Recursively Defined Sequences
In Exercises 101–108, assu...

Question

Recursively Defined Sequences

In Exercises 101–108, assume that each sequence converges and find its limit.

a₁ = 5,aₙ₊₁ = √(5aₙ)

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Let C1 be equal to 3 and C subscript N + 1 be equal to the square root of 3 multiplied by CN. If the sequence converges, what is its limit? OK. So it appears for this particular prompt we are asked to take all the information that is provided to us and we are asked to determine that if this particular sequence converges and if it does converge we are asked to determine what its limit is and that is what we are ultimately trying to solve for. So if we determine that the sequence converges, we will need to determine what the limit is. So now that we know what we're ultimately trying to solve for. What we need to consider first is we need to note based on what we're told by the prom itself and what we should recall from our prior knowledge, we will need to solve this particular problem analytically. We will first need to establish that the sequence converges by using the monotone convergence theorem. And this will require us to prove that the sequence is monotonic, meaning it's non-discreasing. And because it's non-decreasing, and we need to note that it's bounded above. So once again, we need to prove that the sequence will be monotonic, meaning it's non-decreasing and is bounded above, and after we prove convergence exists, we will need to solve for the limit L. By substituting the limit property back into our reoccurrence relation. So this is a rough overview of what we're going to do to solve this problem. So keeping that overview in mind, our first step is to focus our attention on determining proof of convergence. So what we need to do is we need to recall and use algebraic properties in order to prove whether or not the sequence is both bounded and monotonic. Focusing our attention on boundedness, we need to observe that C1 CC subscript 1, which I'll call C1 for short, is equal to 3. And if we assume a term CN C subscript N is less than or equal to 3. That will mean that the next term C subscript N plus 1 will be equal to the square root of 3 multiplied by cn, and this must be less than or equal to the square root of 3 multiplied by 3 in parentheses, which will be 3, fantastic. So which is 3. You know, therefore, without a doubt, we can say that the sequence is bounded above by 3. Now we need to consider monoticity, so we need to compare C subscript N + 1 to Cn by examining their squares. So we need to take C to the power of 2 subscript N + 1. Is equal to 3 multiplied by CN. And we need to note that if songs again. If CN is less than or equal to 3, that will mean then that 3 multiplied by CN is greater than or equal to c to the power of 2 subscript n. And we achieve this by multiplying both sides of 3 is greater than or equal to cn by CN. So once again, we're multiplying both sides by 3 is greater than or equal to cn by CN. And we'll put that in parentheses because that's a note to ourselves. So therefore we can go ahead and write that c to the power of 2 subscript n + 1 is greater than or equal to c 2 subscript n, which will mean that c subscript n + 1 is greater than or equal to cn. And this proves that the sequence is non-decreasing for all n, which will mean, in conclusion, since the sequence is indeed monotonic and is bounded above, as a result, it must converge to a finite limit l according to the monotone convergence theorem. Moving on to our 2nd step, we now need to perform our analytical evaluation of the limit itself. And because the sequence converges, that means as a result, the limit of cn as n approaches infinity will be the same as the limit of C subscript n + 1. So let us state that the limit as n approaches infinity of cn will be equal to l. And when we take the limit of both sides of C subscript N + 1 is equal to the square root of 3 multiplied by CN, we will find that we will receive a result of L is equal to the square root of 3 multiplied by L. And now we need to rearrange this expression to isolate and solve for L. So when we begin to do that, we will have L2 equal to 3 multiplied by L. Which will mean that when we isolate in sulfur L, or rather when we set our expression equal to 0, we will get that L to the power of 2 minus 3L is equal to 0, and when we factor out our common variable, which is L, we will get L multiplied by L minus 3. In parentheses is equal to 0, so that will mean, as a result, the algebraic roots will be L is equal to 0 and L is equal to 3. So now our third step is we need to provide justification for L is equal to 3. So considering our initial condition first, we need to note that the sequence starts at C1 is equal to 3, so this is our start point. So with that in mind, we now need to consider the direction of growth. So we proved in step one that the sequence is non-decreasing. So that means we've established that C subscript n + 1 is greater than or equal to cn. Which will mean that we can consider the following exclusion. A sequence that starts at 3, that never decreases, can never reach or approach 0, so that means as a result, L equals 0 is an extraneous solution, so that's why it doesn't count. So I'll put an X next to L equals 0. So that means the only limit that is consistent with the sequence's analytical properties is L is equal to 3. So that's our only correct answer. So without a doubt, our correct and final answer will be L is equal to 3. Hooray, we did it. So therefore, without a doubt, the sequence converges to the limit L equals 3. So going back to look at the prom itself and to recap what we've learned, we could safely say without a doubt that our final answer for the limit is L is equal to 3, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Recursively Defined SequencesIn Exercises 101–108, assume that each sequence converges and find its limit.a₁ = 5,aₙ₊₁ = √(5aₙ)

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Recursively Defined Sequences
In Exercises 101–108, assume that each sequence converges and find its limit.
a₁ = 5,aₙ₊₁ = √(5aₙ)