[Toán 11 ] Dãy số

D

dien0709

$U_1=\sqrt{2016}>0$

$\dfrac{U_{n+1}}{U_n}=\sqrt{\dfrac{2016U_n^2+3U_n+2015}{U_n^2}}>1\to U_n$ tăng

$\to$ khi $n\to$ \infty thì $U_n\to $ \infty

$\to \lim\dfrac{U_{n+1}}{U_n}=\lim\sqrt{2016+\dfrac{3}{U_n}+\dfrac{2015}{U_n^2}}=\sqrt{2016}$
 
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