Limit of a sum using complex analysis.

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I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.

complex-analysis limits complex-numbers trigonometric-series

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asked 1 hour ago

FuzzyPixelz

191111

add a comment | 

up vote
2
down vote

favorite

1

I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.

complex-analysis limits complex-numbers trigonometric-series

share|cite|improve this question

asked 1 hour ago

FuzzyPixelz

191111

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.

complex-analysis limits complex-numbers trigonometric-series

share|cite|improve this question

asked 1 hour ago

FuzzyPixelz

191111

I'm trying to find the limit of this sum: $$S_n =frac1nleft(frac12+sum_k=1^ncos(kx)right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $cot(xover 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = frac12nleft(cotleft(fracx2right)sin(nx)+cos(nx)right)$$
Thanks in advance.

complex-analysis limits complex-numbers trigonometric-series

complex-analysis limits complex-numbers trigonometric-series

share|cite|improve this question

asked 1 hour ago

FuzzyPixelz

191111

share|cite|improve this question

asked 1 hour ago

FuzzyPixelz

191111

share|cite|improve this question

share|cite|improve this question

asked 1 hour ago

FuzzyPixelz

191111

asked 1 hour ago

FuzzyPixelz

191111

asked 1 hour ago

FuzzyPixelz

191111

191111

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
3
down vote



accepted

You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$

Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$

So $$left| sum_k=1^n cos(kx) right| leq frac2$$

And therefore you get $S_n rightarrow 0$.

If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.

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answered 56 mins ago

TheSilverDoe

1,10511

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
  – TheSilverDoe
  50 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, this was very helpful.
  – FuzzyPixelz
  48 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Just write $cos (kx)$ as the real part of $e^ikx$.

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answered 1 hour ago

AlexL

992

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I already did that but it didn't work.
  – FuzzyPixelz
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then the limit is 0. Why do you think the answer is a cotan ?
  – AlexL
  57 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
  – FuzzyPixelz
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You must have forgotten something.
  – AlexL
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
  – FuzzyPixelz
  45 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Another view of the question.

Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:

$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$

When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.

---

For other $x$'s, i.e. when $x=2pi m$,

$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$

share|cite|improve this answer

edited 11 mins ago

answered 18 mins ago

peterwhy

11.4k21127

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
3
down vote



accepted

You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$

Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$

So $$left| sum_k=1^n cos(kx) right| leq frac2$$

And therefore you get $S_n rightarrow 0$.

If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.

share|cite|improve this answer

answered 56 mins ago

TheSilverDoe

1,10511

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
  – TheSilverDoe
  50 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, this was very helpful.
  – FuzzyPixelz
  48 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$

Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$

So $$left| sum_k=1^n cos(kx) right| leq frac2$$

And therefore you get $S_n rightarrow 0$.

If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.

share|cite|improve this answer

answered 56 mins ago

TheSilverDoe

1,10511

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
  – TheSilverDoe
  50 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, this was very helpful.
  – FuzzyPixelz
  48 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$

Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$

So $$left| sum_k=1^n cos(kx) right| leq frac2$$

And therefore you get $S_n rightarrow 0$.

If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.

share|cite|improve this answer

answered 56 mins ago

TheSilverDoe

1,10511

You have, for all $x neq 2kpi$,
$$sum_k=1^n cos(kx) = mathrmReleft( sum_k=1^n e^ikxright) = mathrmReleft( e^ix frac1-e^inx1-e^ix right)$$

Moreover you have
$$left|e^ix frac1-e^inx1-e^ixright| leq frac2$$

So $$left| sum_k=1^n cos(kx) right| leq frac2$$

And therefore you get $S_n rightarrow 0$.

If $x = 2kpi$ for $k in mathbbZ$, you easily have $S_n rightarrow 1$.

share|cite|improve this answer

answered 56 mins ago

TheSilverDoe

1,10511

share|cite|improve this answer

share|cite|improve this answer

answered 56 mins ago

TheSilverDoe

1,10511

answered 56 mins ago

TheSilverDoe

1,10511

answered 56 mins ago

TheSilverDoe

1,10511

1,10511

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
  – TheSilverDoe
  50 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, this was very helpful.
  – FuzzyPixelz
  48 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
  – TheSilverDoe
  50 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, this was very helpful.
  – FuzzyPixelz
  48 mins ago
  

  
  
  
  

Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago

Remember that $|e^iz|=1$ for all $z in mathbbR$. So by the triangle inequality, $|1-e^inx| leq 2$.
– TheSilverDoe
50 mins ago

Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago

Thank you very much, this was very helpful.
– FuzzyPixelz
48 mins ago

add a comment | 

up vote
2
down vote

Just write $cos (kx)$ as the real part of $e^ikx$.

