Mixing
$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]-left[1^{1^p}2^{2^p}cdots n^{n^p}right]$ converge?
This problem comes from the Titu Andreescu's book Problems in Real Analysis - Chapter 1, page 9.
Let $p$ be a nonnegative real number. Study the convergence of the sequence
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$$
Where $n$ is a positive integer.
Maybe it is useful to know:
$$ lim_{n to infty} frac{left(1^{1^p}2^{2^p}cdots n^{n^p}right)^{1/n^{p+1}}}{n^{1/(p+1)}} = e^{-1/(p+1)^2}label{1}tag{1}$$
Attempt
Just before this exercise, the book solves the case $p =0$ in an example. Using the same reasoning of the example, I did the following:
Define $a_n= left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$ and $b_n = a_{n+1}/a_n$, then
$$x_n = a_{n+1}-a_{n} = a_n(b_n-1) = frac{a_n}{n^{1/(p+1)}}frac{b_n-1}{ln b_n}ln b_n^{n^{1/(p+1)}}$$
But by ref{1}, we have
$$ lim_{n to infty} frac{a_n}{n^{1/(p+1)}} = e^{-1/(p+1)^2}$$
And we also have
$$ lim_{n to infty} b_n = lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}frac{(n+1)^{1/(p+1)}}{n^{1/(p+1)}} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}left(1+frac{1}{n}right)^{1/(p+1)} = 1$$
Thus,
$$ lim_{n to infty} frac{b_n-1}{ln b_n} = lim_{n to infty} left(frac{ln b_n}{b_n-1}right)^{-1} = left[left(frac{d}{dx}ln x right)|_{x=1}right]^{-1}= 1 $$
Since $b_n to 1$. So, we just need to analyze the convergence of $ln b_n^{n^{1/(p+1)}}$, what I couldn't do! However, in the case $p=0$, the book does the following
$$lim_{n to infty} b_n^n = lim_{n to infty} left(frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}right)^n = lim_{n to infty} frac{(n+1)!}{n!}frac{1}{(n+1)!^{1/(n+1)}} = lim_{n to infty} frac{n+1}{(n+1)!^{1/(n+1)}} = e $$
Where we used ref{1} in the last equality.
real-analysis sequences-and-series limits contest-math
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
add a comment |
This problem comes from the Titu Andreescu's book Problems in Real Analysis - Chapter 1, page 9.
Let $p$ be a nonnegative real number. Study the convergence of the sequence
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$$
Where $n$ is a positive integer.
Maybe it is useful to know:
$$ lim_{n to infty} frac{left(1^{1^p}2^{2^p}cdots n^{n^p}right)^{1/n^{p+1}}}{n^{1/(p+1)}} = e^{-1/(p+1)^2}label{1}tag{1}$$
Attempt
Just before this exercise, the book solves the case $p =0$ in an example. Using the same reasoning of the example, I did the following:
Define $a_n= left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$ and $b_n = a_{n+1}/a_n$, then
$$x_n = a_{n+1}-a_{n} = a_n(b_n-1) = frac{a_n}{n^{1/(p+1)}}frac{b_n-1}{ln b_n}ln b_n^{n^{1/(p+1)}}$$
But by ref{1}, we have
$$ lim_{n to infty} frac{a_n}{n^{1/(p+1)}} = e^{-1/(p+1)^2}$$
And we also have
$$ lim_{n to infty} b_n = lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}frac{(n+1)^{1/(p+1)}}{n^{1/(p+1)}} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}left(1+frac{1}{n}right)^{1/(p+1)} = 1$$
Thus,
$$ lim_{n to infty} frac{b_n-1}{ln b_n} = lim_{n to infty} left(frac{ln b_n}{b_n-1}right)^{-1} = left[left(frac{d}{dx}ln x right)|_{x=1}right]^{-1}= 1 $$
Since $b_n to 1$. So, we just need to analyze the convergence of $ln b_n^{n^{1/(p+1)}}$, what I couldn't do! However, in the case $p=0$, the book does the following
$$lim_{n to infty} b_n^n = lim_{n to infty} left(frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}right)^n = lim_{n to infty} frac{(n+1)!}{n!}frac{1}{(n+1)!^{1/(n+1)}} = lim_{n to infty} frac{n+1}{(n+1)!^{1/(n+1)}} = e $$
Where we used ref{1} in the last equality.
real-analysis sequences-and-series limits contest-math
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
1
You can try Cesaro-Stolz on $(log b_{n})/n^{-a}$ where $a=1/(p+1)leq 1$.
