A difficulty in understanding a step in a solution.












0












$begingroup$


Here is the solution:



enter image description here



But I could not understand how the last term in the fourth line came from the line before it, could anyone explain this for me please?



EDIT:
I have highlighted the problem:



enter image description here






real-analysis sequences-and-series proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    The expression in brackets is not part of $ln n$.
    $endgroup$
    – Chrystomath
    Jan 29 at 9:02












  • $begingroup$
    Could you be kind enough to tell us what $c_n$ and $a_n$ denote ? (I known we can guess.)
    $endgroup$
    – Yves Daoust
    Jan 29 at 9:06












  • $begingroup$
    yes it is not part @Chrystomath this is why I am confused why he multiplied the n
    $endgroup$
    – hopefully
    Jan 29 at 9:07










  • $begingroup$
    $c_{n} = n ln n $@YvesDaoust
    $endgroup$
    – hopefully
    Jan 29 at 9:08












  • $begingroup$
    @YvesDaoust I have edited the problem
    $endgroup$
    – hopefully
    Jan 29 at 9:18
















0












$begingroup$


Here is the solution:



enter image description here



But I could not understand how the last term in the fourth line came from the line before it, could anyone explain this for me please?



EDIT:
I have highlighted the problem:



enter image description here






real-analysis sequences-and-series proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    The expression in brackets is not part of $ln n$.
    $endgroup$
    – Chrystomath
    Jan 29 at 9:02












  • $begingroup$
    Could you be kind enough to tell us what $c_n$ and $a_n$ denote ? (I known we can guess.)
    $endgroup$
    – Yves Daoust
    Jan 29 at 9:06












  • $begingroup$
    yes it is not part @Chrystomath this is why I am confused why he multiplied the n
    $endgroup$
    – hopefully
    Jan 29 at 9:07










  • $begingroup$
    $c_{n} = n ln n $@YvesDaoust
    $endgroup$
    – hopefully
    Jan 29 at 9:08












  • $begingroup$
    @YvesDaoust I have edited the problem
    $endgroup$
    – hopefully
    Jan 29 at 9:18














0












0








0





$begingroup$


Here is the solution:



enter image description here



But I could not understand how the last term in the fourth line came from the line before it, could anyone explain this for me please?



EDIT:
I have highlighted the problem:



enter image description here






real-analysis sequences-and-series proof-explanation







share|cite|improve this question











$endgroup$




Here is the solution:



enter image description here



But I could not understand how the last term in the fourth line came from the line before it, could anyone explain this for me please?



EDIT:
I have highlighted the problem:



enter image description here






real-analysis sequences-and-series proof-explanation




real-analysis sequences-and-series proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 9:22









YuiTo Cheng

2,1862937




2,1862937










asked Jan 29 at 8:59









hopefullyhopefully

299214




299214












  • $begingroup$
    The expression in brackets is not part of $ln n$.
    $endgroup$
    – Chrystomath
    Jan 29 at 9:02












  • $begingroup$
    Could you be kind enough to tell us what $c_n$ and $a_n$ denote ? (I known we can guess.)
    $endgroup$
    – Yves Daoust
    Jan 29 at 9:06












  • $begingroup$
    yes it is not part @Chrystomath this is why I am confused why he multiplied the n
    $endgroup$
    – hopefully
    Jan 29 at 9:07










  • $begingroup$
    $c_{n} = n ln n $@YvesDaoust
    $endgroup$
    – hopefully
    Jan 29 at 9:08












  • $begingroup$
    @YvesDaoust I have edited the problem
    $endgroup$
    – hopefully
    Jan 29 at 9:18


















  • $begingroup$
    The expression in brackets is not part of $ln n$.
    $endgroup$
    – Chrystomath
    Jan 29 at 9:02












  • $begingroup$
    Could you be kind enough to tell us what $c_n$ and $a_n$ denote ? (I known we can guess.)
    $endgroup$
    – Yves Daoust
    Jan 29 at 9:06












  • $begingroup$
    yes it is not part @Chrystomath this is why I am confused why he multiplied the n
    $endgroup$
    – hopefully
    Jan 29 at 9:07










  • $begingroup$
    $c_{n} = n ln n $@YvesDaoust
    $endgroup$
    – hopefully
    Jan 29 at 9:08












  • $begingroup$
    @YvesDaoust I have edited the problem
    $endgroup$
    – hopefully
    Jan 29 at 9:18
















$begingroup$
The expression in brackets is not part of $ln n$.
$endgroup$
– Chrystomath
Jan 29 at 9:02






$begingroup$
The expression in brackets is not part of $ln n$.
$endgroup$
– Chrystomath
Jan 29 at 9:02














$begingroup$
Could you be kind enough to tell us what $c_n$ and $a_n$ denote ? (I known we can guess.)
$endgroup$
– Yves Daoust
Jan 29 at 9:06






