I need to study the limit behavior of xn+1=xnxn+1,xo=1 and if the limit exists, compute the limit. I observed the first few terms and it seemed that the sequence was decreasing so I decided to show that the sequence was monotonic:
Need to show xn+1≤xn which is equivalent to showing xn+1−xn≤0
xn+1−xn=xnxn+1−xn=xn−xn(xn+1)xn+1=−(xn)2xn+1=−(xn)2xn+1≤0 for all n if xn≥0 which can be proved by induction:
xo=1≥0x1=1/2≥0
xn=>xn+1
xn+1=xnxn+1>1xn+1>0 because xn>0. (I don't know if that was the proper way to do the induction. Any confirmation?)
Since it was proved that xn≥0, xn+1−xn=−(xn)2xn+1≤0, thus the sequence is monotone and decreasing. The sequence is also bounded:
Since the sequence is decreasing it is bounded above by 1, and because xn≥0 the sequence id bounded below by 0.
The boundedness and monotonicity of the sequence implies that a limit exists:
Let lim. Because \lim x_n=\lim x_{n+1}, x=\frac{x}{x+1}<=>x(x+1)=x<=>x+1=1=>x=0 So 0 is the limit.
I'm not sure if there are problem in the work that I did and any help would be greatly appreciated.
I don't know if I should start a new question for this, so I've included it here anyways:
Since one was able to tell that the \lim x_n=\lim \frac{1}{1+n} for the above sequence, should one try to do the same thing with the sequence x_{n+1}= \frac{(x_n)^2}{x_n+1}, x_o=1?
Answer
There’s a small error in your proof that x_n\ge 0 for all n: it’s not true that
\frac{x_n}{x_n+1}>\frac1{x_n+1}\;,
since in fact it turns out that x_n\le 1 for all n. However, given the induction hypothesis that x_n\ge 0, you certainly have\
x_{n+1}=\frac{x_n}{x_n+1}\ge 0\;,
which is all you need here. Otherwise it looks fine.
Note that if you calculate the first few values, you find that x_1=\frac12, x_2=\frac13, and x_3=\frac14, suggesting that in general x_n=\frac1{n+1}. An alternative approach would be to show by induction that this is true for all n\ge 0; this is not at all difficult.
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