Partial sums of the series $sumlimits_kgeq1frac1sqrt2k+sqrt4k^2-1$

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The series $sumlimits_k=1^inftyfrac1sqrt2k+sqrt4k^2-1$ is divergent. I am interested in its partial sums to do some computations based on them.

I tried to multiply $sqrt2k+sqrt4k^2-1$ by $sqrt2k-sqrt4k^2-1$ and to divide by the same term, but I have got no telescoping sum for evaluation. Maybe I should use others technics.

Now my question is what is the closed form of the sum $$sum_k=1^nfrac1sqrt2k+sqrt4k^2-1 ?$$

sequences-and-series summation

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edited Jul 22 at 20:50

Did

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asked Jul 22 at 15:11

zeraoulia rafik

2,1071823

add a comment | 

up vote
3
down vote

favorite

2

The series $sumlimits_k=1^inftyfrac1sqrt2k+sqrt4k^2-1$ is divergent. I am interested in its partial sums to do some computations based on them.

I tried to multiply $sqrt2k+sqrt4k^2-1$ by $sqrt2k-sqrt4k^2-1$ and to divide by the same term, but I have got no telescoping sum for evaluation. Maybe I should use others technics.

Now my question is what is the closed form of the sum $$sum_k=1^nfrac1sqrt2k+sqrt4k^2-1 ?$$

sequences-and-series summation

share|cite|improve this question

edited Jul 22 at 20:50

Did

242k23208443

asked Jul 22 at 15:11

zeraoulia rafik

2,1071823

add a comment | 

up vote
3
down vote

favorite

2

up vote
3
down vote

favorite

2

2

The series $sumlimits_k=1^inftyfrac1sqrt2k+sqrt4k^2-1$ is divergent. I am interested in its partial sums to do some computations based on them.

I tried to multiply $sqrt2k+sqrt4k^2-1$ by $sqrt2k-sqrt4k^2-1$ and to divide by the same term, but I have got no telescoping sum for evaluation. Maybe I should use others technics.

Now my question is what is the closed form of the sum $$sum_k=1^nfrac1sqrt2k+sqrt4k^2-1 ?$$

sequences-and-series summation

share|cite|improve this question

edited Jul 22 at 20:50

Did

242k23208443

asked Jul 22 at 15:11

zeraoulia rafik

2,1071823

The series $sumlimits_k=1^inftyfrac1sqrt2k+sqrt4k^2-1$ is divergent. I am interested in its partial sums to do some computations based on them.

I tried to multiply $sqrt2k+sqrt4k^2-1$ by $sqrt2k-sqrt4k^2-1$ and to divide by the same term, but I have got no telescoping sum for evaluation. Maybe I should use others technics.

Now my question is what is the closed form of the sum $$sum_k=1^nfrac1sqrt2k+sqrt4k^2-1 ?$$

sequences-and-series summation

share|cite|improve this question

edited Jul 22 at 20:50

Did

242k23208443

asked Jul 22 at 15:11

zeraoulia rafik

2,1071823

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 20:50

Did

242k23208443

edited Jul 22 at 20:50

Did

242k23208443

edited Jul 22 at 20:50

Did

242k23208443

242k23208443

asked Jul 22 at 15:11

zeraoulia rafik

2,1071823

asked Jul 22 at 15:11

zeraoulia rafik

2,1071823

asked Jul 22 at 15:11

zeraoulia rafik

2,1071823

2,1071823

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
9
down vote



accepted

Using the identity

$$sqrta+sqrtb=sqrtfrac12 left(a+sqrta^2-bright)+sqrtfrac12 left(a-sqrta^2-bright)$$

where: $a = 2 k$ and $b=4 k^2-1$, along with evaluating a telescoping series, we find that

$$beginalign
colorredsum _k=1^n frac1sqrt2 k+sqrt4 k^2-1&=sum _k=1^n fracsqrt2sqrt-1+2 k+sqrt1+2 k\\
&=sum _k=1^n fracsqrt2
left(sqrt2 k+1-sqrt2 k-1right)left(sqrt-1+2 k+sqrt1+2 kright) left(sqrt2 k+1-sqrt2 k-1right)\\
&=fracsum _k=1^n
left(sqrt2 k+1-sqrt2 k-1right)sqrt2\\
&=frac-1+sqrt1+2 nsqrt2\\
&=colorred-frac1sqrt2+fracsqrt1+2 nsqrt2
endalign$$

