Autumn tournament

Question

Let x_0, x_1, be a sequence of real numbers such that x_0 = 1 and x_n+1 = (x_n) + 2 - 1 for all n 0. Show that the sequence converges and find its limit. (Kristyan Vasilev)

Accepted Answer

We will first prove that the sequence is strictly increasing. Note that for each

n

we have

x_{n} \leq \frac{π}{2}

, since

sin x \leq 1

for each

x \in R

. Besides we have that the function

f (x) = sin x - x

is decreasing for

x \in R

, because

f^{'} (x) = cos x - 1 \leq 0

for each

x \in R

.

Therefore, for

x \in (- \infty, \frac{π}{2})

we have

f (x) \geq sin \frac{π}{2} - \frac{π}{2}

. So we get that

sin x_{n} \geq x_{n} + 1 - \frac{π}{2}

, which is equivalent to

x_{n + 1} \geq x_{n}

. Therefore, the series

(x_{n})_{n = 1}^{\infty}

is increasing and since it is bounded it follows that it is convergent. If

l

is its limit, then for

l

it is true that

l = sin l + \frac{π}{2} - 1

, i.e.

f (l) = f (\frac{π}{2})

. The function

f

is decreasing, which means that

l = \frac{π}{2}

.

\frac{π}{2}