Adding always the half of the previous number to a number beginning with 1…

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up vote
2
down vote

favorite

This question blows my mind.

Let's say we start at 1 and keep adding always 0.5 * the previous number.

1 + 0.5 + 0.25 + 0.125 + .....

The question is: will it reach infinity?

I really don't know, kept thinking about this alot.

arithmetic infinity

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asked Jul 25 at 18:46

Squareoot

1185

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do you know what a geometric progression is?
  – asdf
  Jul 25 at 18:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/…
  – Vasya
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It will never reach $2$. At each step, so go half the remaining distance to $2$, so you never get there.
  – saulspatz
  Jul 25 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you familiar with proof by induction?
  – Noah Schweber
  Jul 25 at 21:58
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

This question blows my mind.

Let's say we start at 1 and keep adding always 0.5 * the previous number.

1 + 0.5 + 0.25 + 0.125 + .....

The question is: will it reach infinity?

I really don't know, kept thinking about this alot.

arithmetic infinity

share|cite|improve this question

asked Jul 25 at 18:46

Squareoot

1185

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do you know what a geometric progression is?
  – asdf
  Jul 25 at 18:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/…
  – Vasya
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It will never reach $2$. At each step, so go half the remaining distance to $2$, so you never get there.
  – saulspatz
  Jul 25 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you familiar with proof by induction?
  – Noah Schweber
  Jul 25 at 21:58
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

This question blows my mind.

Let's say we start at 1 and keep adding always 0.5 * the previous number.

1 + 0.5 + 0.25 + 0.125 + .....

The question is: will it reach infinity?

I really don't know, kept thinking about this alot.

arithmetic infinity

share|cite|improve this question

asked Jul 25 at 18:46

Squareoot

1185

This question blows my mind.

Let's say we start at 1 and keep adding always 0.5 * the previous number.

1 + 0.5 + 0.25 + 0.125 + .....

The question is: will it reach infinity?

I really don't know, kept thinking about this alot.

arithmetic infinity

share|cite|improve this question

asked Jul 25 at 18:46

Squareoot

1185

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 18:46

Squareoot

1185

asked Jul 25 at 18:46

Squareoot

1185

asked Jul 25 at 18:46

Squareoot

1185

1185

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do you know what a geometric progression is?
  – asdf
  Jul 25 at 18:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/…
  – Vasya
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It will never reach $2$. At each step, so go half the remaining distance to $2$, so you never get there.
  – saulspatz
  Jul 25 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you familiar with proof by induction?
  – Noah Schweber
  Jul 25 at 21:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do you know what a geometric progression is?
  – asdf
  Jul 25 at 18:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/…
  – Vasya
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It will never reach $2$. At each step, so go half the remaining distance to $2$, so you never get there.
  – saulspatz
  Jul 25 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you familiar with proof by induction?
  – Noah Schweber
  Jul 25 at 21:58
  

  
  
  
  

2

2

Do you know what a geometric progression is?
– asdf
Jul 25 at 18:49

Do you know what a geometric progression is?
– asdf
Jul 25 at 18:49

2

2

en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/…
– Vasya
Jul 25 at 18:54

en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/…
– Vasya
Jul 25 at 18:54

It will never reach $2$. At each step, so go half the remaining distance to $2$, so you never get there.
– saulspatz
Jul 25 at 19:10

It will never reach $2$. At each step, so go half the remaining distance to $2$, so you never get there.
– saulspatz
Jul 25 at 19:10

Are you familiar with proof by induction?
– Noah Schweber
Jul 25 at 21:58

Are you familiar with proof by induction?
– Noah Schweber
Jul 25 at 21:58

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

Rewrite it with fractions and it might be a bit clearer:

$1 + frac12 + frac14 + frac18 + frac116 ...$

Now calculate the partial sums:

$1$, $1 frac12$, $1 frac34$, $1 frac78$, $1 frac1516$, ...

See a pattern? Can you see this going over $2$? How close do you think it might get?

share|cite|improve this answer

answered Jul 25 at 19:16

badjohn

3,4321618

add a comment | 

up vote
1
down vote

Consider calculating the sum in steps, at each step adding the next term. At the $n^th$ step, you're adding $1/2^n$ (adding $1$ is the $0^th$ step). Then at the $n^th$ step, the total sum is the following:

$$
s_n = 1 + frac12 + frac14 + ... + frac12^n
$$

Now remark that you can perform the following calculation:

beginalign*
1 + frac12 s_n &= 1 + frac12
left(
1 + frac12 + frac14 + ... + frac12^n
right)
\
&=
1 + frac12 + frac14 + ... + frac12^n + frac12^n+1
\
&= s_n + frac12^n+1
endalign*

