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How to calculate complex expression in WolframAlpha

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1

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I need to evaluate the limit:

$$lim_{ntoinfty}prod_{k=1}^infty left(1-frac{n}{left(frac{n+sqrt{n^2+4}}{2}right)^k+frac{n+sqrt{n^2+4}}{2}}right).$$

I could not type into WolframAlpha and find its value. Can someone help me?

summation wolfram-alpha-queries

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asked 8 hours ago

bilgamishbilgamish

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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The way you could type that into WolframAlpha, or Mathematica is Limit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),{k,1,Infinity}],n->Infinity] but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simpler Limit[Product[1-n/(n^k+n),{k,1,Infinity}],n->Infinity] Perhaps you can think of a way to simplify your problem.
  $endgroup$
  – Bill
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I need to evaluate the limit:

$$lim_{ntoinfty}prod_{k=1}^infty left(1-frac{n}{left(frac{n+sqrt{n^2+4}}{2}right)^k+frac{n+sqrt{n^2+4}}{2}}right).$$

I could not type into WolframAlpha and find its value. Can someone help me?

summation wolfram-alpha-queries

share|improve this question

asked 8 hours ago

bilgamishbilgamish

611 bronze badge

New contributor

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

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The way you could type that into WolframAlpha, or Mathematica is Limit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),{k,1,Infinity}],n->Infinity] but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simpler Limit[Product[1-n/(n^k+n),{k,1,Infinity}],n->Infinity] Perhaps you can think of a way to simplify your problem.
  $endgroup$
  – Bill
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I need to evaluate the limit:

$$lim_{ntoinfty}prod_{k=1}^infty left(1-frac{n}{left(frac{n+sqrt{n^2+4}}{2}right)^k+frac{n+sqrt{n^2+4}}{2}}right).$$

I could not type into WolframAlpha and find its value. Can someone help me?

summation wolfram-alpha-queries

share|improve this question

asked 8 hours ago

bilgamishbilgamish

611 bronze badge

New contributor

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

$endgroup$

share|improve this question

asked 8 hours ago

bilgamishbilgamish

611 bronze badge

New contributor

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

share|improve this question

asked 8 hours ago

bilgamishbilgamish

611 bronze badge

New contributor

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

asked 8 hours ago

bilgamishbilgamish

611 bronze badge

New contributor

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

asked 8 hours ago

bilgamishbilgamish

611 bronze badge

2 Answers
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With Mathematica I have:

func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), {k, 1, Infinity}]

Table[func[10^n], {n, 1, 10}]



(*{0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5} *)

It looks like the limit is: 1/2

share|improve this answer

answered 7 hours ago

Mariusz IwaniukMariusz Iwaniuk

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add a comment | 

1

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Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.

share|improve this answer

answered 4 hours ago

mikadomikado

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2

$begingroup$

With Mathematica I have:

func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), {k, 1, Infinity}]

Table[func[10^n], {n, 1, 10}]



(*{0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5} *)

It looks like the limit is: 1/2

share|improve this answer

answered 7 hours ago

Mariusz IwaniukMariusz Iwaniuk

7,23011 gold badge1212 silver badges3030 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

With Mathematica I have:

func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), {k, 1, Infinity}]

Table[func[10^n], {n, 1, 10}]



(*{0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5} *)

It looks like the limit is: 1/2

share|improve this answer

answered 7 hours ago

Mariusz IwaniukMariusz Iwaniuk

7,23011 gold badge1212 silver badges3030 bronze badges

$endgroup$

add a comment | 

$begingroup$

With Mathematica I have:

func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), {k, 1, Infinity}]

Table[func[10^n], {n, 1, 10}]



(*{0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5} *)

It looks like the limit is: 1/2

share|improve this answer

answered 7 hours ago

Mariusz IwaniukMariusz Iwaniuk

7,23011 gold badge1212 silver badges3030 bronze badges

$endgroup$

func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), {k, 1, Infinity}]

Table[func[10^n], {n, 1, 10}]



(*{0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5} *)

share|improve this answer

answered 7 hours ago

Mariusz IwaniukMariusz Iwaniuk

7,23011 gold badge1212 silver badges3030 bronze badges

answered 7 hours ago

Mariusz IwaniukMariusz Iwaniuk

7,23011 gold badge1212 silver badges3030 bronze badges

answered 7 hours ago

Mariusz IwaniukMariusz Iwaniuk

7,23011 gold badge1212 silver badges3030 bronze badges

1

$begingroup$

Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.

share|improve this answer

answered 4 hours ago

mikadomikado

7,39211 gold badge99 silver badges2929 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.

share|improve this answer

answered 4 hours ago

mikadomikado

7,39211 gold badge99 silver badges2929 bronze badges

$endgroup$

add a comment | 

$begingroup$

Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.

share|improve this answer

answered 4 hours ago

mikadomikado

7,39211 gold badge99 silver badges2929 bronze badges

$endgroup$

share|improve this answer

answered 4 hours ago

mikadomikado

7,39211 gold badge99 silver badges2929 bronze badges

answered 4 hours ago

mikadomikado

7,39211 gold badge99 silver badges2929 bronze badges

answered 4 hours ago

mikadomikado

7,39211 gold badge99 silver badges2929 bronze badges