Prove that the sequence converges to 0

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In summary: Since e_n is increasing and bounded above, it is convergent. Now, let L be the limit of e_n. Then, in the limit as n approaches infinity, e_{n+1} = e_n/(e_n+2) becomes L = L/(L+2). Solving for L, we get L = 0. Thus, the limit of the sequence e_n is 0. In summary, the sequence {e_n} defined inductively as e_{n+1} = e_n/(e_n+2) is proven to be convergent if -1 < e_0 < 0, with a limit of 0. This is shown by proving that the sequence is
  • #1
alexmahone
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e_{n+1} = e_n/(e_n+2)

If -1 < e_0 < 0, prove that the sequence {e_n} converges to 0.

PS: I haven't learned things like sup and inf yet, so please don't use them.
 
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  • #2
Alexmahone said:
e_{n+1} = e_n/(e_n+2)

If -1 < e_0 < 0, prove that the sequence {e_n} converges to 0.

PS: I haven't learned things like sup and inf yet, so please don't use them.

1) Show that the sequence defined inductively is convergent.

2) Let L be the limit of this sequence e_n. Then L will also be the limit of the sequence e_{n+1}. As e_{n+1} = e_n/(e_n+2) it means (in the limit) that L = L/(L+2). Solve for L and show that L=0.
 
  • #3
ThePerfectHacker said:
1) Show that the sequence defined inductively is convergent.

How do I do that? By showing that it is increasing and bounded above?

Let L be the limit of this sequence e_n. Then L will also be the limit of the sequence e_{n+1}. As e_{n+1} = e_n/(e_n+2) it means (in the limit) that L = L/(L+2). Solve for L and show that L=0.

L^2+2L=L

L^2+L=0

L(L+1)=0

L=0 or -1

Clearly, L can't be -1 because e_0 > -1 and the sequence is increasing. So, L = 0.
 
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  • #4
Alexmahone said:
How do I do that? By showing that it is increasing and bounded above?

Exactly.

Let -1 < x < 0, show that -1 < x/(x+2)< 0.
Also show that x < x/(x+2).

Combining these two facts together this means that whatever e_0 is e_1 will be larger and still between -1 and 0. Then by repeating the same argument e_2 will be larger than e_1 and still between -1 and 0. And so forth. Therefore, e_n is increasing and bounded above by 0.
 
  • #5


I understand the importance of providing evidence and proof to support a claim. In this case, the claim is that the sequence {e_n} converges to 0, and we must prove it using the given information.

First, let's rewrite the sequence using index notation:
e_1 = e_0/(e_0+2)
e_2 = e_1/(e_1+2)
e_3 = e_2/(e_2+2)
...
e_n = e_{n-1}/(e_{n-1}+2)

We can see that each term in the sequence is a fraction where the numerator is equal to the previous term and the denominator is the previous term plus 2. This means that as n increases, the denominator will also increase, making the fraction smaller and approaching 0.

Now, let's look at the given information: -1 < e_0 < 0. This means that the initial term, e_0, is a negative number between -1 and 0. When we plug this into the sequence, we get a negative fraction, which will always be smaller than the previous term. This shows that the sequence is decreasing.

Furthermore, since the sequence is decreasing and the denominator is increasing, the sequence is bounded below by 0. This means that the sequence will not go below 0, and as n increases, it will get closer and closer to 0.

Therefore, we can conclude that the sequence {e_n} converges to 0, as the terms get smaller and approach 0, and the sequence is bounded below by 0. This is supported by the given information of -1 < e_0 < 0.
 

FAQ: Prove that the sequence converges to 0

What does it mean for a sequence to converge to 0?

When a sequence converges to 0, it means that the terms of the sequence get closer and closer to 0 as the sequence progresses. In other words, as we continue to add more terms to the sequence, the difference between each term and 0 becomes smaller and smaller until it eventually becomes infinitesimally small.

How can we prove that a sequence converges to 0?

There are a few different methods for proving that a sequence converges to 0. One way is to show that the terms of the sequence get arbitrarily close to 0 as the sequence progresses. Another way is to use the epsilon-N definition of convergence, where we can choose any small positive number (epsilon) and find a point in the sequence where all subsequent terms are within epsilon of 0.

What is the importance of proving that a sequence converges to 0?

Proving that a sequence converges to 0 is important because it allows us to make conclusions about the behavior of the sequence as a whole. For example, if a sequence converges to 0, we can say that the terms are getting smaller and smaller and that the sequence is approaching 0 as its limit. This can have implications in various fields of science, such as physics, engineering, and economics.

Can a sequence converge to 0 from both above and below?

Yes, a sequence can converge to 0 from both above and below. This means that the terms of the sequence can approach 0 from either positive or negative values. For example, the sequence (1, -0.5, 0.25, -0.125, ...) converges to 0 from above, while the sequence (-1, 0.5, -0.25, 0.125, ...) converges to 0 from below.

Are there any common types of sequences that always converge to 0?

Yes, there are a few common types of sequences that always converge to 0. These include geometric sequences with a common ratio between -1 and 1, harmonic sequences (where each term is the reciprocal of the natural numbers), and telescoping sequences (where the terms cancel each other out as the sequence progresses). However, it is important to note that not all sequences of these types will converge to 0, so it is still necessary to prove convergence in each specific case.

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