Then if you solve for $s_n$ using elementary algebra (which you can do since $s_n$ is finite), you end up with:

$$
s_n = 2 - frac12^n < 2
$$
What we have just shown is that $s_n leq 2$ for any $n$ that we choose. In other words, no matter how many terms we add, the resulting sum will always be less than $2$. So, the sum will not reach infinity.

share|cite|improve this answer

answered Jul 26 at 19:08

Sambo

1,2121427

add a comment | 

up vote
-1
down vote

So what you're looking for is the value of this series, which will call it $S$.

$$S=sum_n=0^infty left( frac12 right)^n=1+frac12+frac14+frac18+...$$

If we divide this value $S$ by $2$ we get

$$fracS2=frac12+frac14+frac18+...=S-1 Longrightarrow S=2S-1$$

so now with simple algebra we get the result that the sum of these numbers $S$ tends to $2$, which is not $infty$.

share|cite|improve this answer

answered Jul 25 at 18:59

Mr. Nobody

11

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In order to perform algebraic manipulations with $S$ you need to assume that it is a finite number, which defeats the purpose. Rather, once we've established that $S$ is finite, we can use this argument to conclude $S=2$.
  – Sambo
  Jul 25 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know, but for that I would have to have proved that the corresponding sequence converges, and I don't think it needed this much detail. I know it wasn't totally mathematically rigorous but OK if you want to downvote good for you.
  – Mr. Nobody
  Jul 25 at 19:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question asked whether the sum converged. You didn't show that wasn't the case, so I don't believe you answered the question in a satisfactory way.
  – Sambo
  Jul 25 at 19:34
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Rewrite it with fractions and it might be a bit clearer:

$1 + frac12 + frac14 + frac18 + frac116 ...$

Now calculate the partial sums:

$1$, $1 frac12$, $1 frac34$, $1 frac78$, $1 frac1516$, ...

See a pattern? Can you see this going over $2$? How close do you think it might get?

share|cite|improve this answer

answered Jul 25 at 19:16

badjohn

3,4321618

add a comment | 

up vote
2
down vote

Rewrite it with fractions and it might be a bit clearer:

$1 + frac12 + frac14 + frac18 + frac116 ...$

Now calculate the partial sums:

$1$, $1 frac12$, $1 frac34$, $1 frac78$, $1 frac1516$, ...

See a pattern? Can you see this going over $2$? How close do you think it might get?

share|cite|improve this answer

answered Jul 25 at 19:16

badjohn

3,4321618

add a comment | 

up vote
2
down vote

up vote
2
down vote

Rewrite it with fractions and it might be a bit clearer:

$1 + frac12 + frac14 + frac18 + frac116 ...$

Now calculate the partial sums:

$1$, $1 frac12$, $1 frac34$, $1 frac78$, $1 frac1516$, ...

See a pattern? Can you see this going over $2$? How close do you think it might get?

share|cite|improve this answer

answered Jul 25 at 19:16

badjohn

3,4321618

Rewrite it with fractions and it might be a bit clearer:

$1 + frac12 + frac14 + frac18 + frac116 ...$

Now calculate the partial sums:

$1$, $1 frac12$, $1 frac34$, $1 frac78$, $1 frac1516$, ...

See a pattern? Can you see this going over $2$? How close do you think it might get?

share|cite|improve this answer

answered Jul 25 at 19:16

badjohn

3,4321618

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 19:16

badjohn

3,4321618

answered Jul 25 at 19:16

badjohn

3,4321618

answered Jul 25 at 19:16

badjohn

3,4321618

3,4321618

add a comment | 

add a comment | 

up vote
1
down vote

Consider calculating the sum in steps, at each step adding the next term. At the $n^th$ step, you're adding $1/2^n$ (adding $1$ is the $0^th$ step). Then at the $n^th$ step, the total sum is the following:

$$
s_n = 1 + frac12 + frac14 + ... + frac12^n
$$

Now remark that you can perform the following calculation:

beginalign*
1 + frac12 s_n &= 1 + frac12
left(
1 + frac12 + frac14 + ... + frac12^n
right)
\
&=
1 + frac12 + frac14 + ... + frac12^n + frac12^n+1
\
&= s_n + frac12^n+1
endalign*

Then if you solve for $s_n$ using elementary algebra (which you can do since $s_n$ is finite), you end up with:

$$
s_n = 2 - frac12^n < 2
$$
What we have just shown is that $s_n leq 2$ for any $n$ that we choose. In other words, no matter how many terms we add, the resulting sum will always be less than $2$. So, the sum will not reach infinity.

