Does this series converges to some constant? [closed]

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Let $S = sum_i=1^log n fraci2^i$

Question : What is $S$ when $n$ tends to infinity? Is it a constant or some function of $n$?

sequences-and-series divergent-series

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edited Aug 2 at 12:34

BCLC

6,98221973

asked Aug 2 at 11:58

lovw

224

closed as unclear what you're asking by Jam, Simply Beautiful Art, user 108128, RRL, Key Flex Aug 2 at 14:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To me it appears to be something like $O(fraclognn)$.
  – old
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  If we let $n$ tend to infinity, the resulting value can't really depend on $n$.
  – Arthur
  Aug 2 at 12:01
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The upper limit should be a natural number. Do you mean perhaps $lfloor log nrfloor$ ?
  – Peter
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am fine with Order notation also.
  – lovw
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$.
  – Eff
  Aug 2 at 12:05
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

3

Let $S = sum_i=1^log n fraci2^i$

Question : What is $S$ when $n$ tends to infinity? Is it a constant or some function of $n$?

sequences-and-series divergent-series

share|cite|improve this question

edited Aug 2 at 12:34

BCLC

6,98221973

asked Aug 2 at 11:58

lovw

224

closed as unclear what you're asking by Jam, Simply Beautiful Art, user 108128, RRL, Key Flex Aug 2 at 14:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To me it appears to be something like $O(fraclognn)$.
  – old
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  If we let $n$ tend to infinity, the resulting value can't really depend on $n$.
  – Arthur
  Aug 2 at 12:01
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The upper limit should be a natural number. Do you mean perhaps $lfloor log nrfloor$ ?
  – Peter
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am fine with Order notation also.
  – lovw
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$.
  – Eff
  Aug 2 at 12:05
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

3

up vote
0
down vote

favorite

3

3

Let $S = sum_i=1^log n fraci2^i$

Question : What is $S$ when $n$ tends to infinity? Is it a constant or some function of $n$?

sequences-and-series divergent-series

share|cite|improve this question

edited Aug 2 at 12:34

BCLC

6,98221973

asked Aug 2 at 11:58

lovw

224

Let $S = sum_i=1^log n fraci2^i$

Question : What is $S$ when $n$ tends to infinity? Is it a constant or some function of $n$?

sequences-and-series divergent-series

share|cite|improve this question

edited Aug 2 at 12:34

BCLC

6,98221973

asked Aug 2 at 11:58

lovw

224

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 12:34

BCLC

6,98221973

edited Aug 2 at 12:34

BCLC

6,98221973

edited Aug 2 at 12:34

BCLC

6,98221973

6,98221973

asked Aug 2 at 11:58

lovw

224

asked Aug 2 at 11:58

lovw

224

asked Aug 2 at 11:58

lovw

224

224

closed as unclear what you're asking by Jam, Simply Beautiful Art, user 108128, RRL, Key Flex Aug 2 at 14:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

closed as unclear what you're asking by Jam, Simply Beautiful Art, user 108128, RRL, Key Flex Aug 2 at 14:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To me it appears to be something like $O(fraclognn)$.
  – old
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  If we let $n$ tend to infinity, the resulting value can't really depend on $n$.
  – Arthur
  Aug 2 at 12:01
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The upper limit should be a natural number. Do you mean perhaps $lfloor log nrfloor$ ?
  – Peter
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am fine with Order notation also.
  – lovw
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$.
  – Eff
  Aug 2 at 12:05
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To me it appears to be something like $O(fraclognn)$.
  – old
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  If we let $n$ tend to infinity, the resulting value can't really depend on $n$.
  – Arthur
  Aug 2 at 12:01
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The upper limit should be a natural number. Do you mean perhaps $lfloor log nrfloor$ ?
  – Peter
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am fine with Order notation also.
  – lovw
  Aug 2 at 12:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$.
  – Eff
  Aug 2 at 12:05
  

  
  
  
  

To me it appears to be something like $O(fraclognn)$.
– old
Aug 2 at 12:01

To me it appears to be something like $O(fraclognn)$.
– old
Aug 2 at 12:01

3

3

If we let $n$ tend to infinity, the resulting value can't really depend on $n$.
– Arthur
Aug 2 at 12:01

If we let $n$ tend to infinity, the resulting value can't really depend on $n$.
– Arthur
Aug 2 at 12:01

2

2

The upper limit should be a natural number. Do you mean perhaps $lfloor log nrfloor$ ?
– Peter
Aug 2 at 12:01

The upper limit should be a natural number. Do you mean perhaps $lfloor log nrfloor$ ?
– Peter
Aug 2 at 12:01

