Evaluating a limit from a recursive sequence

How do we find the limit of a sequence if we are given the recursive formula? Note: this method might not always work. We have to know if the sequence converges or not first. This question can be for precalculus or calculus 2.

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#justcalculus

[Vladivostok]

Any chance we will get some Calc 3 on this channel? Thanks for all the content man

[ricardoguzman]

The red fractions in the denominator also converge to the square root of 2. 3/2, 7/5, 17/12, 41/29, ... (√2 + 1)^n = a + b√2----> this formula also produces the coefficients of the fractions in the denominator which are a/b. Also, notice the sequence of integers in the fractions. The sum of numerator and denominator in each fraction is the denominator of the next fraction, and the sum of the numerator and twice the denominator in each fraction becomes the numerator of the following fraction.
Example: 3/2--3 +2 = 5, so 5 is the denominator of the next fraction which is 7/5. 3 + 2x2 = 7, which is the numerator of the next fraction.
Finally, the sequence produces all numbers that are both simultaneously square and triangular. The first fraction is technically 1/1. Now 1^2 x 1^2 = 1, which is the first number that is both square and triangular. The second fraction is 3/2, so we get 3^2 x 2^2 = 9 x 4 = 36, which is indeed the second number that is simultaneously square and triangular. Third fraction, 7/5. 7^2 x 5^2 = 49x25=1, 225 which is the third one. Etc. Very cool sequence you picked.

[todr]

Thanks for the help. I really had no idea how to solve this problem.

[EngMorvan]

The negative solution for L²=2 is for the limit when n tends to minus infinity.

It's easy to see that if you know the formula for the general term a_n:
a_n =

And the process to find the above formula could be a nice video as well. 😉

[nitayweksler]

How do you prove that it conv tho? I know how to do it for a series but for this one its wierd cus i cant say wether an+1< or > an for n>n0

[jorgelenny]

If we assume that the limit exists, we can simply solve for x = 1 + 1/(x+1) => x - 1 = 1/(x+1) => (x-1)(x+1) = 1 => x^2 - 1 = 1 => x^2 = 2
Now then the question is whether there is any starting a_1 for which the limit approaches the negative branch of sqrt2

[axeldaliramirezgonzalez]

you're such a genius always helping me with my doubts

[NewtonMD]

Coincidentally, we had the limit of a recursive sequence on my further maths test today. But it was way duckin harder

[imtiazursyed]

Math is beautiful. You proved it one more time. This proof is genius. Love it.

[jaywyn]

I was lost trying to figure this one out. Perfect explanation. Well done.

[sujitsivadanam]

Just by the nature of this recursive definition, you can see that "infinity" is not even possible, because the left hand side would approach infinity while the right hand side will approach 1, which is a clear contradiction.

[kepler]

Something my teacher in school said is that when we try to calculate recursive sequences, we take the U(n+1) as a function f(Un+1)= f(x) and then solve for f(x)=x.
I tried it on this question and it gave 2 answers, sqr(2) and -sqr(2)

[saharhaimyaccov]

Can u use more video's equation like this?

[ChrisKoyo]

Another technique (my favorite) is to express the sequence in terms of n. Then solving like a normal function.

[JayTemple]

This is similar to how I worked out the asymptotic limit of the ratios of consecutive numbers in the Fibonacci sequence, although I didn't know to prove that there WAS a limit.

[panPetrff]

When I noticed the values a(n) oscillated around the resulting limits, I tried to express the members of the sequence using an alternating series:
a(n)= 1 + ( 1/2 - 1/10 + 1/60 - 1/348 + 1/2030 - . . .1/M(n-1)) = 1 + SUM_(k=1)^(n-1) (-1)^(k+1)/M(k) ...for n>1; a(1)=1
where M(k) can be expressed from denominators in the fractions: 1/1, 3/2, 7/5, 17/12, 41/29 ==> 1*2, 2*5, 5*12, 12*29....
M(k) = 1/8*((1+√2)^(2k+1) + (1-√2)^(2k+1) - 2*(-1)^k)
How to prove that members in the series have to be integers reciprocal ?

[pneujai]

wow another nice way to calculate sqrt2

[맹맛초코]

But how can we proof that the sequence converges?

[tayserbinjafor]

The common term should have either sqrt(2)+1 or sqrt(2)-1 for this sequence.

[gogo-pjlm]

Could show the convergence of odd and even subsequences first, then show two subsequences converge to the same limit, and hence the whole sequence converges.