share|cite|improve this answer

answered Jul 26 at 19:08

Sambo

1,2121427

add a comment | 

up vote
1
down vote

Consider calculating the sum in steps, at each step adding the next term. At the $n^th$ step, you're adding $1/2^n$ (adding $1$ is the $0^th$ step). Then at the $n^th$ step, the total sum is the following:

$$
s_n = 1 + frac12 + frac14 + ... + frac12^n
$$

Now remark that you can perform the following calculation:

beginalign*
1 + frac12 s_n &= 1 + frac12
left(
1 + frac12 + frac14 + ... + frac12^n
right)
\
&=
1 + frac12 + frac14 + ... + frac12^n + frac12^n+1
\
&= s_n + frac12^n+1
endalign*

Then if you solve for $s_n$ using elementary algebra (which you can do since $s_n$ is finite), you end up with:

$$
s_n = 2 - frac12^n < 2
$$
What we have just shown is that $s_n leq 2$ for any $n$ that we choose. In other words, no matter how many terms we add, the resulting sum will always be less than $2$. So, the sum will not reach infinity.

share|cite|improve this answer

answered Jul 26 at 19:08

Sambo

1,2121427

add a comment | 

up vote
1
down vote

up vote
1
down vote

Consider calculating the sum in steps, at each step adding the next term. At the $n^th$ step, you're adding $1/2^n$ (adding $1$ is the $0^th$ step). Then at the $n^th$ step, the total sum is the following:

$$
s_n = 1 + frac12 + frac14 + ... + frac12^n
$$

Now remark that you can perform the following calculation:

beginalign*
1 + frac12 s_n &= 1 + frac12
left(
1 + frac12 + frac14 + ... + frac12^n
right)
\
&=
1 + frac12 + frac14 + ... + frac12^n + frac12^n+1
\
&= s_n + frac12^n+1
endalign*

Then if you solve for $s_n$ using elementary algebra (which you can do since $s_n$ is finite), you end up with:

$$
s_n = 2 - frac12^n < 2
$$
What we have just shown is that $s_n leq 2$ for any $n$ that we choose. In other words, no matter how many terms we add, the resulting sum will always be less than $2$. So, the sum will not reach infinity.

share|cite|improve this answer

answered Jul 26 at 19:08

Sambo

1,2121427

Consider calculating the sum in steps, at each step adding the next term. At the $n^th$ step, you're adding $1/2^n$ (adding $1$ is the $0^th$ step). Then at the $n^th$ step, the total sum is the following:

$$
s_n = 1 + frac12 + frac14 + ... + frac12^n
$$

Now remark that you can perform the following calculation:

beginalign*
1 + frac12 s_n &= 1 + frac12
left(
1 + frac12 + frac14 + ... + frac12^n
right)
\
&=
1 + frac12 + frac14 + ... + frac12^n + frac12^n+1
\
&= s_n + frac12^n+1
endalign*

Then if you solve for $s_n$ using elementary algebra (which you can do since $s_n$ is finite), you end up with:

$$
s_n = 2 - frac12^n < 2
$$
What we have just shown is that $s_n leq 2$ for any $n$ that we choose. In other words, no matter how many terms we add, the resulting sum will always be less than $2$. So, the sum will not reach infinity.

share|cite|improve this answer

answered Jul 26 at 19:08

Sambo

1,2121427

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 19:08

Sambo

1,2121427

answered Jul 26 at 19:08

Sambo

1,2121427

answered Jul 26 at 19:08

Sambo

1,2121427

1,2121427

add a comment | 

add a comment | 

up vote
-1
down vote

So what you're looking for is the value of this series, which will call it $S$.

$$S=sum_n=0^infty left( frac12 right)^n=1+frac12+frac14+frac18+...$$

If we divide this value $S$ by $2$ we get

$$fracS2=frac12+frac14+frac18+...=S-1 Longrightarrow S=2S-1$$

so now with simple algebra we get the result that the sum of these numbers $S$ tends to $2$, which is not $infty$.

share|cite|improve this answer

answered Jul 25 at 18:59

Mr. Nobody

11

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In order to perform algebraic manipulations with $S$ you need to assume that it is a finite number, which defeats the purpose. Rather, once we've established that $S$ is finite, we can use this argument to conclude $S=2$.
  – Sambo
  Jul 25 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know, but for that I would have to have proved that the corresponding sequence converges, and I don't think it needed this much detail. I know it wasn't totally mathematically rigorous but OK if you want to downvote good for you.
  – Mr. Nobody
  Jul 25 at 19:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question asked whether the sum converged. You didn't show that wasn't the case, so I don't believe you answered the question in a satisfactory way.
  – Sambo
  Jul 25 at 19:34
  

  
  
  
  

add a comment | 

up vote
-1
down vote

So what you're looking for is the value of this series, which will call it $S$.

