# Proof that a recursive sequence converges

## Homework Statement

Prove that ##\displaystyle t_{n+1} = (1 - \frac{1}{4n^2}) t_n## where ##t_1=1## converges.

## The Attempt at a Solution

First, we must prove that the sequence is bounded below. We will prove that it is bounded below by 0. ##t_1 = 1 \ge 0##, so the base case holds. Now, suppose that ##t_k \ge 0##. Then ##t_{k+1} = (1 - \frac{1}{4k^2}) t_k \ge (1 - \frac{1}{4k^2}) (0) = 0##. So the induction is complete, and the sequence is bounded below by 0.

Now we must show that the sequence is decreasing. ##t_2 = 15/16 \le 1 = t_1##, so the base case holds. Now, suppose that ##t_{k+1} \le t_k##. Then ##t_{k+2} = (1 - \frac{1}{4(k+1)^2})t_{k+1} \le t_{k+1}##. So the induction is complete, and the sequence is decreasing.

By the Monotone Convergence Theorem, the sequence converges.

Related Calculus and Beyond Homework Help News on Phys.org
MathematicalPhysicist
Gold Member
Seems ok, btw you can actually solve this recursive equation by iteration, and see that it's decreasing explicitly.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Prove that ##\displaystyle t_{n+1} = (1 - \frac{1}{4n^2}) t_n## where ##t_1=1## converges.

## The Attempt at a Solution

First, we must prove that the sequence is bounded below. We will prove that it is bounded below by 0. ##t_1 = 1 \ge 0##, so the base case holds. Now, suppose that ##t_k \ge 0##. Then ##t_{k+1} = (1 - \frac{1}{4k^2}) t_k \ge (1 - \frac{1}{4k^2}) (0) = 0##. So the induction is complete, and the sequence is bounded below by 0.

Now we must show that the sequence is decreasing. ##t_2 = 15/16 \le 1 = t_1##, so the base case holds. Now, suppose that ##t_{k+1} \le t_k##. Then ##t_{k+2} = (1 - \frac{1}{4(k+1)^2})t_{k+1} \le t_{k+1}##. So the induction is complete, and the sequence is decreasing.

By the Monotone Convergence Theorem, the sequence converges.
Another, perhaps easier way is to note that
$$|t_{n+1}-t_n| = \frac{1}{4n^2} |t_n| \leq \frac{1}{4} |t_n|,$$
so we have (at least) geometrically fast convergence.

One question. In showing that the sequence decreases, why didn't I have to use the inductive hypothesis?

Ray Vickson
Homework Helper
Dearly Missed
One question. In showing that the sequence decreases, why didn't I have to use the inductive hypothesis?
$$\frac{t_{n+1}}{t_n }= 1 - \frac{1}{4n^2} ,$$
so if you know that ##t_n > 0## then you have ##0 < t_{n+1}/t_n < 1,## hence a positive strictly decreasing sequence.

pasmith
Homework Helper
For $n \geq 1$, it follows from $$\frac34 \leq 1 - \frac{1}{4n^2} < 1$$ that $$0 < |t_{n+1}| = \left|1 - \frac{1}{4n^2}\right||t_n| < |t_n|$$ and so $|t_n|$ is strictly decreasing. Since each $t_n$ is by construction a product of strictly positive reals we can immediately remove the absolute value signs.