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2

1

$begingroup$

How would one prove that$$lim_ntoinftyfrac1sqrtnsum_k=1^nfrac1sqrt2k+sqrt2k-1$$ converges (rather slowly) to $frac 1sqrt2$, which appears obvious from numerical computation.

sequences-and-series limits analysis summation telescopic-series

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edited 7 hours ago

Michael Rozenberg

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asked 8 hours ago

HonzaHonza

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$endgroup$

add a comment | 

2

1

$begingroup$

How would one prove that$$lim_ntoinftyfrac1sqrtnsum_k=1^nfrac1sqrt2k+sqrt2k-1$$ converges (rather slowly) to $frac 1sqrt2$, which appears obvious from numerical computation.

sequences-and-series limits analysis summation telescopic-series

share|cite|improve this question

edited 7 hours ago

Michael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

asked 8 hours ago

HonzaHonza

4766 bronze badges

$endgroup$

add a comment | 

$begingroup$

How would one prove that$$lim_ntoinftyfrac1sqrtnsum_k=1^nfrac1sqrt2k+sqrt2k-1$$ converges (rather slowly) to $frac 1sqrt2$, which appears obvious from numerical computation.

sequences-and-series limits analysis summation telescopic-series

share|cite|improve this question

edited 7 hours ago

Michael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

asked 8 hours ago

HonzaHonza

4766 bronze badges

$endgroup$

share|cite|improve this question

edited 7 hours ago

Michael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

asked 8 hours ago

HonzaHonza

4766 bronze badges

share|cite|improve this question

edited 7 hours ago

Michael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

asked 8 hours ago

HonzaHonza

4766 bronze badges

edited 7 hours ago

Michael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

edited 7 hours ago

Michael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

asked 8 hours ago

HonzaHonza

4766 bronze badges

asked 8 hours ago

HonzaHonza

4766 bronze badges

2 Answers
2

active

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7

$begingroup$

$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)<frac1sqrt2left(sqrtk+sqrtk-1right)=frac1sqrt2left(sqrtk-sqrtk-1right),$$
$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)>frac1sqrt2left(sqrtk+1+sqrtkright)=frac1sqrt2left(sqrtk+1-sqrtkright)$$ and use the telescoping summation.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RozenbergMichael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't telescope. You always add roots of even numbers and subtract roots of odd numbers.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes, that seems to be it, thanks!
  $endgroup$
  – Honza
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome! @saulspatz Thank you very much! I fixed.
  $endgroup$
  – Michael Rozenberg
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You are most welcome. Nice answer. I was getting nowhere with integrals.
  $endgroup$
  – saulspatz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz, if you are interested I got somewhere with integrals ;)
  $endgroup$
  – Yuriy S
  7 hours ago
  

  
  
  
  

 | 
show 5 more comments

0

$begingroup$

Another interesting way to prove the lower bound.

Consider:

$$A=lim_ntoinftyfrac1sqrtnsum_k=1^nfrac1sqrt2k+sqrt2k-1$$

$$I_k=frac1sqrt2k+sqrt2k-1=frac2sqrtpiint_0^infty e^-(4k-1)x^2 e^-4k x^2 sqrt1-1/(2k) dx$$

Since $(sqrt2k+sqrt2k-1)^2=4 k-1+4 sqrtk(k-1/2)$.

$$sqrt1-frac12k= 1-frac14k-frac132k^2-O left( frac18k^3 right)<1-frac14k$$

We have:

$$I_k > frac2sqrtpi int_0^infty e^-8kx^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1sqrtnsum_k=1^n int_0^infty e^-8kx^2 dx= \ =frac2sqrtpi lim_ntoinftyfrac1sqrtn int_0^infty frace^-8 x^2-e^-8(n+1) x^21-e^-8 x^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frace^-8 y^2/n-e^-8(1+1/n) y^21-e^-8 y^2/n dy$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frac1-e^-8 y^2e^8 y^2/n-1 dy=frac14 sqrtpiint_0^infty frac1-e^-8 y^2y^2 dy$$

$$A geq frac1sqrt2 piint_0^infty frac1-e^-z^2z^2 dz= frac1sqrt2$$

The last integral is easy to prove using integration by parts.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Yuriy SYuriy S

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add a comment | 

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7

$begingroup$

$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)<frac1sqrt2left(sqrtk+sqrtk-1right)=frac1sqrt2left(sqrtk-sqrtk-1right),$$
$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)>frac1sqrt2left(sqrtk+1+sqrtkright)=frac1sqrt2left(sqrtk+1-sqrtkright)$$ and use the telescoping summation.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RozenbergMichael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't telescope. You always add roots of even numbers and subtract roots of odd numbers.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes, that seems to be it, thanks!
  $endgroup$
  – Honza
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome! @saulspatz Thank you very much! I fixed.
  $endgroup$
  – Michael Rozenberg
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You are most welcome. Nice answer. I was getting nowhere with integrals.
  $endgroup$
  – saulspatz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz, if you are interested I got somewhere with integrals ;)
  $endgroup$
  – Yuriy S
  7 hours ago
  

  
  
