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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

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    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 $2.25$. – 2017-02-19
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    Thats what happens if you do too much maths... Thank you – 2017-02-19
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    Isn't showing an upper bound for the sequence required? – 2017-02-19
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    Didnt i just show its imit is the upper bound? – 2017-02-19
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    You still have to prove the sequence is bounded, otherwise the limit might be $+\infty$ – 2017-02-19
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    so $\lim a_{n} = \lim a_{n+1} = \lim \sqrt{a_n} + \frac{15}{4} = \sqrt{6.25} + \frac{15}{4} = 6.25$ ? – 2017-02-19
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    @Situ You cannot assume, a priori, that the sequence has a limit. – 2017-02-19

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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 $a_{n+1}=\sqrt{a_n} + \frac{15}{4} \le \frac 5 2 + \frac {15} 4 = \frac {25} 4$

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    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! – 2017-02-19