Hovedinnhold

# Converting explicit series terms to summation notation (n ≥ 2)

## Videotranskripsjon

Let's say that we're told that this sum right over here, where our index starts at 2 and we go all the way to infinity, that this infinite series is negative 8/5 plus 16/7 minus 32/9 plus-- and we just keep going on and on forever. And so what I want to do is to explicitly define what a sub n is here. So right now we just say, hey, if you take the sum of a sub n from n equals 2 to infinity, it turns out you get this sum right over here. But let's think about what a sub n-- how we can actually define it in terms of n. And I encourage you to pause the video right now and try it on your own. So the first thing that you might realize is, well, this is the number that we're going to get. Let me write it this way. a sub 2 is equal to negative 8/5. a sub 3 is equal to 16/7. a sub 4 is equal to negative 32/9. And I'm just giving the sign to the number in the numerator. Negative 8/5 is the same thing as negative 8 over 5. Let me make that a little bit clearer. So I'll make that a little bit clearer. So this is negative 8/5. Obviously, this is positive, so I don't have to really worry about it too much. And then here, I'm just saying negative 32/9, so it's the same thing as negative 32 over 9. So let's see if we can first find a pattern in the numerator. So when we go from negative 8 to 16, what's happening? Well, we're multiplying by negative 2. Now, to go from 16 to negative 32, we're multiplying by negative 2 again. So you might say, OK, well, whatever we have in the numerator must be a power of negative 2. And, all right, if you say, well, maybe this is negative 2 squared, well, you know that negative 8 isn't negative 2 squared. Negative 2 squared is equal to positive 4. Negative 8-- this right over here. Negative 8, that is equal to negative 2 to the third power. 16 is equal to negative 2 to the fourth power. Negative 32 is equal to negative 2 to the fifth power. So notice, our exponent on the negative 2 is always going to be one more than our index. Our index is 2, our exponent is 3. Our index is 3, our exponent is 4. Our index is 4, our exponent is 5. So that gives a sense that at least the numerator is going to be-- whatever our index is, it's going to be-- so let me write this down. So a sub n is equal to-- well, it's going to be negative 2 to whatever index we're at, to that index plus 1 power. So that's a reasonable way to think about our numerator. Now let's think about our denominators. So over-- So we go from 5, so when n is 2, we're at 5. When n is 3, we're at 7. When n is 4, we're at 9. So notice, 5 is 2 times 2 plus 1. This right over here is 2 times 3 plus 1. This right over here is 2 times 4 plus 1. And you should just kind of play around with different patterns in your head until you say, hey, well, look, this is increasing by 2 every time. Notice, this increases by 2 every time. But these aren't exactly multiples of 2. These seem to be off by one more than the multiples of 2, which is a good sign that this is going to be 2 times our index plus 1. So we could write this as 2 times our index plus 1. And we're done. That's what a sub n is. And if we wanted to write this series in sigma notation, we would write this as the sum from n equals 2 to infinity of negative 2 to the n plus 1 power over 2n plus 1. And that would equal this series right over here.