$begingroup$
Could you be kind enough to tell us what $c_n$ and $a_n$ denote ? (I known we can guess.)
$endgroup$
– Yves Daoust
Jan 29 at 9:06














$begingroup$
yes it is not part @Chrystomath this is why I am confused why he multiplied the n
$endgroup$
– hopefully
Jan 29 at 9:07




$begingroup$
yes it is not part @Chrystomath this is why I am confused why he multiplied the n
$endgroup$
– hopefully
Jan 29 at 9:07












$begingroup$
$c_{n} = n ln n $@YvesDaoust
$endgroup$
– hopefully
Jan 29 at 9:08






$begingroup$
$c_{n} = n ln n $@YvesDaoust
$endgroup$
– hopefully
Jan 29 at 9:08














$begingroup$
@YvesDaoust I have edited the problem
$endgroup$
– hopefully
Jan 29 at 9:18




$begingroup$
@YvesDaoust I have edited the problem
$endgroup$
– hopefully
Jan 29 at 9:18










2 Answers
2






active

oldest

votes


















1












$begingroup$

begin{align*}
operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} &= operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} - operatorname{ln}n + operatorname{ln} n \
&= operatorname{ln}nleft(frac{4n^2+5n}{4n^2+8n+4} - 1right) + operatorname{ln}n
end{align*}




Edit: For the new edited question note that



$$noperatorname{ln}nfrac{4n+5}{4n^2+8n+4} = operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4}$$



This is the same situation as $$xyfrac{a}{b} = yfrac{xa}{b}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30






  • 1




    $begingroup$
    @hopefully See the edited answer above
    $endgroup$
    – Perturbative
    Jan 29 at 9:36










  • $begingroup$
    another small question taking limit as n tends to infinity at the last line ..... what happens to the term $ln n $...... it was multiplied by 0 after taking the limit ..... is the effect of a polynomial function more rapid than the effect of the ln function?
    $endgroup$
    – hopefully
    Jan 29 at 9:56





















1












$begingroup$

It's the same as in $xy = x(y-1)+x$. Here $x=ln n$ and $y$ is the fraction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30










  • $begingroup$
    Ah, the third line ... I was wondering why you ask about such an easy thing :D
    $endgroup$
    – Stockfish
    Jan 29 at 12:54












Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

begin{align*}
operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} &= operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} - operatorname{ln}n + operatorname{ln} n \
&= operatorname{ln}nleft(frac{4n^2+5n}{4n^2+8n+4} - 1right) + operatorname{ln}n
end{align*}




Edit: For the new edited question note that



$$noperatorname{ln}nfrac{4n+5}{4n^2+8n+4} = operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4}$$



This is the same situation as $$xyfrac{a}{b} = yfrac{xa}{b}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30






  • 1




    $begingroup$
    @hopefully See the edited answer above
    $endgroup$
    – Perturbative
    Jan 29 at 9:36










  • $begingroup$
    another small question taking limit as n tends to infinity at the last line ..... what happens to the term $ln n $...... it was multiplied by 0 after taking the limit ..... is the effect of a polynomial function more rapid than the effect of the ln function?
    $endgroup$
    – hopefully
    Jan 29 at 9:56


















1












$begingroup$

begin{align*}
operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} &= operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} - operatorname{ln}n + operatorname{ln} n \
&= operatorname{ln}nleft(frac{4n^2+5n}{4n^2+8n+4} - 1right) + operatorname{ln}n
end{align*}




Edit: For the new edited question note that



$$noperatorname{ln}nfrac{4n+5}{4n^2+8n+4} = operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4}$$



This is the same situation as $$xyfrac{a}{b} = yfrac{xa}{b}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30






  • 1




    $begingroup$
    @hopefully See the edited answer above
    $endgroup$
    – Perturbative
    Jan 29 at 9:36










  • $begingroup$
    another small question taking limit as n tends to infinity at the last line ..... what happens to the term $ln n $...... it was multiplied by 0 after taking the limit ..... is the effect of a polynomial function more rapid than the effect of the ln function?
    $endgroup$
    – hopefully
    Jan 29 at 9:56
















1












1








1





$begingroup$

begin{align*}
operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} &= operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} - operatorname{ln}n + operatorname{ln} n \
&= operatorname{ln}nleft(frac{4n^2+5n}{4n^2+8n+4} - 1right) + operatorname{ln}n
end{align*}




Edit: For the new edited question note that



$$noperatorname{ln}nfrac{4n+5}{4n^2+8n+4} = operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4}$$



This is the same situation as $$xyfrac{a}{b} = yfrac{xa}{b}$$






share|cite|improve this answer











$endgroup$



begin{align*}
operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} &= operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4} - operatorname{ln}n + operatorname{ln} n \
&= operatorname{ln}nleft(frac{4n^2+5n}{4n^2+8n+4} - 1right) + operatorname{ln}n
end{align*}




