Prove that the sequence converges to 0

[Alexmahone]

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 learnt things like sup and inf yet, so please don't use them.

 

[ThePerfectHacker]

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 learnt 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.

 

[Alexmahone]

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.

 

[ThePerfectHacker]

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.