1

1

I am fine with Order notation also.
– lovw
Aug 2 at 12:01

I am fine with Order notation also.
– lovw
Aug 2 at 12:01

3

3

I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$.
– Eff
Aug 2 at 12:05

I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$.
– Eff
Aug 2 at 12:05

 | 
show 1 more comment

3 Answers
3

active

oldest

votes

up vote
4
down vote

If you mean$$sum_i=1^lfloor log nrfloorfrac i2^i,$$then what you haee is a subsequence of the sequence$$left(sum_i=1^nfrac i2^iright)_ninmathbb N,$$whose sum is $2$. Therefore, the limit of your sequence is $2$, too.

share|cite|improve this answer

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$. However, I might be mistaken.
  – Eff
  Aug 2 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Elf Let us wait and see what the OP has to say about that.
  – José Carlos Santos
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

up vote
1
down vote

Consider for $|x|<1$

$$frac ddxsum_i=0^infty x^i=sum_i=1^infty ix^(i-1)=frac ddxleft(frac11-xright)=frac1(1-x)^2$$

then

$$sum_i=1^infty ix^i=xsum_i=1^infty ix^(i-1)=fracx(1-x)^2$$

therefore

$$sum_i=1^inftyfrac i2^i=2$$

and then

$$lim_nto infty sum_i=1^lfloor log nrfloorfrac i2^i=2$$

For the asymptotic grow we have that

$$frac ddxsum_i=0^N x^i=sum_i=1^N ix^(i-1)=frac ddxleft(frac1-x^N+11-xright)=frac1(1-x)^2-fracx^N(1-x)^2(1+N-Nx)$$

and then

$$sum_i=1^N ix^i=xsum_i=1^N ix^(i-1)=fracx(1-x)^2-fracx(1-x)^2x^N(1+N-Nx)$$

and then for $x=frac12$

$$sum_i=1^Nfrac i2^i=2-2frac12^Nleft(1+frac N 2right)=2-frac22^N-fracN2^N=2+Oleft(fracN2^Nright)$$

share|cite|improve this answer

edited Aug 2 at 13:19

answered Aug 2 at 12:01

gimusi

63.8k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lovw If the value of the limit is not what you want, then you should be clearer in the question
  – Calvin Khor
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint. Show by induction that for $Ngeq 1$,
$$sum_i=1^N fraci2^i=2-fracN+22^N.$$
Then the asymptotic growth is
$$S(n)=sum_i=1^lfloor log nrfloor fraci2^i=2-fraclfloor log nrfloor+22^lfloor log nrfloorsim 2-fraclog nn^log 2.$$

share|cite|improve this answer

edited Aug 2 at 14:39

answered Aug 2 at 12:35

Robert Z

83.5k954122

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

If you mean$$sum_i=1^lfloor log nrfloorfrac i2^i,$$then what you haee is a subsequence of the sequence$$left(sum_i=1^nfrac i2^iright)_ninmathbb N,$$whose sum is $2$. Therefore, the limit of your sequence is $2$, too.

share|cite|improve this answer

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$. However, I might be mistaken.
  – Eff
  Aug 2 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Elf Let us wait and see what the OP has to say about that.
  – José Carlos Santos
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

up vote
4
down vote

If you mean$$sum_i=1^lfloor log nrfloorfrac i2^i,$$then what you haee is a subsequence of the sequence$$left(sum_i=1^nfrac i2^iright)_ninmathbb N,$$whose sum is $2$. Therefore, the limit of your sequence is $2$, too.

share|cite|improve this answer

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$. However, I might be mistaken.
  – Eff
  Aug 2 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Elf Let us wait and see what the OP has to say about that.
  – José Carlos Santos
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

If you mean$$sum_i=1^lfloor log nrfloorfrac i2^i,$$then what you haee is a subsequence of the sequence$$left(sum_i=1^nfrac i2^iright)_ninmathbb N,$$whose sum is $2$. Therefore, the limit of your sequence is $2$, too.

share|cite|improve this answer

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

If you mean$$sum_i=1^lfloor log nrfloorfrac i2^i,$$then what you haee is a subsequence of the sequence$$left(sum_i=1^nfrac i2^iright)_ninmathbb N,$$whose sum is $2$. Therefore, the limit of your sequence is $2$, too.

share|cite|improve this answer

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

answered Aug 2 at 12:03

José Carlos Santos

112k1696172

112k1696172

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$. However, I might be mistaken.
  – Eff
  Aug 2 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Elf Let us wait and see what the OP has to say about that.
  – José Carlos Santos
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$. However, I might be mistaken.
  – Eff
  Aug 2 at 12:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Elf Let us wait and see what the OP has to say about that.
  – José Carlos Santos
  Aug 2 at 12:07
  

