Prove by induction that sum(1/2^i)<1 i =1 to n

[Eliva]

Homework Statement

prove by mathematical induction that it is true:


$$\frac{1}{2}$$+$$\frac{1}{2^{2}}$$+$$\frac{1}{2^{3}}$$+...+$$\frac{1}{2^{n}}$$ < 1

 

[morphism]

Well, where's your attempt?

 

[carltouss619]

Is this prove by induction: like first proving that the statement is true for n=1

then prove its true for n+1?

 

[Dick]

Change the induction hypothesis to sum(i=1,n)(1/2^i)=1-1/2^n. That will prove what you want and will make your life a lot easier.

 

[Mark44]

Is this prove by induction: like first proving that the statement is true for n=1

then prove its true for n+1?

Not quite. First, assume that the statement is true for n = k. Then if the statement is true for n = k, show that it must also be true for n = k + 1.

In the context of the OP's problem,
a) assume the statement is true for n = k. IOW, that
$$\frac{1}{2}$$ + $$\frac{1}{{2}^2}$$ + ... + $$\frac{1}{{2}^k}$$ < 1

b) Now prove that
$$\frac{1}{2}$$ + $$\frac{1}{{2}^2}$$ + ... + $$\frac{1}{{2}^k}$$ + $$\frac{1}{{2}^k+1}$$< 1

It's not immediately obvious what you need to do to carry out the proof, unless you can come up with a different expression for the sum of the k terms, the expression on the left in a).


As a hint, look at the sequence of sums for
1/2
1/2 + 1/4
1/2 + 1/4 + 1/8

and so on, and see if you can discover a pattern that you can extend to the sum with k terms in it. If you can do that, you can prove the inequality marked (*).
Mark

 

[Mark44]

Is this prove by induction: like first proving that the statement is true for n=1

then prove its true for n+1?

Not quite. First, assume that the statement is true for n = k. Then if the statement is true for n = k, show that it must also be true for n = k + 1.

In the context of the OP's problem,
a) assume the statement is true for n = k. IOW, that
$$\frac{1}{2}$$ + $$\frac{1}{{2}^2}$$ + ... + $$\frac{1}{{2}^k}$$ < 1

b) Now prove that
$$\frac{1}{2}$$ + $$\frac{1}{{2}^2}$$ + ... + $$\frac{1}{{2}^k}$$ + $$\frac{1}{{2}^k+1}$$< 1

It's not immediately obvious what you need to do to carry out the proof, unless you can come up with a different expression for the sum of the k terms, the expression on the left in a).


As a hint, look at the sequence of sums for
1/2
1/2 + 1/4
1/2 + 1/4 + 1/8

and so on, and see if you can discover a pattern that you can extend to the sum with k terms in it. If you can do that, you can prove the inequality in b).
Mark

 

[carltouss619]

Not quite. First, assume that the statement is true for n = k. Then if the statement is true for n = k, show that it must also be true for n = k + 1.

In the context of the OP's problem,
a) assume the statement is true for n = k. IOW, that
$$\frac{1}{2}$$ + $$\frac{1}{{2}^2}$$ + ... + $$\frac{1}{{2}^k}$$ < 1

b) Now prove that
$$\frac{1}{2}$$ + $$\frac{1}{{2}^2}$$ + ... + $$\frac{1}{{2}^k}$$ + $$\frac{1}{{2}^k+1}$$< 1

It's not immediately obvious what you need to do to carry out the proof, unless you can come up with a different expression for the sum of the k terms, the expression on the left in a).


As a hint, look at the sequence of sums for
1/2
1/2 + 1/4
1/2 + 1/4 + 1/8

and so on, and see if you can discover a pattern that you can extend to the sum with k terms in it. If you can do that, you can prove the inequality marked (*).
Mark

But isn't he supposed to prove by induction? And that's the general definition of induction that I know of. But I do see where you're coming from though.

 

[Mark44]

What I have described in a) and b) is how you do a proof by induction. The part at the end was just some advice on proving the statement in b).
Mark