Edit: For the new edited question note that



$$noperatorname{ln}nfrac{4n+5}{4n^2+8n+4} = operatorname{ln}nfrac{4n^2+5n}{4n^2+8n+4}$$



This is the same situation as $$xyfrac{a}{b} = yfrac{xa}{b}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 29 at 9:36

























answered Jan 29 at 9:03









PerturbativePerturbative

4,51621554




4,51621554












  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30






  • 1




    $begingroup$
    @hopefully See the edited answer above
    $endgroup$
    – Perturbative
    Jan 29 at 9:36










  • $begingroup$
    another small question taking limit as n tends to infinity at the last line ..... what happens to the term $ln n $...... it was multiplied by 0 after taking the limit ..... is the effect of a polynomial function more rapid than the effect of the ln function?
    $endgroup$
    – hopefully
    Jan 29 at 9:56




















  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30






  • 1




    $begingroup$
    @hopefully See the edited answer above
    $endgroup$
    – Perturbative
    Jan 29 at 9:36










  • $begingroup$
    another small question taking limit as n tends to infinity at the last line ..... what happens to the term $ln n $...... it was multiplied by 0 after taking the limit ..... is the effect of a polynomial function more rapid than the effect of the ln function?
    $endgroup$
    – hopefully
    Jan 29 at 9:56


















$begingroup$
please read the question carefully
$endgroup$
– hopefully
Jan 29 at 9:05




$begingroup$
please read the question carefully
$endgroup$
– hopefully
Jan 29 at 9:05












$begingroup$
I have clarified my question sorry
$endgroup$
– hopefully
Jan 29 at 9:30




$begingroup$
I have clarified my question sorry
$endgroup$
– hopefully
Jan 29 at 9:30




1




1




$begingroup$
@hopefully See the edited answer above
$endgroup$
– Perturbative
Jan 29 at 9:36




$begingroup$
@hopefully See the edited answer above
$endgroup$
– Perturbative
Jan 29 at 9:36












$begingroup$
another small question taking limit as n tends to infinity at the last line ..... what happens to the term $ln n $...... it was multiplied by 0 after taking the limit ..... is the effect of a polynomial function more rapid than the effect of the ln function?
$endgroup$
– hopefully
Jan 29 at 9:56






$begingroup$
another small question taking limit as n tends to infinity at the last line ..... what happens to the term $ln n $...... it was multiplied by 0 after taking the limit ..... is the effect of a polynomial function more rapid than the effect of the ln function?
$endgroup$
– hopefully
Jan 29 at 9:56













1












$begingroup$

It's the same as in $xy = x(y-1)+x$. Here $x=ln n$ and $y$ is the fraction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30










  • $begingroup$
    Ah, the third line ... I was wondering why you ask about such an easy thing :D
    $endgroup$
    – Stockfish
    Jan 29 at 12:54
















1












$begingroup$

It's the same as in $xy = x(y-1)+x$. Here $x=ln n$ and $y$ is the fraction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30










  • $begingroup$
    Ah, the third line ... I was wondering why you ask about such an easy thing :D
    $endgroup$
    – Stockfish
    Jan 29 at 12:54














1












1








1





$begingroup$

It's the same as in $xy = x(y-1)+x$. Here $x=ln n$ and $y$ is the fraction.






share|cite|improve this answer









$endgroup$



It's the same as in $xy = x(y-1)+x$. Here $x=ln n$ and $y$ is the fraction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 29 at 9:01









StockfishStockfish

62726




62726












  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30










  • $begingroup$
    Ah, the third line ... I was wondering why you ask about such an easy thing :D
    $endgroup$
    – Stockfish
    Jan 29 at 12:54


















  • $begingroup$
    please read the question carefully
    $endgroup$
    – hopefully
    Jan 29 at 9:05










  • $begingroup$
    I have clarified my question sorry
    $endgroup$
    – hopefully
    Jan 29 at 9:30










  • $begingroup$
    Ah, the third line ... I was wondering why you ask about such an easy thing :D
    $endgroup$
    – Stockfish
    Jan 29 at 12:54
















$begingroup$
please read the question carefully
$endgroup$
– hopefully
Jan 29 at 9:05




$begingroup$
please read the question carefully
$endgroup$
– hopefully
Jan 29 at 9:05












$begingroup$
I have clarified my question sorry
$endgroup$
– hopefully
Jan 29 at 9:30




$begingroup$
I have clarified my question sorry
$endgroup$
– hopefully
Jan 29 at 9:30












$begingroup$
Ah, the third line ... I was wondering why you ask about such an easy thing :D
$endgroup$
– Stockfish
Jan 29 at 12:54




$begingroup$
Ah, the third line ... I was wondering why you ask about such an easy thing :D
$endgroup$
– Stockfish
Jan 29 at 12:54


















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