$$S=sum_n=0^infty left( frac12 right)^n=1+frac12+frac14+frac18+...$$

If we divide this value $S$ by $2$ we get

$$fracS2=frac12+frac14+frac18+...=S-1 Longrightarrow S=2S-1$$

so now with simple algebra we get the result that the sum of these numbers $S$ tends to $2$, which is not $infty$.

share|cite|improve this answer

answered Jul 25 at 18:59

Mr. Nobody

11

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In order to perform algebraic manipulations with $S$ you need to assume that it is a finite number, which defeats the purpose. Rather, once we've established that $S$ is finite, we can use this argument to conclude $S=2$.
  – Sambo
  Jul 25 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know, but for that I would have to have proved that the corresponding sequence converges, and I don't think it needed this much detail. I know it wasn't totally mathematically rigorous but OK if you want to downvote good for you.
  – Mr. Nobody
  Jul 25 at 19:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question asked whether the sum converged. You didn't show that wasn't the case, so I don't believe you answered the question in a satisfactory way.
  – Sambo
  Jul 25 at 19:34
  

  
  
  
  

add a comment | 

up vote
-1
down vote

up vote
-1
down vote

So what you're looking for is the value of this series, which will call it $S$.

$$S=sum_n=0^infty left( frac12 right)^n=1+frac12+frac14+frac18+...$$

If we divide this value $S$ by $2$ we get

$$fracS2=frac12+frac14+frac18+...=S-1 Longrightarrow S=2S-1$$

so now with simple algebra we get the result that the sum of these numbers $S$ tends to $2$, which is not $infty$.

share|cite|improve this answer

answered Jul 25 at 18:59

Mr. Nobody

11

So what you're looking for is the value of this series, which will call it $S$.

$$S=sum_n=0^infty left( frac12 right)^n=1+frac12+frac14+frac18+...$$

If we divide this value $S$ by $2$ we get

$$fracS2=frac12+frac14+frac18+...=S-1 Longrightarrow S=2S-1$$

so now with simple algebra we get the result that the sum of these numbers $S$ tends to $2$, which is not $infty$.

share|cite|improve this answer

answered Jul 25 at 18:59

Mr. Nobody

11

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 18:59

Mr. Nobody

11

answered Jul 25 at 18:59

Mr. Nobody

11

answered Jul 25 at 18:59

Mr. Nobody

11

11

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In order to perform algebraic manipulations with $S$ you need to assume that it is a finite number, which defeats the purpose. Rather, once we've established that $S$ is finite, we can use this argument to conclude $S=2$.
  – Sambo
  Jul 25 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know, but for that I would have to have proved that the corresponding sequence converges, and I don't think it needed this much detail. I know it wasn't totally mathematically rigorous but OK if you want to downvote good for you.
  – Mr. Nobody
  Jul 25 at 19:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question asked whether the sum converged. You didn't show that wasn't the case, so I don't believe you answered the question in a satisfactory way.
  – Sambo
  Jul 25 at 19:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In order to perform algebraic manipulations with $S$ you need to assume that it is a finite number, which defeats the purpose. Rather, once we've established that $S$ is finite, we can use this argument to conclude $S=2$.
  – Sambo
  Jul 25 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know, but for that I would have to have proved that the corresponding sequence converges, and I don't think it needed this much detail. I know it wasn't totally mathematically rigorous but OK if you want to downvote good for you.
  – Mr. Nobody
  Jul 25 at 19:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question asked whether the sum converged. You didn't show that wasn't the case, so I don't believe you answered the question in a satisfactory way.
  – Sambo
  Jul 25 at 19:34
  

  
  
  
  

3

3

In order to perform algebraic manipulations with $S$ you need to assume that it is a finite number, which defeats the purpose. Rather, once we've established that $S$ is finite, we can use this argument to conclude $S=2$.
– Sambo
Jul 25 at 19:05

In order to perform algebraic manipulations with $S$ you need to assume that it is a finite number, which defeats the purpose. Rather, once we've established that $S$ is finite, we can use this argument to conclude $S=2$.
– Sambo
Jul 25 at 19:05

I know, but for that I would have to have proved that the corresponding sequence converges, and I don't think it needed this much detail. I know it wasn't totally mathematically rigorous but OK if you want to downvote good for you.
– Mr. Nobody
Jul 25 at 19:24

I know, but for that I would have to have proved that the corresponding sequence converges, and I don't think it needed this much detail. I know it wasn't totally mathematically rigorous but OK if you want to downvote good for you.
– Mr. Nobody
Jul 25 at 19:24

1

1

The question asked whether the sum converged. You didn't show that wasn't the case, so I don't believe you answered the question in a satisfactory way.
– Sambo
Jul 25 at 19:34

The question asked whether the sum converged. You didn't show that wasn't the case, so I don't believe you answered the question in a satisfactory way.
– Sambo
Jul 25 at 19:34

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