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 $\displaystyle \frac{t_{k+1}}{t_{k}}=1-\frac{1}{(n+1)^2}.$

Re: Prove a sequence is nonincreasing by induction

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**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.**

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.

Re: Prove a sequence is nonincreasing by induction

Quote:

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

Re: Prove a sequence is nonincreasing by induction

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**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} $

Re: Prove a sequence is nonincreasing by induction

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**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.

Re: Prove a sequence is nonincreasing by induction

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**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 $\displaystyle t_{k+2}\le t_{k+1}$.

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?

Re: Prove a sequence is nonincreasing by induction

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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?

Re: Prove a sequence is nonincreasing by induction

Quote:

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.

Re: Prove a sequence is nonincreasing by induction

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**JSB1917** Aside from it's negative, not sure.

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

Re: Prove a sequence is nonincreasing by induction

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**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.

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