share|cite|improve this answer

edited Jul 23 at 21:44

Mark Viola

126k1172167

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much for the slick answer
  – zeraoulia rafik
  Jul 22 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zeraouliarafik . You're welcome :)
  – Mariusz Iwaniuk
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $sqrt(2k+1)-sqrt(2k-1)$ to yield a telescoping series.
  – Mark Viola
  Jul 22 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola. Thanks for HINT.
  – Mariusz Iwaniuk
  Jul 22 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome. My pleasure.
  – Mark Viola
  Jul 23 at 21:43
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
9
down vote



accepted

Using the identity

$$sqrta+sqrtb=sqrtfrac12 left(a+sqrta^2-bright)+sqrtfrac12 left(a-sqrta^2-bright)$$

where: $a = 2 k$ and $b=4 k^2-1$, along with evaluating a telescoping series, we find that

$$beginalign
colorredsum _k=1^n frac1sqrt2 k+sqrt4 k^2-1&=sum _k=1^n fracsqrt2sqrt-1+2 k+sqrt1+2 k\\
&=sum _k=1^n fracsqrt2
left(sqrt2 k+1-sqrt2 k-1right)left(sqrt-1+2 k+sqrt1+2 kright) left(sqrt2 k+1-sqrt2 k-1right)\\
&=fracsum _k=1^n
left(sqrt2 k+1-sqrt2 k-1right)sqrt2\\
&=frac-1+sqrt1+2 nsqrt2\\
&=colorred-frac1sqrt2+fracsqrt1+2 nsqrt2
endalign$$

share|cite|improve this answer

edited Jul 23 at 21:44

Mark Viola

126k1172167

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much for the slick answer
  – zeraoulia rafik
  Jul 22 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zeraouliarafik . You're welcome :)
  – Mariusz Iwaniuk
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $sqrt(2k+1)-sqrt(2k-1)$ to yield a telescoping series.
  – Mark Viola
  Jul 22 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola. Thanks for HINT.
  – Mariusz Iwaniuk
  Jul 22 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome. My pleasure.
  – Mark Viola
  Jul 23 at 21:43
  

  
  
  
  

add a comment | 

up vote
9
down vote



accepted

Using the identity

$$sqrta+sqrtb=sqrtfrac12 left(a+sqrta^2-bright)+sqrtfrac12 left(a-sqrta^2-bright)$$

where: $a = 2 k$ and $b=4 k^2-1$, along with evaluating a telescoping series, we find that

$$beginalign
colorredsum _k=1^n frac1sqrt2 k+sqrt4 k^2-1&=sum _k=1^n fracsqrt2sqrt-1+2 k+sqrt1+2 k\\
&=sum _k=1^n fracsqrt2
left(sqrt2 k+1-sqrt2 k-1right)left(sqrt-1+2 k+sqrt1+2 kright) left(sqrt2 k+1-sqrt2 k-1right)\\
&=fracsum _k=1^n
left(sqrt2 k+1-sqrt2 k-1right)sqrt2\\
&=frac-1+sqrt1+2 nsqrt2\\
&=colorred-frac1sqrt2+fracsqrt1+2 nsqrt2
endalign$$

share|cite|improve this answer

edited Jul 23 at 21:44

Mark Viola

126k1172167

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much for the slick answer
  – zeraoulia rafik
  Jul 22 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zeraouliarafik . You're welcome :)
  – Mariusz Iwaniuk
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $sqrt(2k+1)-sqrt(2k-1)$ to yield a telescoping series.
  – Mark Viola
  Jul 22 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola. Thanks for HINT.
  – Mariusz Iwaniuk
  Jul 22 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome. My pleasure.
  – Mark Viola
  Jul 23 at 21:43
  

  
  
