How to find the limit of a sequence an

[mae_tfb015]

The question is this:

The sequence {a_n} is defined by a₁ = 2, and


a_n+1= 1/2(a_n +2/a_n),

for n is greater than or equal to 1. Assuming that {a_n} converges, find its limit.


Hint: let a = lim a_n (where n goes to infinity). Then, since a_n+1 = 1/2(a_n + 2/a_n), we have a = 1/2(a + 2/a). Now solve for a.

 

[Zermelo]

Well, you just have to solve the equation. Multiply both skdes by 2, and then by a. The only problem with this is that you cant really be sure that a exists, you have to prove that the sequence is convergent first. Try proving that its bounded and monotonous

 

[pka]

The question is this: The sequence {a_n} is defined by a₁ = 2, and
a_n+1= 1/2(a_n +2/a_n),
for n is greater than or equal to 1. Assuming that {a_n} converges, find its limit.

If you have \(L=\dfrac{1}{2}\left(L+\dfrac{2}{L}\right)\) could you solve for \(L~?\)

 

[Dr.Peterson]

Well, you just have to solve the equation. Multiply both sides by 2, and then by a. The only problem with this is that you cant really be sure that a exists, you have to prove that the sequence is convergent first. Try proving that its bounded and monotonous

Interestingly, the problem says "Assuming that {a_n} converges, find its limit." So they are asking the student to bypass the important math and do the easy part!

 

[HallsofIvy]

Actually, it is not a bad idea to start by assuming such a sequence converges to find a "possible limit" then use that to prove that it does converge!

Here we given that \(\displaystyle a_{n+1}= (1/2)(a_n+ 2/a_n)\).

Assuming that it does converge, to limit L, and taking the limit of both sides, we have \(\displaystyle \lim_{n\to\infty} a_{n+1}= (1/2)(\lim_{n\to\infty} a_n+ 2/\lim_{n\to\infty} a_n)\) and, replacing each limit with L,
\(\displaystyle L= (1/2)(L+ 2/L)\) as pka said.

Multiplying both sides by 2L, we have \(\displaystyle 2L^2= L^2+ 2\) or \(\displaystyle L^2= 2\). The roots of that equation are \(\displaystyle L= \pm\sqrt{2}\) but here, the initial value was \(\displaystyle a_1= 2\) so that every term in \(\displaystyle a_{n+1}= (1/2)(a_n+ 2/a_n)\) is always positive. L must be positive so IF THE SEQUENCE CONVERGES, it converges to \(\displaystyle \sqrt{2}\).

The second part of the problem is then to show that this sequence DOES converge to \(\displaystyle \sqrt{2}\).

 

[mae_tfb015]

Actually, it is not a bad idea to start by assuming such a sequence converges to find a "possible limit" then use that to prove that it does converge!

Here we given that \(\displaystyle a_{n+1}= (1/2)(a_n+ 2/a_n)\).

Assuming that it does converge, to limit L, and taking the limit of both sides, we have \(\displaystyle \lim_{n\to\infty} a_{n+1}= (1/2)(\lim_{n\to\infty} a_n+ 2/\lim_{n\to\infty} a_n)\) and, replacing each limit with L,
\(\displaystyle L= (1/2)(L+ 2/L)\) as pka said.

Multiplying both sides by 2L, we have \(\displaystyle 2L^2= L^2+ 2\) or \(\displaystyle L^2= 2\). The roots of that equation are \(\displaystyle L= \pm\sqrt{2}\) but here, the initial value was \(\displaystyle a_1= 2\) so that every term in \(\displaystyle a_{n+1}= (1/2)(a_n+ 2/a_n)\) is always positive. L must be positive so IF THE SEQUENCE CONVERGES, it converges to \(\displaystyle \sqrt{2}\).

The second part of the problem is then to show that this sequence DOES converge to \(\displaystyle \sqrt{2}\).

I'm sorry if I'm slow, but can you please explain why we have to multiply it by 2L on both sides? And how would you, in general, know that you have to multiply or divide a situation like this one by something in order to find the answer?

 

[Deleted member 4993]

L=(1/2)(L+2/L)L=(1/2)(L+2/L)\displaystyle L= (1/2)(L+ 2/L)

@mae_tfb015

You ask:

why we have to multiply it by 2L on both sides?

Note:

L=(1/2)(L+2/L) →

\(\displaystyle L = \frac{1}{2} * (L + \frac{2}{L})\)

\(\displaystyle L = \frac{L^2 + 2*L}{2*L}\)

When you multiply by 2L

the fractional term evaporates, and you get

\(\displaystyle 2 * L^2 = L^2 + 2*L\)

 

[Dr.Peterson]

I'm sorry if I'm slow, but can you please explain why we have to multiply it by 2L on both sides? And how would you, in general, know that you have to multiply or divide a situation like this one by something in order to find the answer?

You can also take it in two steps, as described in #2: multiply first by 2, then by L (which is there called a).

But this is basic algebra: just do, one by one, whatever you see that will help! Multiplying by 2 clears one fraction, and once that is gone, you can see another to clear by multiplying by its denominator. The goal is to get an equation you can solve for L, by whatever means you see.

 

[Steven G]

Algebra is a prerequisite for Calculus

 

[mae_tfb015]

@mae_tfb015

You ask:

why we have to multiply it by 2L on both sides?

Note:

L=(1/2)(L+2/L) →

\(\displaystyle L = \frac{1}{2} * (L + \frac{2}{L})\)

\(\displaystyle L = \frac{L^2 + 2*L}{2*L}\)

When you multiply by 2L

the fractional term evaporates, and you get

\(\displaystyle 2 * L^2 = L^2 + 2*L\)

Ooohh okay I get it now!! ? thank you very much!!?

 

[HallsofIvy]

I'm sorry if I'm slow, but can you please explain why we have to multiply it by 2L on both sides? And how would you, in general, know that you have to multiply or divide a situation like this one by something in order to find the answer?

We don't "have to" but it is a logical thing to do. We have "L=(1/2)(L+2/L)" with a "2" and an "L" in the denominators of fractions, and I really don't like fractions!

We could, of course, multiply by 2 first to get 2L= L+ 2/L which no longer has "1/2" but still has "2/L" so we multiply by L. Or we could do it the other way around, multiply first by L to get L^2= (1/2)(L^2+ 2) and then multiply by 2.

Or, if you like working with fractions, you could have combined the fractions first:
L= (1/2)(L+ 2/L)= (1/2)(L^2/L+ 2/L)= (1/2)(L^2+ 2)/L= (L^2+ 2)/2L.

L- (L^2+ 2)/2L= 2L^2/2L- (L^2+ 2)/2L= (2L^2- L^2- 2)/2L= (L^2- 2L)/2L= 0.

Now, a fraction is equal to 0 if and only if the numerator is 0 so we must have L^2- 2L= 0. (Or multiply both sides by 2L now!)

I just think working with those fractions is more work than necessary if you get rid of them at the beginning.