IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

hsbhatt

Well one of the ways to prove convergence of a sequence is to prove that the sequence is monotonically increasing/decreasing and has a upper/lower bound respectively.

So, consider the sequence, assuming a non-negative

$x_1 = \sqrt a; x_2 = \sqrt{a+\sqrt a}; x_3 = \sqrt{a+\sqrt{a+\sqrt a}}} ...$

Here happily two things come together at once.

1. We can prove that this is an increasing sequence

2. It is bounded above

We prove the second result first

Let us choose $M = \frac{1+\sqrt{1+4a}}{2}$

We can prove that M is an upper bound of this sequence (in fact the least upper bound) as we have

$x_1 < M \Rightarrow x_2 = \sqrt {a+x_1} < \sqrt{a+M} = M$ and thus we can inductively prove that each x_n<M

Using this, we can prove that it is an increasing sequence

We have $x_{n+1} = \sqrt{a+x_n}$

Thus

$x_{n+1} > x_n \text{if} \ \sqrt{a+x_n}> x_n \\ \\<br/>\text{if} \ x_n^2 - x_n - a<0 \\ \\<br/>\text{if} \ x_n < \frac{1+\sqrt{1+4a}}{2}$

But, we have proved that it is true for all n that $x_n < \frac{1+\sqrt{1+4a}}{2}$

Hence, we have that the sequence is monotonically inreasing and is bounded above and is hence convergent in fact to M

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