# Thread: Prove a sequence is nonincreasing by induction

1. ## Prove a sequence is nonincreasing by induction

I'm stuck on this....

Problem:
Show that (tn) is a nonincreasing sequence. That is, use induction to prove that $\displaystyle t_{n} \geq t_{n+1}$ where $\displaystyle t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n}.$

I have the base case so, $\displaystyle 1 \geq \frac{3}{4}$. Then assume $\displaystyle t_{k} \geq t_{k+1}$.
But I'm stuck at this part. I don't know how to arrive at $\displaystyle t_{k+1} \geq t_{k+2}$. I tried messing with the inequalities and stuff, but I'm still not sure.

Edit: The only thing I could really come up with was .....

$\displaystyle (1-\frac{1}{(k+1)^2}) \geq (1-\frac{1}{(k+1)^2})(1-\frac{1}{(k+2)^2}$ then use the fact $\displaystyle t_{k+1}=1-\frac{1}{(k+1)^2}$ and so $\displaystyle \frac{t_{k+1}}{t_{k}}=1-\frac{1}{(n+1)^2}.$

2. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by JSB1917
Problem:
Show that (tn) is a nonincreasing sequence. That is, use induction to prove that $\displaystyle t_{n} \geq t_{n+1}$ where $\displaystyle t_{n+1}=1-\frac{1}{(n+1)^2}.$
You better double check the problem!
$\displaystyle t_{n+1}=1-\frac{1}{(n+1)^2}$ is an increasing sequence.

3. ## Re: Prove a sequence is nonincreasing by induction

Yeah, sorry, it's actually $\displaystyle t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n}$ I forgot the $\displaystyle t_{n}$ at the end. I put that in the original post now.

4. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by JSB1917
Yeah, sorry, it's actually $\displaystyle t_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)t_{n}$ I forgot the $\displaystyle t_{n}$ at the end. I put that in the original post now.

Suppose you know that $\displaystyle t_k\ge t_{k+1}$.
Then multiply: $\displaystyle t_k\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)\ge t_{k+1}\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)$
But $\displaystyle \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)$
So $\displaystyle t_{k+1}\ge t_{k+2}$

5. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by Plato
Suppose you know that $\displaystyle t_k\ge t_{k+1}$.
Then multiply: $\displaystyle t_k\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)\ge t_{k+1}\left( {1 - \frac{1}{{(k + 1)^2 }}} \right)$
But $\displaystyle \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)$
So $\displaystyle t_{k+1}\ge t_{k+2}$
Yeah, but I don't think it's correct that $\displaystyle \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)$

Because....
$\displaystyle (k+1)^2 \leq (k+2)^2$, then
$\displaystyle -\frac{1}{(k+2)^2} \geq -\frac{1}{(k+1)^2}$ and so
$\displaystyle 1-\frac{1}{(k+2)^2} \geq 1-\frac{1}{(k+1)^2}$

6. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by JSB1917
Yeah, but I don't think it's correct that $\displaystyle \left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)$
Because....
$\displaystyle (k+1)^2 \leq (k+2)^2$, then
$\displaystyle -\frac{1}{(k+2)^2} \geq -\frac{1}{(k+1)^2}$ and so
$\displaystyle 1-\frac{1}{(k+2)^2} \geq 1-\frac{1}{(k+1)^2}$
You are correct about that. Sorry, I got lost in my own notation.
I still have your OP in my head. It is odd how the brains holds on to things.

7. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by Plato
You are correct about that. Sorry, I got lost in my own notation.
I know that we need to work backwards from $\displaystyle t_{k+2}\le t_{k+1}$.
I think using $\displaystyle \left(1-\frac{1}{(k+2)^2}\right) \geq \left(1-\frac{1}{(k+1)^2}\right)$ that then

$\displaystyle \left(1-\frac{1}{(k+2)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+2)^2}\right)t_{k+1}$ and so,

$\displaystyle \left(1-\frac{1}{(k+2)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+1)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+2)^2}\right)t_{k+1}$ and so

$\displaystyle t_{k+1} \geq t_{k+2}$

That seems right to me. Thoughts?

8. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by JSB1917
$\displaystyle \left(1-\frac{1}{(k+2)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+1)^2}\right)t_{k} \geq \left(1-\frac{1}{(k+2)^2}\right)t_{k+1}$ and so Thoughts?
I do not follow that. As I said above, the brain just get stuck on an idea and will not let it go.

Think about this. We know $\displaystyle t_{k + 1} = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k$.

Consider $\displaystyle t_{k + 1} - t_k = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k - t_k = \left( { - \frac{{t_k }}{{(k + 1)^2 }}} \right) < 0$.

What does that tell us?

9. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by Plato
I do not follow that. As I said above, the brain just get stuck on an idea and will not let it go.

Think about this. We know $\displaystyle t_{k + 1} = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k$.

Consider $\displaystyle t_{k + 1} - t_k = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k - t_k = \left( { - \frac{{t_k }}{{(k + 1)^2 }}} \right) < 0$.

What does that tell us?
Aside from it's negative, not sure.

10. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by JSB1917
Aside from it's negative, not sure.
Surely $\displaystyle t_{k+1}-t_k<0$ says $\displaystyle t_{k+1}<t_k$

11. ## Re: Prove a sequence is nonincreasing by induction

Originally Posted by Plato
Surely $\displaystyle t_{k+1}-t_k<0$ says $\displaystyle t_{k+1}<t_k$
yeah...I know that, but where is this leading? I'm just not seeing it.

12. ## Re: Prove a sequence is nonincreasing by induction

If we have:

$\displaystyle t_{k + 1} = \left( {1 - \frac{1}{{(k + 1)^2 }}} \right)t_k$

then since $\displaystyle 1 - \frac{1}{{(k + 1)^2 }}<1$, $\displaystyle t_{k + 1} < t_k$ and therefore the sequence is decreasing.

- Hollywood