Question Video: Identifying Whether a Given Geometric Series Is Convergent or Divergent | Nagwa Question Video: Identifying Whether a Given Geometric Series Is Convergent or Divergent | Nagwa

# Question Video: Identifying Whether a Given Geometric Series Is Convergent or Divergent Mathematics • Higher Education

Is the series 884 + 884/9 + 884/81 + ＿ convergent or divergent?

04:11

### Video Transcript

Is the series 884 plus 884 over nine plus 884 over 81 plus so on and so on convergent or divergent?

Well first, let’s think about our series. Our first term is 884, second term is 884 over nine, and the third term is 884 over 81. Well I could rewrite the first term as 884 over one, and then we can notice that each time the denominator is getting multiplied by nine. One times nine is nine. Nine times nine is 81. Well this is basically the pattern you get when you’re looking at a geometric series.

If we call our first term 𝑎, then we can see that our first term is 884 over one or just 884, and the common ratio 𝑟, but we’re multiplying the denominator by nine each time. So we’re actually multiplying the whole term by one over nine, a ninth. So we’ve got a geometric series with a first term of 884 and a common ratio of one-ninth.

Now the word convergent in this context means that if we were to sum up an infinite number of terms in that series, then that sum is not gonna exceed a certain value. And the word divergent in this context means that if we sum up an infinite number of terms in our series, then that sum is just gonna get bigger and bigger and bigger the more terms we add. So we’re interested in summing terms of series. Now if we had 𝑛 terms in our series and we were to sum them up for this geometric series, the sum of 𝑛 terms 𝑆𝑛 is equal to the first term 𝑎 times one minus the common ratio 𝑟 to the power of 𝑛 all over one minus the common ratio.

So in our case, the sum of 𝑛 terms is gonna be 884 times one minus a ninth to the power of 𝑛 all over one minus a ninth. Well one minus a ninth is eight-ninths. The denominator of eight-ninths, that means the denominator divided by eight-ninths which we can rearrange like this. Now nine times 884 is 7956. And the 7956 and the eight actually cancel down as well. So the sum of 𝑛 terms is 1989 times one minus a ninth to the power of 𝑛 all over two.

And the big question is what happens as 𝑛 tends to infinity as we get an infinite number of terms in our sequence, what would the sum to infinity become? Well the more times that we multiply a ninth by itself, the closer to zero that gets. So this term here as 𝑛 becomes infinite is gonna become closer and closer and closer to zero. So this term here one minus zero is just gonna become one. So our sum to infinity is gonna get closer and closer to 1989 over two. That means it’s gonna be a convergent series.

So that’s our answer, convergent series. Now another way of looking at this with geometric series is to just think about the value of 𝑟. Because of the format of this sum of 𝑛 terms of a geometric series, if 𝑟 is between negative one and positive one, then it’s definitely going to be a convergent series. But if 𝑟 is less than negative one or 𝑟 is greater than one, then it’s going to be divergent.

In fact, if 𝑟 is equal to one, then it’s also gonna be divergent. And it’s gonna be a sort of an oscillating sum if 𝑟 is equal to negative one. Because you’re going to be depending on how many terms you take, you’re gonna be taking term and then subtracting that same term and then adding that term and then subtracting the same term. Anyway in this case, 𝑟 is clearly in between negative one and positive one, so it’s a convergent series.

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