how does sqrt(n)/n equal 1/sqrt(n)?












0












$begingroup$


I'm working through a probability book where the following is stated (chapter 3):




Definition 3.3



Let $a_n$ and $b_n$ be two sequences of numbers.



We say $a_n$ is asymptotically equal to $b_n$, and write $a_n sim b_n$, if ...



$lim_{n to infty} frac{a_n}{b_n} = 1$



Example 3.4



If $a_n = n + sqrt{n}$ and $b_n = n$ then, since $frac{a_n}{b_n} = 1 + 1/sqrt{n}$ and this ratio tends to 1 as $n$ tends to infinity, we have $a_n sim b_n$




My math is very rusty (15+ years), so I'm confused how the following is
arrived at:



$frac{a_n}{b_n} = 1 + frac{1}{sqrt{n}}$



I expected it to be the following ...



$frac{a_n}{b_n} = frac{sqrt{n} + n}{n} = 1 + frac{sqrt{n}}{n}$




Simple answers will be preferred over complex answers.




Edit:



Checking with sympy, it seems that both the book and my version are the same:



>>> a, b, n = symbols('a b n')

>>> a = n + sqrt(n)

>>> b = n    

>>> simplify( (a/b) - (1 + 1/sqrt(n)) ) == 0

True

>>> simplify( (a/b) - (1 + sqrt(n)/n) ) == 0

True



# Double check equality

>>> simplify( (1 + sqrt(n)/n) - (1 + 1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) - (1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) )

1 / sqrt(n)



# Double check I'm not doing something stupid

>>> simplify( (a/b) - (1) ) == 0

False


So my question becomes how can I derive my equation from the book equation.






asymptotics stirling-numbers transpose







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $n=sqrt{n}cdotsqrt{n}$
    $endgroup$
    – Wojowu
    Jan 11 at 18:57
















0












$begingroup$


I'm working through a probability book where the following is stated (chapter 3):




Definition 3.3



Let $a_n$ and $b_n$ be two sequences of numbers.



We say $a_n$ is asymptotically equal to $b_n$, and write $a_n sim b_n$, if ...



$lim_{n to infty} frac{a_n}{b_n} = 1$



Example 3.4



If $a_n = n + sqrt{n}$ and $b_n = n$ then, since $frac{a_n}{b_n} = 1 + 1/sqrt{n}$ and this ratio tends to 1 as $n$ tends to infinity, we have $a_n sim b_n$




My math is very rusty (15+ years), so I'm confused how the following is
arrived at:



$frac{a_n}{b_n} = 1 + frac{1}{sqrt{n}}$



I expected it to be the following ...



$frac{a_n}{b_n} = frac{sqrt{n} + n}{n} = 1 + frac{sqrt{n}}{n}$




Simple answers will be preferred over complex answers.




Edit:



Checking with sympy, it seems that both the book and my version are the same:



>>> a, b, n = symbols('a b n')

>>> a = n + sqrt(n)

>>> b = n    

>>> simplify( (a/b) - (1 + 1/sqrt(n)) ) == 0

True

>>> simplify( (a/b) - (1 + sqrt(n)/n) ) == 0

True



# Double check equality

>>> simplify( (1 + sqrt(n)/n) - (1 + 1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) - (1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) )

1 / sqrt(n)



# Double check I'm not doing something stupid

>>> simplify( (a/b) - (1) ) == 0

False


So my question becomes how can I derive my equation from the book equation.






asymptotics stirling-numbers transpose







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $n=sqrt{n}cdotsqrt{n}$
    $endgroup$
    – Wojowu
    Jan 11 at 18:57














0












0








0





$begingroup$


I'm working through a probability book where the following is stated (chapter 3):




Definition 3.3



Let $a_n$ and $b_n$ be two sequences of numbers.



We say $a_n$ is asymptotically equal to $b_n$, and write $a_n sim b_n$, if ...



$lim_{n to infty} frac{a_n}{b_n} = 1$



Example 3.4



If $a_n = n + sqrt{n}$ and $b_n = n$ then, since $frac{a_n}{b_n} = 1 + 1/sqrt{n}$ and this ratio tends to 1 as $n$ tends to infinity, we have $a_n sim b_n$




My math is very rusty (15+ years), so I'm confused how the following is
arrived at:



$frac{a_n}{b_n} = 1 + frac{1}{sqrt{n}}$



I expected it to be the following ...



$frac{a_n}{b_n} = frac{sqrt{n} + n}{n} = 1 + frac{sqrt{n}}{n}$




Simple answers will be preferred over complex answers.




Edit:



Checking with sympy, it seems that both the book and my version are the same:



>>> a, b, n = symbols('a b n')

>>> a = n + sqrt(n)

>>> b = n    

>>> simplify( (a/b) - (1 + 1/sqrt(n)) ) == 0

True

>>> simplify( (a/b) - (1 + sqrt(n)/n) ) == 0

True



# Double check equality

>>> simplify( (1 + sqrt(n)/n) - (1 + 1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) - (1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) )

1 / sqrt(n)



# Double check I'm not doing something stupid

>>> simplify( (a/b) - (1) ) == 0

False


So my question becomes how can I derive my equation from the book equation.






asymptotics stirling-numbers transpose







share|cite|improve this question











$endgroup$




I'm working through a probability book where the following is stated (chapter 3):




Definition 3.3



Let $a_n$ and $b_n$ be two sequences of numbers.



We say $a_n$ is asymptotically equal to $b_n$, and write $a_n sim b_n$, if ...