  
  

add a comment | 

up vote
9
down vote



accepted

up vote
9
down vote



accepted

Using the identity

$$sqrta+sqrtb=sqrtfrac12 left(a+sqrta^2-bright)+sqrtfrac12 left(a-sqrta^2-bright)$$

where: $a = 2 k$ and $b=4 k^2-1$, along with evaluating a telescoping series, we find that

$$beginalign
colorredsum _k=1^n frac1sqrt2 k+sqrt4 k^2-1&=sum _k=1^n fracsqrt2sqrt-1+2 k+sqrt1+2 k\\
&=sum _k=1^n fracsqrt2
left(sqrt2 k+1-sqrt2 k-1right)left(sqrt-1+2 k+sqrt1+2 kright) left(sqrt2 k+1-sqrt2 k-1right)\\
&=fracsum _k=1^n
left(sqrt2 k+1-sqrt2 k-1right)sqrt2\\
&=frac-1+sqrt1+2 nsqrt2\\
&=colorred-frac1sqrt2+fracsqrt1+2 nsqrt2
endalign$$

share|cite|improve this answer

edited Jul 23 at 21:44

Mark Viola

126k1172167

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

Using the identity

$$sqrta+sqrtb=sqrtfrac12 left(a+sqrta^2-bright)+sqrtfrac12 left(a-sqrta^2-bright)$$

where: $a = 2 k$ and $b=4 k^2-1$, along with evaluating a telescoping series, we find that

$$beginalign
colorredsum _k=1^n frac1sqrt2 k+sqrt4 k^2-1&=sum _k=1^n fracsqrt2sqrt-1+2 k+sqrt1+2 k\\
&=sum _k=1^n fracsqrt2
left(sqrt2 k+1-sqrt2 k-1right)left(sqrt-1+2 k+sqrt1+2 kright) left(sqrt2 k+1-sqrt2 k-1right)\\
&=fracsum _k=1^n
left(sqrt2 k+1-sqrt2 k-1right)sqrt2\\
&=frac-1+sqrt1+2 nsqrt2\\
&=colorred-frac1sqrt2+fracsqrt1+2 nsqrt2
endalign$$

share|cite|improve this answer

edited Jul 23 at 21:44

Mark Viola

126k1172167

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 21:44

Mark Viola

126k1172167

edited Jul 23 at 21:44

Mark Viola

126k1172167

edited Jul 23 at 21:44

Mark Viola

126k1172167

126k1172167

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

answered Jul 22 at 15:56

Mariusz Iwaniuk

1,6511615

1,6511615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much for the slick answer
  – zeraoulia rafik
  Jul 22 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zeraouliarafik . You're welcome :)
  – Mariusz Iwaniuk
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $sqrt(2k+1)-sqrt(2k-1)$ to yield a telescoping series.
  – Mark Viola
  Jul 22 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola. Thanks for HINT.
  – Mariusz Iwaniuk
  Jul 22 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome. My pleasure.
  – Mark Viola
  Jul 23 at 21:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks so much for the slick answer
  – zeraoulia rafik
  Jul 22 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zeraouliarafik . You're welcome :)
  – Mariusz Iwaniuk
  Jul 22 at 17:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $sqrt(2k+1)-sqrt(2k-1)$ to yield a telescoping series.
  – Mark Viola
  Jul 22 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola. Thanks for HINT.
  – Mariusz Iwaniuk
  Jul 22 at 18:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome. My pleasure.
  – Mark Viola
  Jul 23 at 21:43
  

  
  
  
  

Thanks so much for the slick answer
– zeraoulia rafik
Jul 22 at 17:09

Thanks so much for the slick answer
– zeraoulia rafik
Jul 22 at 17:09

@zeraouliarafik . You're welcome :)
– Mariusz Iwaniuk
Jul 22 at 17:20

@zeraouliarafik . You're welcome :)
– Mariusz Iwaniuk
Jul 22 at 17:20

1

1

(+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $sqrt(2k+1)-sqrt(2k-1)$ to yield a telescoping series.
– Mark Viola
Jul 22 at 18:18

(+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $sqrt(2k+1)-sqrt(2k-1)$ to yield a telescoping series.
– Mark Viola
Jul 22 at 18:18

@MarkViola. Thanks for HINT.
– Mariusz Iwaniuk
Jul 22 at 18:40

@MarkViola. Thanks for HINT.
– Mariusz Iwaniuk
Jul 22 at 18:40

You're welcome. My pleasure.
– Mark Viola
Jul 23 at 21:43

You're welcome. My pleasure.
– Mark Viola
Jul 23 at 21:43

add a comment | 

 

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