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Numerade Educator



Problem 22 Easy Difficulty

Calculate, to four decimal places, the first ten terms of the sequence and use them to plot the graph of the sequence by hand. Does the sequence appear to have a limit? If so, calculate it. If not, explain why.
$ a_n = 1 + \frac{10^n}{9^n} $




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Video Transcript

Let's find up to four decimal places the first ten terms of the sequence. So it's a one a two, all the way up to a ten. So I'll go to a calculator and plug this in and goto four decimal places so you can see. Here's the table. Here's our and one plus ten end over nine to the end and five digits. So that means we're going for places after the decimal and you could pause the screen and go ahead and record these values. So I'll go back to the original and then used them to plot a graph by hand so you could use the table to drop the graph or come over here to. Does Mose hear about already planted the graph? Here's the formula for Anne and then here's the graph of it, and we could see the graph is increasing and it looks like this increasing at an increasing rate. So does the sequence appear to have a limit? I say No. It appears to diverge. No, it's actually explain why. So let's look at and which we can write is one plus ten over nine to the end. Now I claim that the limit is n goes to infinity of a n is infinity Since the limit as n goes to infinity of ten over nine to the end equals infinity Now, any time you take a number bigger than one and you keep multiplying by itself, you're going to get larger. But if you like more convincing argument of this last fast over here, what you can do is the following. First of all, it's maybe use X here instead of the end. You have ten over nine to the ex. You can rewrite this as e to the X natural log ten over nine. So here I'm just using the fact that we can write any number. Why is eat to the Ellen? Why? Due to the fact that he and Eleanor in Versace and now this is just e infinity Ellen ten over nine. But this is just either the infinity since ln ten over nine is positive and eat of infinity equals infinity. Therefore, since Ann is obtained by adding one to this term here, we've also shown and never just to infinity. So the sequence appears to diverge. And then we just proved that the sequence and diverges So that's your final answer