  
  
  
  

4

4

I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$. However, I might be mistaken.
– Eff
Aug 2 at 12:06

I think the OP is not looking for the limit per se, but the asymptotic growth of $S(n)$, i.e. with respect to $n$. However, I might be mistaken.
– Eff
Aug 2 at 12:06

2

2

@Elf Let us wait and see what the OP has to say about that.
– José Carlos Santos
Aug 2 at 12:07

@Elf Let us wait and see what the OP has to say about that.
– José Carlos Santos
Aug 2 at 12:07

add a comment | 

up vote
1
down vote

Consider for $|x|<1$

$$frac ddxsum_i=0^infty x^i=sum_i=1^infty ix^(i-1)=frac ddxleft(frac11-xright)=frac1(1-x)^2$$

then

$$sum_i=1^infty ix^i=xsum_i=1^infty ix^(i-1)=fracx(1-x)^2$$

therefore

$$sum_i=1^inftyfrac i2^i=2$$

and then

$$lim_nto infty sum_i=1^lfloor log nrfloorfrac i2^i=2$$

For the asymptotic grow we have that

$$frac ddxsum_i=0^N x^i=sum_i=1^N ix^(i-1)=frac ddxleft(frac1-x^N+11-xright)=frac1(1-x)^2-fracx^N(1-x)^2(1+N-Nx)$$

and then

$$sum_i=1^N ix^i=xsum_i=1^N ix^(i-1)=fracx(1-x)^2-fracx(1-x)^2x^N(1+N-Nx)$$

and then for $x=frac12$

$$sum_i=1^Nfrac i2^i=2-2frac12^Nleft(1+frac N 2right)=2-frac22^N-fracN2^N=2+Oleft(fracN2^Nright)$$

share|cite|improve this answer

edited Aug 2 at 13:19

answered Aug 2 at 12:01

gimusi

63.8k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lovw If the value of the limit is not what you want, then you should be clearer in the question
  – Calvin Khor
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

up vote
1
down vote

Consider for $|x|<1$

$$frac ddxsum_i=0^infty x^i=sum_i=1^infty ix^(i-1)=frac ddxleft(frac11-xright)=frac1(1-x)^2$$

then

$$sum_i=1^infty ix^i=xsum_i=1^infty ix^(i-1)=fracx(1-x)^2$$

therefore

$$sum_i=1^inftyfrac i2^i=2$$

and then

$$lim_nto infty sum_i=1^lfloor log nrfloorfrac i2^i=2$$

For the asymptotic grow we have that

$$frac ddxsum_i=0^N x^i=sum_i=1^N ix^(i-1)=frac ddxleft(frac1-x^N+11-xright)=frac1(1-x)^2-fracx^N(1-x)^2(1+N-Nx)$$

and then

$$sum_i=1^N ix^i=xsum_i=1^N ix^(i-1)=fracx(1-x)^2-fracx(1-x)^2x^N(1+N-Nx)$$

and then for $x=frac12$

$$sum_i=1^Nfrac i2^i=2-2frac12^Nleft(1+frac N 2right)=2-frac22^N-fracN2^N=2+Oleft(fracN2^Nright)$$

share|cite|improve this answer

edited Aug 2 at 13:19

answered Aug 2 at 12:01

gimusi

63.8k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lovw If the value of the limit is not what you want, then you should be clearer in the question
  – Calvin Khor
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Consider for $|x|<1$

$$frac ddxsum_i=0^infty x^i=sum_i=1^infty ix^(i-1)=frac ddxleft(frac11-xright)=frac1(1-x)^2$$

then

$$sum_i=1^infty ix^i=xsum_i=1^infty ix^(i-1)=fracx(1-x)^2$$

therefore

$$sum_i=1^inftyfrac i2^i=2$$

and then

$$lim_nto infty sum_i=1^lfloor log nrfloorfrac i2^i=2$$

For the asymptotic grow we have that

$$frac ddxsum_i=0^N x^i=sum_i=1^N ix^(i-1)=frac ddxleft(frac1-x^N+11-xright)=frac1(1-x)^2-fracx^N(1-x)^2(1+N-Nx)$$

and then

$$sum_i=1^N ix^i=xsum_i=1^N ix^(i-1)=fracx(1-x)^2-fracx(1-x)^2x^N(1+N-Nx)$$

and then for $x=frac12$

$$sum_i=1^Nfrac i2^i=2-2frac12^Nleft(1+frac N 2right)=2-frac22^N-fracN2^N=2+Oleft(fracN2^Nright)$$