  
  

 | 
show 5 more comments

7

$begingroup$

$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)<frac1sqrt2left(sqrtk+sqrtk-1right)=frac1sqrt2left(sqrtk-sqrtk-1right),$$
$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)>frac1sqrt2left(sqrtk+1+sqrtkright)=frac1sqrt2left(sqrtk+1-sqrtkright)$$ and use the telescoping summation.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RozenbergMichael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't telescope. You always add roots of even numbers and subtract roots of odd numbers.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes, that seems to be it, thanks!
  $endgroup$
  – Honza
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome! @saulspatz Thank you very much! I fixed.
  $endgroup$
  – Michael Rozenberg
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You are most welcome. Nice answer. I was getting nowhere with integrals.
  $endgroup$
  – saulspatz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz, if you are interested I got somewhere with integrals ;)
  $endgroup$
  – Yuriy S
  7 hours ago
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)<frac1sqrt2left(sqrtk+sqrtk-1right)=frac1sqrt2left(sqrtk-sqrtk-1right),$$
$$frac1sqrt2k+sqrt2k-1=frac1sqrt2left(sqrtk+sqrtk-frac12right)>frac1sqrt2left(sqrtk+1+sqrtkright)=frac1sqrt2left(sqrtk+1-sqrtkright)$$ and use the telescoping summation.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RozenbergMichael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RozenbergMichael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

answered 8 hours ago

Michael RozenbergMichael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

answered 8 hours ago

Michael RozenbergMichael Rozenberg

121k2020 gold badges104104 silver badges210210 bronze badges

0

$begingroup$

Another interesting way to prove the lower bound.

Consider:

$$A=lim_ntoinftyfrac1sqrtnsum_k=1^nfrac1sqrt2k+sqrt2k-1$$

$$I_k=frac1sqrt2k+sqrt2k-1=frac2sqrtpiint_0^infty e^-(4k-1)x^2 e^-4k x^2 sqrt1-1/(2k) dx$$

Since $(sqrt2k+sqrt2k-1)^2=4 k-1+4 sqrtk(k-1/2)$.

$$sqrt1-frac12k= 1-frac14k-frac132k^2-O left( frac18k^3 right)<1-frac14k$$

We have:

$$I_k > frac2sqrtpi int_0^infty e^-8kx^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1sqrtnsum_k=1^n int_0^infty e^-8kx^2 dx= \ =frac2sqrtpi lim_ntoinftyfrac1sqrtn int_0^infty frace^-8 x^2-e^-8(n+1) x^21-e^-8 x^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frace^-8 y^2/n-e^-8(1+1/n) y^21-e^-8 y^2/n dy$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frac1-e^-8 y^2e^8 y^2/n-1 dy=frac14 sqrtpiint_0^infty frac1-e^-8 y^2y^2 dy$$

$$A geq frac1sqrt2 piint_0^infty frac1-e^-z^2z^2 dz= frac1sqrt2$$

The last integral is easy to prove using integration by parts.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Yuriy SYuriy S

17.6k44 gold badges3434 silver badges123123 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

Another interesting way to prove the lower bound.

Consider:

$$A=lim_ntoinftyfrac1sqrtnsum_k=1^nfrac1sqrt2k+sqrt2k-1$$

$$I_k=frac1sqrt2k+sqrt2k-1=frac2sqrtpiint_0^infty e^-(4k-1)x^2 e^-4k x^2 sqrt1-1/(2k) dx$$

Since $(sqrt2k+sqrt2k-1)^2=4 k-1+4 sqrtk(k-1/2)$.

$$sqrt1-frac12k= 1-frac14k-frac132k^2-O left( frac18k^3 right)<1-frac14k$$

We have:

$$I_k > frac2sqrtpi int_0^infty e^-8kx^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1sqrtnsum_k=1^n int_0^infty e^-8kx^2 dx= \ =frac2sqrtpi lim_ntoinftyfrac1sqrtn int_0^infty frace^-8 x^2-e^-8(n+1) x^21-e^-8 x^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frace^-8 y^2/n-e^-8(1+1/n) y^21-e^-8 y^2/n dy$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frac1-e^-8 y^2e^8 y^2/n-1 dy=frac14 sqrtpiint_0^infty frac1-e^-8 y^2y^2 dy$$

$$A geq frac1sqrt2 piint_0^infty frac1-e^-z^2z^2 dz= frac1sqrt2$$

The last integral is easy to prove using integration by parts.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Yuriy SYuriy S

17.6k44 gold badges3434 silver badges123123 bronze badges

$endgroup$

add a comment | 

$begingroup$

Another interesting way to prove the lower bound.

Consider:

$$A=lim_ntoinftyfrac1sqrtnsum_k=1^nfrac1sqrt2k+sqrt2k-1$$

$$I_k=frac1sqrt2k+sqrt2k-1=frac2sqrtpiint_0^infty e^-(4k-1)x^2 e^-4k x^2 sqrt1-1/(2k) dx$$

Since $(sqrt2k+sqrt2k-1)^2=4 k-1+4 sqrtk(k-1/2)$.

$$sqrt1-frac12k= 1-frac14k-frac132k^2-O left( frac18k^3 right)<1-frac14k$$

We have:

$$I_k > frac2sqrtpi int_0^infty e^-8kx^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1sqrtnsum_k=1^n int_0^infty e^-8kx^2 dx= \ =frac2sqrtpi lim_ntoinftyfrac1sqrtn int_0^infty frace^-8 x^2-e^-8(n+1) x^21-e^-8 x^2 dx$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frace^-8 y^2/n-e^-8(1+1/n) y^21-e^-8 y^2/n dy$$

$$A geq frac2sqrtpi lim_ntoinftyfrac1n int_0^infty frac1-e^-8 y^2e^8 y^2/n-1 dy=frac14 sqrtpiint_0^infty frac1-e^-8 y^2y^2 dy$$

$$A geq frac1sqrt2 piint_0^infty frac1-e^-z^2z^2 dz= frac1sqrt2$$

The last integral is easy to prove using integration by parts.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Yuriy SYuriy S

17.6k44 gold badges3434 silver badges123123 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Yuriy SYuriy S

17.6k44 gold badges3434 silver badges123123 bronze badges

answered 7 hours ago

Yuriy SYuriy S

17.6k44 gold badges3434 silver badges123123 bronze badges

answered 7 hours ago

Yuriy SYuriy S

17.6k44 gold badges3434 silver badges123123 bronze badges