share|cite|improve this answer

answered 1 hour ago

AlexL

992

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I already did that but it didn't work.
  – FuzzyPixelz
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then the limit is 0. Why do you think the answer is a cotan ?
  – AlexL
  57 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
  – FuzzyPixelz
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You must have forgotten something.
  – AlexL
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
  – FuzzyPixelz
  45 mins ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

Just write $cos (kx)$ as the real part of $e^ikx$.

share|cite|improve this answer

answered 1 hour ago

AlexL

992

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I already did that but it didn't work.
  – FuzzyPixelz
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then the limit is 0. Why do you think the answer is a cotan ?
  – AlexL
  57 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
  – FuzzyPixelz
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You must have forgotten something.
  – AlexL
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
  – FuzzyPixelz
  45 mins ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Just write $cos (kx)$ as the real part of $e^ikx$.

share|cite|improve this answer

answered 1 hour ago

AlexL

992

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Just write $cos (kx)$ as the real part of $e^ikx$.

share|cite|improve this answer

answered 1 hour ago

AlexL

992

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

AlexL

992

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 1 hour ago

AlexL

992

answered 1 hour ago

AlexL

992

992

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

AlexL is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I already did that but it didn't work.
  – FuzzyPixelz
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then the limit is 0. Why do you think the answer is a cotan ?
  – AlexL
  57 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
  – FuzzyPixelz
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You must have forgotten something.
  – AlexL
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
  – FuzzyPixelz
  45 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I already did that but it didn't work.
  – FuzzyPixelz
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then the limit is 0. Why do you think the answer is a cotan ?
  – AlexL
  57 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
  – FuzzyPixelz
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You must have forgotten something.
  – AlexL
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
  – FuzzyPixelz
  45 mins ago
  
  

  
  
  
  

I already did that but it didn't work.
– FuzzyPixelz
1 hour ago

I already did that but it didn't work.
– FuzzyPixelz
1 hour ago

1

1

Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago

Then the limit is 0. Why do you think the answer is a cotan ?
– AlexL
57 mins ago

I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago

I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$
– FuzzyPixelz
56 mins ago

1

1

You must have forgotten something.
– AlexL
54 mins ago

You must have forgotten something.
– AlexL
54 mins ago

TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago

TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $cot$, my bad.
– FuzzyPixelz
45 mins ago

add a comment | 

up vote
0
down vote

Another view of the question.

Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:

$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$

When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.

---

For other $x$'s, i.e. when $x=2pi m$,

$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$

share|cite|improve this answer

edited 11 mins ago

answered 18 mins ago

peterwhy

11.4k21127

add a comment | 

up vote
0
down vote

Another view of the question.

Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:

$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$

When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.

---

For other $x$'s, i.e. when $x=2pi m$,

$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$

share|cite|improve this answer

edited 11 mins ago

answered 18 mins ago

peterwhy

11.4k21127

add a comment | 

up vote
0
down vote

up vote
0
down vote

Another view of the question.

Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:

$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$

When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.

---

For other $x$'s, i.e. when $x=2pi m$,

$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$

share|cite|improve this answer

edited 11 mins ago

answered 18 mins ago

peterwhy

11.4k21127

Another view of the question.

Instead of taking the real part of $e^ikx$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:

$$beginalign*
cos kx &= frace^-ikx+e^ikx2\
frac12+sum_k=1^ncos kx &= frac12sum_k=-n^ne^ikx\
&= frac12 cdot e^-inxfrace^i(2n+1)x-1e^ix-1\
&= frac12 cdot frace^i(n+1)x-e^-inxe^ix-1\
&= frac12 cdot frace^ileft(n+frac12right)x-e^-ileft(n+frac12right)xe^ifrac12x-e^-ifrac12x\
&= frac12 cdot fracsinleft(n+frac12right)xsinfrac12x\
S_n&= fracsinleft(n+frac12right)x2nsinfrac12x\
endalign*$$

When $e^ix ne 1$, i.e. $xne 2pi m$ for integer $m$, $S_n to 0$.

---

For other $x$'s, i.e. when $x=2pi m$,

$$beginalign*
cos kx = cos 2pi km &= 1\
sum_k=1^ncos kx &= n\
frac12 + sum_k=1^ncos kx &= frac12+ n\
S_n &= frac1nleft(frac12 + nright)\
&= frac12n + 1\
&to 1
endalign*$$

share|cite|improve this answer

edited 11 mins ago

answered 18 mins ago

peterwhy

11.4k21127

share|cite|improve this answer

share|cite|improve this answer

edited 11 mins ago

edited 11 mins ago

edited 11 mins ago

answered 18 mins ago

peterwhy

11.4k21127

answered 18 mins ago

peterwhy

11.4k21127

answered 18 mins ago

peterwhy

11.4k21127

11.4k21127

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