limit of implicit sequences
$begingroup$
In a HS problem, we study the functions
$$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$
First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.
From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.
The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.
I'm stuck with this last question.
calculus sequences-and-series
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
$endgroup$
add a comment |
$begingroup$
In a HS problem, we study the functions
$$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$
First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.
From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.
The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.
I'm stuck with this last question.
calculus sequences-and-series
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
$endgroup$
add a comment |
$begingroup$
In a HS problem, we study the functions
$$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$
First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.
From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.
The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.
I'm stuck with this last question.
calculus sequences-and-series
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
$endgroup$
In a HS problem, we study the functions
$$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$
First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.
From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.
The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.
I'm stuck with this last question.
calculus sequences-and-series
calculus sequences-and-series
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
edited Dec 27 '18 at 10:27
Oussama Sarih
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
asked Dec 27 '18 at 8:49
Oussama SarihOussama Sarih
48327
48327
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the first part note that
$$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$
Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.
For the second part
$$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$
answered Dec 27 '18 at 9:09
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
This could be a stupid answer.
You look for the zero of
$$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
$$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
$$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first part note that
$$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$
Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.
For the second part
$$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$
answered Dec 27 '18 at 9:09
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
For the first part note that
$$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$
Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.
For the second part
$$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$
answered Dec 27 '18 at 9:09
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
For the first part note that
$$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$
Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.
For the second part
$$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$
answered Dec 27 '18 at 9:09
ODFODF
1,486510
$endgroup$
For the first part note that
$$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$
Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.
For the second part
$$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$
answered Dec 27 '18 at 9:09
ODFODF
1,486510
answered Dec 27 '18 at 9:09
ODFODF
1,486510
answered Dec 27 '18 at 9:09
ODFODF
1,486510
answered Dec 27 '18 at 9:09
ODFODF
1,486510
1,486510
add a comment |
add a comment |
$begingroup$
This could be a stupid answer.
You look for the zero of
$$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
$$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
$$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
$endgroup$
add a comment |
$begingroup$
This could be a stupid answer.
You look for the zero of
$$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
$$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
$$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
$endgroup$
add a comment |
$begingroup$
This could be a stupid answer.
You look for the zero of
$$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
$$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
$$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
$endgroup$
This could be a stupid answer.
You look for the zero of
$$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
$$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
$$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
answered Dec 27 '18 at 9:32
Claude LeiboviciClaude Leibovici
121k1157134
121k1157134
add a comment |
add a comment |
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