Inequality Proof Homework: Proving 2^k+1 $\geq$ 1/2

[jeff1evesque]

Homework Statement

\frac{1}{2^{k}+1} + \frac{1}{2^{k} +2} + ... + \frac{1}{2^k + 2^k} \geq \frac{1}{2}
2. The attempt at a solution
Not too sure, I am working on a larger proof (not too much difficult) and the above is my attempt to prove the induction step k+1 (since \frac{1}{2^k + 2^k} = \frac{1}{2^{k+1}}).

Should i try to factor out \frac{1}{2^k}?

 

[tiny-tim]

Hi jeff!

Hint: they're all greater than the last one.

 

[jeff1evesque]

Hi jeff!

Hint: they're all greater than the last one.

I still don't see it, i wish it were clear to me.

Thanks though,

Jeffrey Levesque

 

[l'Hôpital]

I still don't see it, i wish it were clear to me.

Thanks though,

Jeffrey Levesque

If a < b < c
a + a + a < a + b + c

Right?

Try something along those lines.

 

[Dick]

I still don't see it, i wish it were clear to me.

Thanks though,

Jeffrey Levesque

Don't you agree with tiny-tim that the last term in the sum is smaller than the rest? How many terms in the sequence are there?

 

[jeff1evesque]

Don't you agree with tiny-tim that the last term in the sum is smaller than the rest? How many terms in the sequence are there?

Yes I agree, but that isn't helping me at all- I just don't know how to formulate a proof for this.

 

[jeff1evesque]

There are 2^{k+1} terms

 

[l'Hôpital]

There are 2^{k+1} terms

Are you sure? Count again. : )

 

[jeff1evesque]

Are you sure? Count again. : )

2^k terms.

 

[jeff1evesque]

never minnd, I actually did this in a different problem- thanks everyone

 

[snipez90]

I also noticed that the last term on the LHS is the smallest, but I was confused for awhile since I thought the denominators were 2^k + 2^0, 2^k + 2^1, ..., 2^k + 2^k, in which case there are k+1 terms, and the estimate fails.

Anyways assuming the progression is what everyone else thinks it is, then basically 2^i \leq 2^k for i = 1, 2, ..., k so
2^k + 2^i \leq 2^k + 2^k = 2^{k+1} \Rightarrow \frac{1}{2^k + 2^i} \geq \frac{1}{2^{k+1}}
for i = 1, 2, ..., k.