$lim_{n to infty} frac{a_n}{b_n} = 1$



Example 3.4



If $a_n = n + sqrt{n}$ and $b_n = n$ then, since $frac{a_n}{b_n} = 1 + 1/sqrt{n}$ and this ratio tends to 1 as $n$ tends to infinity, we have $a_n sim b_n$




My math is very rusty (15+ years), so I'm confused how the following is
arrived at:



$frac{a_n}{b_n} = 1 + frac{1}{sqrt{n}}$



I expected it to be the following ...



$frac{a_n}{b_n} = frac{sqrt{n} + n}{n} = 1 + frac{sqrt{n}}{n}$




Simple answers will be preferred over complex answers.




Edit:



Checking with sympy, it seems that both the book and my version are the same:



>>> a, b, n = symbols('a b n')

>>> a = n + sqrt(n)

>>> b = n    

>>> simplify( (a/b) - (1 + 1/sqrt(n)) ) == 0

True

>>> simplify( (a/b) - (1 + sqrt(n)/n) ) == 0

True



# Double check equality

>>> simplify( (1 + sqrt(n)/n) - (1 + 1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) - (1/sqrt(n)) ) == 0

True



# and ...

>>> simplify( (sqrt(n)/n) )

1 / sqrt(n)



# Double check I'm not doing something stupid

>>> simplify( (a/b) - (1) ) == 0

False


So my question becomes how can I derive my equation from the book equation.






asymptotics stirling-numbers transpose




asymptotics stirling-numbers transpose






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 18:54







Chris Snow

















asked Jan 11 at 18:28









Chris SnowChris Snow

1738




1738








  • 4




    $begingroup$
    $n=sqrt{n}cdotsqrt{n}$
    $endgroup$
    – Wojowu
    Jan 11 at 18:57














  • 4




    $begingroup$
    $n=sqrt{n}cdotsqrt{n}$
    $endgroup$
    – Wojowu
    Jan 11 at 18:57








4




4




$begingroup$
$n=sqrt{n}cdotsqrt{n}$
$endgroup$
– Wojowu
Jan 11 at 18:57




$begingroup$
$n=sqrt{n}cdotsqrt{n}$
$endgroup$
– Wojowu
Jan 11 at 18:57










1 Answer
1






active

oldest

votes


















1












$begingroup$

$n = sqrt{n}^2$



So in your result you can simplify both numerator and denominator by $sqrt{n}$ and get the book's result. Or, starting from the book's result, you just have to multiply both by $sqrt{n}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You use LaTeX exactly the same way as on desktop - put code between dollar signs.
    $endgroup$
    – Wojowu
    Jan 11 at 18:59










  • $begingroup$
    Thanks for that, I'm new on StackExchange so I didn't know it was that simple
    $endgroup$
    – Ashik4ga
    Jan 11 at 19:00











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

$n = sqrt{n}^2$



So in your result you can simplify both numerator and denominator by $sqrt{n}$ and get the book's result. Or, starting from the book's result, you just have to multiply both by $sqrt{n}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You use LaTeX exactly the same way as on desktop - put code between dollar signs.
    $endgroup$
    – Wojowu
    Jan 11 at 18:59










  • $begingroup$
    Thanks for that, I'm new on StackExchange so I didn't know it was that simple
    $endgroup$
    – Ashik4ga
    Jan 11 at 19:00
















1












$begingroup$

$n = sqrt{n}^2$



So in your result you can simplify both numerator and denominator by $sqrt{n}$ and get the book's result. Or, starting from the book's result, you just have to multiply both by $sqrt{n}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You use LaTeX exactly the same way as on desktop - put code between dollar signs.
    $endgroup$
    – Wojowu
    Jan 11 at 18:59










  • $begingroup$
    Thanks for that, I'm new on StackExchange so I didn't know it was that simple
    $endgroup$
    – Ashik4ga
    Jan 11 at 19:00














1












1








1





$begingroup$

$n = sqrt{n}^2$



So in your result you can simplify both numerator and denominator by $sqrt{n}$ and get the book's result. Or, starting from the book's result, you just have to multiply both by $sqrt{n}$.






share|cite|improve this answer











$endgroup$



$n = sqrt{n}^2$



So in your result you can simplify both numerator and denominator by $sqrt{n}$ and get the book's result. Or, starting from the book's result, you just have to multiply both by $sqrt{n}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 19:27









Chris Snow

1738




1738










answered Jan 11 at 18:57









Ashik4gaAshik4ga

463




463












  • $begingroup$
    You use LaTeX exactly the same way as on desktop - put code between dollar signs.
    $endgroup$
    – Wojowu
    Jan 11 at 18:59










  • $begingroup$
    Thanks for that, I'm new on StackExchange so I didn't know it was that simple
    $endgroup$
    – Ashik4ga
    Jan 11 at 19:00


















  • $begingroup$
    You use LaTeX exactly the same way as on desktop - put code between dollar signs.
    $endgroup$
    – Wojowu
    Jan 11 at 18:59










  • $begingroup$
    Thanks for that, I'm new on StackExchange so I didn't know it was that simple
    $endgroup$
    – Ashik4ga
    Jan 11 at 19:00
















$begingroup$
You use LaTeX exactly the same way as on desktop - put code between dollar signs.
$endgroup$
– Wojowu
Jan 11 at 18:59




$begingroup$
You use LaTeX exactly the same way as on desktop - put code between dollar signs.
$endgroup$
– Wojowu
Jan 11 at 18:59












$begingroup$
Thanks for that, I'm new on StackExchange so I didn't know it was that simple
$endgroup$
– Ashik4ga
Jan 11 at 19:00




$begingroup$
Thanks for that, I'm new on StackExchange so I didn't know it was that simple
$endgroup$
– Ashik4ga
Jan 11 at 19:00


















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