share|cite|improve this answer

edited Aug 2 at 13:19

answered Aug 2 at 12:01

gimusi

63.8k73480

Consider for $|x|<1$

$$frac ddxsum_i=0^infty x^i=sum_i=1^infty ix^(i-1)=frac ddxleft(frac11-xright)=frac1(1-x)^2$$

then

$$sum_i=1^infty ix^i=xsum_i=1^infty ix^(i-1)=fracx(1-x)^2$$

therefore

$$sum_i=1^inftyfrac i2^i=2$$

and then

$$lim_nto infty sum_i=1^lfloor log nrfloorfrac i2^i=2$$

For the asymptotic grow we have that

$$frac ddxsum_i=0^N x^i=sum_i=1^N ix^(i-1)=frac ddxleft(frac1-x^N+11-xright)=frac1(1-x)^2-fracx^N(1-x)^2(1+N-Nx)$$

and then

$$sum_i=1^N ix^i=xsum_i=1^N ix^(i-1)=fracx(1-x)^2-fracx(1-x)^2x^N(1+N-Nx)$$

and then for $x=frac12$

$$sum_i=1^Nfrac i2^i=2-2frac12^Nleft(1+frac N 2right)=2-frac22^N-fracN2^N=2+Oleft(fracN2^Nright)$$

share|cite|improve this answer

edited Aug 2 at 13:19

answered Aug 2 at 12:01

gimusi

63.8k73480

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 13:19

edited Aug 2 at 13:19

edited Aug 2 at 13:19

answered Aug 2 at 12:01

gimusi

63.8k73480

answered Aug 2 at 12:01

gimusi

63.8k73480

answered Aug 2 at 12:01

gimusi

63.8k73480

63.8k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lovw If the value of the limit is not what you want, then you should be clearer in the question
  – Calvin Khor
  Aug 2 at 12:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lovw If the value of the limit is not what you want, then you should be clearer in the question
  – Calvin Khor
  Aug 2 at 12:07
  

  
  
  
  

@lovw If the value of the limit is not what you want, then you should be clearer in the question
– Calvin Khor
Aug 2 at 12:07

@lovw If the value of the limit is not what you want, then you should be clearer in the question
– Calvin Khor
Aug 2 at 12:07

add a comment | 

up vote
1
down vote

Hint. Show by induction that for $Ngeq 1$,
$$sum_i=1^N fraci2^i=2-fracN+22^N.$$
Then the asymptotic growth is
$$S(n)=sum_i=1^lfloor log nrfloor fraci2^i=2-fraclfloor log nrfloor+22^lfloor log nrfloorsim 2-fraclog nn^log 2.$$

share|cite|improve this answer

edited Aug 2 at 14:39

answered Aug 2 at 12:35

Robert Z

83.5k954122

add a comment | 

up vote
1
down vote

Hint. Show by induction that for $Ngeq 1$,
$$sum_i=1^N fraci2^i=2-fracN+22^N.$$
Then the asymptotic growth is
$$S(n)=sum_i=1^lfloor log nrfloor fraci2^i=2-fraclfloor log nrfloor+22^lfloor log nrfloorsim 2-fraclog nn^log 2.$$

share|cite|improve this answer

edited Aug 2 at 14:39

answered Aug 2 at 12:35

Robert Z

83.5k954122

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint. Show by induction that for $Ngeq 1$,
$$sum_i=1^N fraci2^i=2-fracN+22^N.$$
Then the asymptotic growth is
$$S(n)=sum_i=1^lfloor log nrfloor fraci2^i=2-fraclfloor log nrfloor+22^lfloor log nrfloorsim 2-fraclog nn^log 2.$$

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edited Aug 2 at 14:39

answered Aug 2 at 12:35

Robert Z

83.5k954122

Hint. Show by induction that for $Ngeq 1$,
$$sum_i=1^N fraci2^i=2-fracN+22^N.$$
Then the asymptotic growth is
$$S(n)=sum_i=1^lfloor log nrfloor fraci2^i=2-fraclfloor log nrfloor+22^lfloor log nrfloorsim 2-fraclog nn^log 2.$$

share|cite|improve this answer

edited Aug 2 at 14:39

answered Aug 2 at 12:35

Robert Z

83.5k954122

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 14:39

edited Aug 2 at 14:39

edited Aug 2 at 14:39

answered Aug 2 at 12:35

Robert Z

83.5k954122

answered Aug 2 at 12:35

Robert Z

83.5k954122

answered Aug 2 at 12:35

Robert Z

83.5k954122

83.5k954122

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