3
$\begingroup$

I am preparing for a test and wanted to ask you

$a_0 = 1; a_{n+1} = \sqrt{a_n} + \frac{15}{4} $

I already showed its strictly monotonically increasing. Now im trying to calculate the limit.

$$\lim a_{n+1} = \lim a_n \Leftrightarrow a = \sqrt{a} + \frac{15}{4} \Leftrightarrow a = (a- \frac{15}{4})^2 \Leftrightarrow 0 = a^2 - \frac{17a}{2} + \frac{225}{16}$$

$$\Longrightarrow a_1 = 2.25 , a_2 = 6.25$$ So you basically take the first limit $a_1 = 2.25$ . Is that correct? Is there better way of calculating the limit? Thank you

$\endgroup$
  • 1
    $\begingroup$ Everything looks good except for the last statement. Each term is no less than $\frac{15}{4}=3.75$, so the limit can't be $2.25$. $\endgroup$ – Nick D. Feb 19 '17 at 20:21
  • $\begingroup$ Thats what happens if you do too much maths... Thank you $\endgroup$ – user391105 Feb 19 '17 at 20:22
  • 2
    $\begingroup$ Isn't showing an upper bound for the sequence required? $\endgroup$ – rookie Feb 19 '17 at 20:24
  • 2
    $\begingroup$ You still have to prove the sequence is bounded, otherwise the limit might be $+\infty$ $\endgroup$ – user261263 Feb 19 '17 at 20:42
  • 2
    $\begingroup$ @Situ You cannot assume, a priori, that the sequence has a limit. $\endgroup$ – Mark Viola Feb 19 '17 at 20:55
3
$\begingroup$

The missing part is $a_n$ has upper bound.

Using induction, we'll show $a_n \le \frac {25} 4, \forall n \ge 1$.

For $n=1$ it's obvious. Suppose it's true for $n$. Then $a_{n+1}=\sqrt{a_n} + \frac{15}{4} \le \frac 5 2 + \frac {15} 4 = \frac {25} 4$

$\endgroup$
  • $\begingroup$ I thought its enough to show its strictly monotonically increasing and it has a limit smaller than infinity to be bounded ( because $a_0 = 1$ ). Thank you! $\endgroup$ – user391105 Feb 19 '17 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy