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# Converting explicit series terms to summation notation

## Video transcript

So I have the series-- negative 5/3 plus 25 over 6 minus 125 over 9 plus-- and it just keeps going on and on and on forever. So this right over here is an infinite sum or an infinite series, and what I want you to do right now is to pause this video and try to express this infinite series using sigma notation. So I'm assuming you've given a go at it, so let's just look at each term of the series and let's see if we can express it with kind of an ever-increasing index. So the first thing that might jump out at you is this oscillating sign that's happening right over here. And whenever we see an oscillating sign, that's a pretty good idea that we could kind of view this as negative 1 to the n-th, where n is our index. So for example, that right over there is negative 1 to the first power. That is-- this right over here is negative 1 to the second power. That right over there is negative 1 to the third power. So it looks like the sign is being defined by raising negative 1 to the index. Now let's look at the other parts of these terms right over here. So we have 5, then we have 25, the we have 125, so these are the powers of 5. So this right over here is 5 to the first power, this right over here is 5 to the second power, this right over here is 5 to the third power. So this part, we're raising 5 to our index. Notice, 1, 1, 2, 2, 3, 3. And then finally, let's look at this. We have 3, 6, and 9. So this literally-- if our index here is 1, this is 3 times 1. If our index here is 2, this is 3 times 2. If our index here is 3, this is 3 times 3. So this is 3 times 1, that is 3 times 2-- let me write it this way-- 3 times 2, that right over there is 3 times 3. So this sets us up pretty well to write this in sigma notation. So let's write it over here just so we can compare. So let me give myself some real estate to work with. So we could write this as the sum-- I'll do it in yellow-- as the sum, so this is our sigma. We can start our index n at 1, from n equals 1, and we're going to keep going on and on forever. We just keep going on and on forever. And so it's negative 1 to the n-th power times 5 to the n-th over-- notice 5 to the n-th-- over 3n is going to be equal to this. And you can verify when n equals 1, it's negative 1 to the nth power-- I'm sorry-- negative 1 to the first power, which is negative 1 times 5 to the first power, which is 5 over 3 times 1. And we can do that for each successive term. And so we're all done.