Geometric Series: Sum of x^n series = 1/(1 - x) if |x| is less than 1

Geometric Series: Sum of x^n series = 1/(1 - x) if |x| is less than 1 by @mes on 3Speak

In this video, I show that the infinite series x n = x 0 + x 1 + x 2 + ... is just a geometric series, and converges if the absolute value of x is less than 1. By common convention, x 0 = 1, so the...

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Transcript

Example 5, and it states, find the sum of the series, and we're given this one here, this one's sigma or summation from n equals 0, so starting from n equals 0 now, to infinity of x the power of n, where absolute value of x is less than 1. In other words, we're getting a geometric series, just in a slightly different format, that's convergent, this is less than 1. So let's look at the solution. So notice that this series starts with n equals 0, and so the first term is x to the power of 0 equals 1. Again, this is by definition. So important note that with this series, or that with series in general, we adopt a convention that x of 0 equals to 1, even when x is 0. So yes, if you have 0 to the power of 0, we're defining it as 1. And yes, no, check out one of my first videos exploring the powers of 0, such as x the power of 0, and 0 to power of 0. And yeah, they're undefined, but for convention, it's usually defined as equaling to 1. Yeah, so thus we have, we were to write this series out, and we'll show that it becomes a geometric series. So let's write this out, x to power of n. So we have, like this, yeah, just write it out, n equals to 0, up to infinity like that, x to the n, so we just write these terms out, so for n equals 0, by definition, again, or convention, x to the power of 0 is just 1. plus, then we have x to the 1, and then plus x to the 2, plus x3, and so on, plus x4, plus keeps going on and on, and notice here, so we have our a is here, a, and now these are separated by, well, x times x equals to x squared, and likewise here, x squared, x squared, so in other words, this is our r, is x. Yeah, so this is a geometric series with A equals 1 and R equals X, and since the absolute value of R is less than, I mean, equals to, so the absolute value of R, which is equal to epsilon of x, is less than 1, it converges, and from example 1 gives, again the summation, S equals 2, summation right here, infinity, and it starts off a 0 by the first time. and then we have x to the n equals 2 just a over 1 minus r where a is equal to 1 r is equal to x so 1 minus x this is a summation summation for x is absolute value of x less than 1

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