Sequences

Jan 2009
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Determine whether the following sequence is convergent. The question and my attempt to the solution is in the PFD l attached to the thread
 

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Oct 2009
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Determine whether the following sequence is convergent. The question and my attempt to the solution is in the PFD l attached to the thread

I'm not sure I understand what you did, though it looks really bad: you wrote \(\displaystyle ...\frac{12+a_n^3}{5}<\frac{10^3+12}{5}\) and then from here, magically (at least for me), you deduce that \(\displaystyle a_{n+1}>10\) ...(Wondering)(Thinking)

You wrote that your sequence \(\displaystyle a_1:=1\,,\,a_{n+1}=\frac{12+a_n^3}{5}\) is monotone increasing, though I can't see proof of this (it, though, is true, imo), so then it converges iff it is bounded.

Now, you can argue as follows: suppose the seq. is bounded and thus it converges (to a finite limit), say \(\displaystyle a_n\xrightarrow [n\to\infty]{}\alpha\) , then from the definition of the seq. and arithmetic of limits we get:

\(\displaystyle \alpha=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{12+a_n^3}{5}=\frac{12+\alpha^3}{5}\iff \alpha^3-5\alpha+12=0\) . But the only real root of this cubic is \(\displaystyle -3\) (why?) and since the seq. is positive (why?)

we get a contradiction.

Tonio