If a 1 = 1, a n+1 = 
for all n ≥ 1. Show that the sequence {a n } is monotonically increasing. Hence show that 1 < a n < 2 for all n > 1.
Text Solution
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Sol. We have,
a n + 1 2 = 2 + a n , a n 2 = 2 + a n–1 ⇒ a n+1 2 –a n 2 = a n –a n–1 …(i)
Now, as a 2 = 
= 
We note that a 2 > a 1 . On putting n = 2 in (i), we get
a 3 2 – a 2 2 = a 2 –a 1 > 0 ⇒ a 3 2 > a 2 2 ⇒ a 3 > a 2
( a 2 , a 3 > 0)
Again putting n = 3 in (i), we get a 4 2 –a 3 2 = a 3 –a 2 > 0, a 4 > a 3 and so on.
Thus a n+1 > a n for all n.
⇒ a n+1 2 > a n 2 ⇒ 2 + a n > a n 2
⇒ a n 2 – a n – 2 < 0 ⇒ (a n – 2) (a n + 1) < 0 ⇒ –1 < a n < 2
But as a 1 = 1 and {a n } is MI all a n ’s are greater than 1. Hence 1 ≤ a n < 2 for all n.
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