Question Video: Verifying Whether the Sum of an Infinite Geometric Series Can Be Found or Not | Nagwa Question Video: Verifying Whether the Sum of an Infinite Geometric Series Can Be Found or Not | Nagwa

Question Video: Verifying Whether the Sum of an Infinite Geometric Series Can Be Found or Not Mathematics

Can the sum of an infinite geometric series be found for any value of π‘Ÿ satisfying |π‘Ÿ| ≀ 1?

02:53

Video Transcript

Can the sum of an infinite geometric series be found for any value of π‘Ÿ satisfying the absolute value of π‘Ÿ is less than or equal to one?

Let’s firstly remind ourselves of what a geometric series is. It’s a series which has a constant ratio between successive terms, such as the series one over two, one over eight, one over 32, one over 128, and so on, where we multiply by one over four to get the next term. So, one over four is called the common ratio of this series. The general form of a geometric series is π‘Ž, π‘Žπ‘Ÿ, π‘Žπ‘Ÿ squared, π‘Žπ‘Ÿ cubed, and so on, where π‘Ž is the first term and π‘Ÿ is the common ratio.

We can see that to get to the next term, we multiply by π‘Ÿ. To find the sum of this geometric series, we would add all the terms together. We’ve been asked whether you can find the sum of an infinite geometric series when the absolute value of π‘Ÿ is less than or equal to one. In other words, is it possible to find the sum of an infinite geometric series when π‘Ÿ is greater than or equal to negative one but less than or equal to one?

Let’s firstly think about π‘Ÿ between negative one and one but not equal to negative one or one. Then, the sum of the infinite geometric series looks like this. And π‘Ÿ will be small but π‘Ÿ squared will be even smaller, and π‘Ÿ cubed will be even smaller still and so on. So, each term being added will get smaller and smaller each time. And so, this sum will approach a specific value.

So, we are able to find the sum of an infinite geometric series for π‘Ÿ between negative one and one. But what if π‘Ÿ is equal to one? Then, with the substitution of π‘Ÿ equals one, the sum would look like this. And of course, one raised to any power is just going to give us one. So, we’ll just be adding π‘Ž each time. And because this is an infinite geometric series, this will increase with no bound. It’s not approaching any specific value. So, we can’t find the sum of an infinite geometric series when π‘Ÿ is equal to one.

And finally, if π‘Ÿ is equal to negative one, then with the substitution of π‘Ÿ equals negative one, the sum would look like this. We then remember, if we raise negative one to an even power, we get one. And if we raise negative one to an odd power, we get negative one. So, when π‘Ÿ is negative one, the sum of the infinite geometric series oscillates between being zero and π‘Ž. So, as this sum doesn’t approach a specific value, we can’t find the sum when π‘Ÿ equals negative one. So, to answer the question, can we have π‘Ÿ greater than or equal to negative one but less than or equal to one? In other words, can we have the absolute value of π‘Ÿ less than or equal to one? The answer is no.

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