going to zero. So if this series is convergent, then the terms themselves are going to zero. So very, very fascinating as you take the limit. Now let's make some notes on this theorem and proof. So with any series summation of A.N, we associate two sequences, the sequence SN of its partial sums, and the sequence A.N of its terms. This is just the sum of all the terms at any given point here, SN, and the sequence there. A.N of its actual terms. So, if the series A.N is convergent or the summation of the terms A.N of the sequence is S, the sum of the series. And as theorem one asserts, the limit of the sequence A.N is zero. So the limit of the sequence, A.N is zero, but the limit of the, yet limit of the sequence A.N is zero, but the limit of this partial sum, sequence is S, the sum of the series. Note 2, the converse of theorem 1 is not true in general, so you have the opposite. For example, if the limit as n approaches infinity of a n is equal to 0, so if the terms are going to 0, That does not mean we cannot conclude that A.N is convergent. Observe that for the harmonic series, the one that's 1 over N like that, we have the terms A.N equals 1 over n, approaching infinity, but we show an example 7 that the harmonic series, yeah, the summation of 1 over n, is divergent. So it's divergent, even though the terms are going to zero. So it's as if the terms are going to zero slowly, hence it diverges away and summing to infinity. It's very, very interesting. And now let's go further into this, and now let's look at the test for divergence. So if we're given, if we have limit as n approaches infinity of A. N. Yeah, so if this limit of these terms does not exist, or if the limit... and approaches infinity of a n is not equal to zero. So if the terms are not approaching zero or they don't exist, then, then the series right here, when we take the summation, so a series is the summation of the sequences. infinity n equals to 1 of a n this is divergent because if the terms does not exist i.e. goes to infinity, or if it's not equal to 0, so then you're going to be adding full numbers. You're going to be adding non-zero, non -small numbers, you can be larger and larger. So then the series, the summation of A -N from n equals 1 to infinity, is divergent. So yes, this test for divergence follows from theorem 1, because if the series is not divergent, then it is convergent, and so the limit has to be equal to 0, the limit of the terms, and approaches infinity of AN is equal to 0. If it is, yeah, convergent, that is, if it's not divergent, so it's very fascinating. So now let's go over example 8, so it states, exemplite states show that the following series diverges. So from the test of divergence, so we have here right here, the summation n equals 1 up to infinity of n squared over 5n squared plus 4. What we have to do is, we'll just check the term and take the limit as end approaches infinity and see if it exists or if it's not equal to 0. So what we'll do is limit using the test of divergence. Limit as N approach infinity of the term A.N, or the sequence A.N, which is just the terms inside there, equals to limit as N approaches infinity of n squared over 5 n squared over 4. And now what I'll do here is, well, we have the highest power here. Recall on my limit videos on limits. to one way of doing simplifying this limit is timesing the top and bottom by 1 over the highest power n squared so that we end up getting something that looks like this limit as an approach to infinity of then this would cancel with this once we have a 1 over 5 this cancels we're going to have a plus 4 over n squared now this over here Yeah, this right here, this 4 over n squared, so this is going to 0. So this, when you plug in infinity there, we have a 4 over infinity, or put a 4 over infinity, which is approaching 0, like that. So this is equals to 1 over 5. So it equals to 1 over 5, but this is not 0. Not 0. So in other words, the term, The terms of the series are approaching a non-zero value. It's approaching 1 over 5. In other words, the summation of 1 over 5 is going to be infinity. So, hence it diverges by the divergence test. So yeah, so the series diverges by the test for divergence, i.e. the sum of the terms keeps getting larger since the terms are not going to 0. It's going to 1 over 5. Note 3 on theorem, if we find that the limit as n approaches infinity of A .N is not equal to 0, we know that, yeah, we know that the series A .N is divergent, but if we find that the limit as N approaches infinity of A .N equals to zero, we know nothing about the convergence or divergence of the series A .N. We just know the opposite. If it's not zero, then it is. divergent but if it is zero it could be or couldn't remember the warning in note 2 if limit and approaches infinity a n equals 0 the series a n might converge or might not diverge as it diverged in the harmonic series even though it reports you so very fascinating