If there is a recursively defined sequence such that

Question

If there is a recursively defined sequence such that

a₁ = sqrt(2)
a_{n + 1} = sqrt(2 + a_n)

Show that a_n < 2 for all n ≥ 1

Answer 1

a1 = √2

a2 = √(2 + √2)
a3 = √(2 + √(2 + √2))
I see a pattern

an = √( 2 + √(2 + ...

let x = √( 2 + √(2 + ...
square both sides, that will drop the left-most √ on the right side

x^2 = 2 + √( 2 + √(2 + ...
x^2 - 2 = √( 2 + √(2 + ...
but the right side is what I originally defined as x
so
x^2 - 2 = x
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x=-1
(clearly x = -1 is an extraneous solution)

so an must have a limit of 2.
and since n is a finite number, term a_n < 2 .

an interesting loop on the calculator is this

1) enter √2
2) press =
3) plus 2
4) press =
5) press √
6) repeat step 2)

You should get appr 2 correct to 5 decimals after about 10 iterations.

Answer 2

You prove that the sequences converges to 2, but you don't necessarily prove that it will never exceed 2...

Thanks for the great help though

Answer 3

I proved that its limit is 2

i.e. it will actually never reach 2, only if I take all of its infinite terms.
so clearly it can never exceed 2 if it can't reach 2.

Answer 4

You are absolutely correct. Thanks so much for the help on this!

I'd love to pay you guys back for this service :)

Answer 5

The only thing missing here is that you must show that the series is monotonously increasing, i.e. that

a_{n+1} > a_{n} (that should be easy). This, combined with the fact that the limit is 2 proves that a_n < 2.

If you have a sequence that is not monotonously increasing (or decreasing), it isn't necesarily true that the limiting value isn't attained for finite n. Take e.g. the sequence

sin(pi n/7)/n

which tends to zero, but it is also equal to zero for n a multiple of 7.

Answer 6

Brianna arrived at county fair at 10:25. she left at 11:40. how long was she at the fair?

Answer 7

To prove that a_n < 2 for all n ≥ 1, we can use mathematical induction.

Step 1: Base Case (n = 1)
For the base case, we need to show that a₁ = sqrt(2) < 2. Since the square root of 2 is approximately 1.414, which is less than 2, the base case holds true.

Step 2: Inductive Hypothesis
Assume that for some positive integer k ≥ 1, a_k < 2 is true.

Step 3: Inductive Step
We need to show that a_k+1 < 2 is also true based on the assumption in Step 2.

Using the recursive formula, we have:
a_k+1 = sqrt(2 + a_k)

Since a_k < 2 (based on the inductive hypothesis), we can substitute a_k < 2 into the formula:
a_k+1 = sqrt(2 + a_k) < sqrt(2 + 2) = sqrt(4) = 2

Therefore, a_k+1 < 2 is true.

Step 4: Conclusion
By mathematical induction, we have shown that a_n < 2 for all n ≥ 1.

If there is a recursively defined sequence such that

a1 = √2

You prove that the sequences converges to 2, but you don't necessarily prove that it will never exceed 2...

I proved that its limit is 2

You are absolutely correct. Thanks so much for the help on this!

The only thing missing here is that you must show that the series is monotonously increasing, i.e. that

Brianna arrived at county fair at 10:25. she left at 11:40. how long was she at the fair?

To prove that an < 2 for all n ≥ 1, we can use mathematical induction.

To prove that a_n < 2 for all n ≥ 1, we can use mathematical induction.