equivalent of $frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$
The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.
By using standard estimations, my book obtains that
$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$
I don't understand the following : they write that the previous expression is equal to
$frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?
calculus
asked Nov 27 '18 at 21:28
trocho
8519
add a comment |
The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.
By using standard estimations, my book obtains that
$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$
I don't understand the following : they write that the previous expression is equal to
$frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?
calculus
asked Nov 27 '18 at 21:28
trocho
8519
add a comment |
The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.
By using standard estimations, my book obtains that
$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$
I don't understand the following : they write that the previous expression is equal to
$frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?
calculus
asked Nov 27 '18 at 21:28
trocho
8519
The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.
By using standard estimations, my book obtains that
$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$
I don't understand the following : they write that the previous expression is equal to
$frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?
calculus
calculus
asked Nov 27 '18 at 21:28
trocho
8519
asked Nov 27 '18 at 21:28
trocho
8519
asked Nov 27 '18 at 21:28
trocho
8519
asked Nov 27 '18 at 21:28
trocho
8519
asked Nov 27 '18 at 21:28
trocho
8519
8519
add a comment |
add a comment |
1 Answer
1
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We have that
$$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$
and
$$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
$$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
answered Nov 27 '18 at 21:46
gimusi
1
Thanks ! I get it.
– trocho
Nov 27 '18 at 21:47
@trocho You are welcome! Bye
– gimusi
Nov 27 '18 at 21:48
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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We have that
$$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$
and
$$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
$$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
answered Nov 27 '18 at 21:46
gimusi
1
Thanks ! I get it.
– trocho
Nov 27 '18 at 21:47
@trocho You are welcome! Bye
– gimusi
Nov 27 '18 at 21:48
add a comment |
We have that
$$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$
and
$$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
$$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
answered Nov 27 '18 at 21:46
gimusi
1
Thanks ! I get it.
– trocho
Nov 27 '18 at 21:47
@trocho You are welcome! Bye
– gimusi
Nov 27 '18 at 21:48
add a comment |
We have that
$$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$
and
$$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
$$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
answered Nov 27 '18 at 21:46
gimusi
1
We have that
$$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$
and
$$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
$$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$
answered Nov 27 '18 at 21:46
gimusi
1
answered Nov 27 '18 at 21:46
gimusi
1
answered Nov 27 '18 at 21:46
gimusi
1
answered Nov 27 '18 at 21:46
gimusi
1
1
Thanks ! I get it.
– trocho
Nov 27 '18 at 21:47
@trocho You are welcome! Bye
– gimusi
Nov 27 '18 at 21:48
add a comment |
Thanks ! I get it.
– trocho
Nov 27 '18 at 21:47
@trocho You are welcome! Bye
– gimusi
Nov 27 '18 at 21:48
Thanks ! I get it.
– trocho
Nov 27 '18 at 21:47
Thanks ! I get it.
– trocho
Nov 27 '18 at 21:47
@trocho You are welcome! Bye
– gimusi
Nov 27 '18 at 21:48
@trocho You are welcome! Bye
– gimusi
Nov 27 '18 at 21:48
add a comment |
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