Prove a sequence is nonincreasing by induction

**JSB1917** · November 13th 2012, 09:59 AM

I'm stuck on this....

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

I have the base case so, $1 \geq \frac{3}{4}$ . Then assume $t_{k} \geq t_{k+1}$ .
But I'm stuck at this part. I don't know how to arrive at $t_{k+1} \geq t_{k+2}$ . 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 .....

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

**Plato** · November 13th 2012, 10:32 AM

Originally Posted by JSB1917

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

You better double check the problem!
$t_{n+1}=1-\frac{1}{(n+1)^2}$ is an increasing sequence.

**JSB1917** · November 13th 2012, 10:52 AM

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

**Plato** · November 13th 2012, 11:34 AM

Originally Posted by JSB1917

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

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

**JSB1917** · November 13th 2012, 11:51 AM

Originally Posted by Plato

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

Yeah, but I don't think it's correct that $\left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)$

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

**Plato** · November 13th 2012, 12:02 PM

Originally Posted by JSB1917

Yeah, but I don't think it's correct that $\left( {1 - \frac{1}{{(k + 1)^2 }}} \right) \ge \left( {1 - \frac{1}{{(k + 2)^2 }}} \right)$
Because....
$(k+1)^2 \leq (k+2)^2$ , then
$-\frac{1}{(k+2)^2} \geq -\frac{1}{(k+1)^2}$ and so
$1-\frac{1}{(k+2)^2} \geq 1-\frac{1}{(k+1)^2}$

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.

**JSB1917** · November 13th 2012, 12:18 PM

Originally Posted by Plato

You are correct about that. Sorry, I got lost in my own notation.
I still have your OP in my head.
I know that we need to work backwards from $t_{k+2}\le t_{k+1}$ .

I think using $\left(1-\frac{1}{(k+2)^2}\right) \geq \left(1-\frac{1}{(k+1)^2}\right)$ that then

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

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

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

That seems right to me. Thoughts?

**Plato** · November 13th 2012, 02:40 PM

Originally Posted by JSB1917

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

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

What does that tell us?

**JSB1917** · November 13th 2012, 03:10 PM

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

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

What does that tell us?

Aside from it's negative, not sure.

**Plato** · November 13th 2012, 03:24 PM

Originally Posted by JSB1917

Aside from it's negative, not sure.

Surely $t_{k+1}-t_k<0$ says $t_{k+1}<t_k$

**JSB1917** · November 13th 2012, 03:28 PM

Originally Posted by Plato

Surely $t_{k+1}-t_k<0$ says $t_{k+1}<t_k$

yeah...I know that, but where is this leading? I'm just not seeing it.

**hollywood** · November 13th 2012, 08:43 PM

If we have:

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

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

- Hollywood

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