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## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 6: Comparison tests

# Worked example: direct comparison test

Using the direct comparison test to determine that the infinite sum of 1/(2ⁿ+n) converges by comparing it to the infinite sum 1/2ⁿ.

## Video transcript

- [Voiceover] Let's think about the infinite series, so we're going to go from n equals one to infinity, of one over two to the n plus n. And what I want to do is see if we can prove whether this thing converges or diverges. And as you can imagine based on the context of where this video shows up on Khan Academy that maybe we will do it using the comparison test. At any point if you feel like you can kinda take this to the finish line, feel free to pause the video and do so. So in order to kind of figure out or get a sense for this series right over here, it never hurts to kind of expand it out a little bit, so let's do that. So this would be equal to when n equals one this is gonna be one over two to the one plus one, so it's gonna be one over two plus one, it's gonna be one third plus that's n equals one. When n equals two it's gonna be one over two squared, which is four plus two plus one over six. Plus let's see we go to three, n equals one, n equals two, n equals three is gonna be one over two to the third, which is eight plus three is 11. So one over 11, maybe I'll do one more term. Two the fourth power is going to be 16 plus four is 20. Plus one over 20 and obviously we just keep going on and on and on. So it looks, it feels like this thing could converge. All of our terms are positive, but they are getting smaller and smaller quite fast. And if we really look at the behavior of the terms as n gets larger and larger, we see that the two to the n in the denominator will grow much, much faster than the n will. So this kind of behaves like one over two to the n, which is a clue of something that we might be able to use for the comparison test. So let me just write that down. So we have one over, so the infinite series from n equals one to infinity of one over two to the n, and so when n equals one this is going to be equal to one half. When n is equal to two this is going to be equal to one fourth. When n is equal to three this is equal to one eighth. When n is equal to four this is equal to one sixteenth. And we go on and on and on and on. And what's interesting about this is we recognize it. This is a geometric series, so let me be clear. This thing right over here, that is the same thing as the sum of from n equals one to infinity of one half to the n power, just writing it in a different way. And since the absolute value of one half, which is just one half, so because the absolute value of one half is less than one we know that this geometric series converges, we know that it converges And actually we even have formulas for finding the exact sum of or to figure out what it converges to. And so we know this thing converges and we see that actually these two series combined meet all of the constraints we need for the comparison test. So let's go back to what we wrote about the comparison test. So the comparison test, we have two series, all of their terms are greater than or equal to zero. All of these terms are greater than or equal to zero. And then for the corresponding terms in one series, all of them are going to be less than or equal to the corresponding terms in the next one. And so if we look over here, we can consider this one, the magenta series, this is kind of our infinite series of dealing with A sub n and that this right over here is, well I already did it in blue this is kind of the blue series. And notice all their terms are nonnegative and the corresponding terms one half is greater than one third, one fourth is greater than one sixth, one eighth is greater than one eleventh. One over two to the n is always going to be greater than one over two to the n plus n for the n that we care about here. And since we know that this converges, since we know that the larger one converges. It's a geometric series where the common ratio the absolute value of the common ratio is less than one, since we know that the larger series converges therefore, the smaller series or the one where every corresponding term is less than the one in the blue one, that one must also converge. So by the comparison test the series in question must also converge.