– Paramanand Singh
Nov 21 '18 at 12:41
@ParamanandSingh How can I use this? I tried and get again the limit $infty.0$ for $p>0$.
– Rafael Deiga
Nov 21 '18 at 13:34
Well Cesaro-Stolz does complicate the expression and I guess one needs some other kind of simplification here.
– Paramanand Singh
Nov 21 '18 at 14:13
add a comment |
This problem comes from the Titu Andreescu's book Problems in Real Analysis - Chapter 1, page 9.
Let $p$ be a nonnegative real number. Study the convergence of the sequence
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$$
Where $n$ is a positive integer.
Maybe it is useful to know:
$$ lim_{n to infty} frac{left(1^{1^p}2^{2^p}cdots n^{n^p}right)^{1/n^{p+1}}}{n^{1/(p+1)}} = e^{-1/(p+1)^2}label{1}tag{1}$$
Attempt
Just before this exercise, the book solves the case $p =0$ in an example. Using the same reasoning of the example, I did the following:
Define $a_n= left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$ and $b_n = a_{n+1}/a_n$, then
$$x_n = a_{n+1}-a_{n} = a_n(b_n-1) = frac{a_n}{n^{1/(p+1)}}frac{b_n-1}{ln b_n}ln b_n^{n^{1/(p+1)}}$$
But by ref{1}, we have
$$ lim_{n to infty} frac{a_n}{n^{1/(p+1)}} = e^{-1/(p+1)^2}$$
And we also have
$$ lim_{n to infty} b_n = lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}frac{(n+1)^{1/(p+1)}}{n^{1/(p+1)}} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}left(1+frac{1}{n}right)^{1/(p+1)} = 1$$
Thus,
$$ lim_{n to infty} frac{b_n-1}{ln b_n} = lim_{n to infty} left(frac{ln b_n}{b_n-1}right)^{-1} = left[left(frac{d}{dx}ln x right)|_{x=1}right]^{-1}= 1 $$
Since $b_n to 1$. So, we just need to analyze the convergence of $ln b_n^{n^{1/(p+1)}}$, what I couldn't do! However, in the case $p=0$, the book does the following
$$lim_{n to infty} b_n^n = lim_{n to infty} left(frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}right)^n = lim_{n to infty} frac{(n+1)!}{n!}frac{1}{(n+1)!^{1/(n+1)}} = lim_{n to infty} frac{n+1}{(n+1)!^{1/(n+1)}} = e $$
Where we used ref{1} in the last equality.
real-analysis sequences-and-series limits contest-math
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
This problem comes from the Titu Andreescu's book Problems in Real Analysis - Chapter 1, page 9.
Let $p$ be a nonnegative real number. Study the convergence of the sequence
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$$
Where $n$ is a positive integer.
Maybe it is useful to know:
$$ lim_{n to infty} frac{left(1^{1^p}2^{2^p}cdots n^{n^p}right)^{1/n^{p+1}}}{n^{1/(p+1)}} = e^{-1/(p+1)^2}label{1}tag{1}$$
Attempt
Just before this exercise, the book solves the case $p =0$ in an example. Using the same reasoning of the example, I did the following:
Define $a_n= left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}$ and $b_n = a_{n+1}/a_n$, then
$$x_n = a_{n+1}-a_{n} = a_n(b_n-1) = frac{a_n}{n^{1/(p+1)}}frac{b_n-1}{ln b_n}ln b_n^{n^{1/(p+1)}}$$
But by ref{1}, we have
$$ lim_{n to infty} frac{a_n}{n^{1/(p+1)}} = e^{-1/(p+1)^2}$$
And we also have
$$ lim_{n to infty} b_n = lim_{n to infty} frac{a_{n+1}}{a_n} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}frac{(n+1)^{1/(p+1)}}{n^{1/(p+1)}} = lim_{n to infty} frac{a_{n+1}}{(n+1)^{1/(p+1)}}frac{n^{1/(p+1)}}{a_n}left(1+frac{1}{n}right)^{1/(p+1)} = 1$$
Thus,
$$ lim_{n to infty} frac{b_n-1}{ln b_n} = lim_{n to infty} left(frac{ln b_n}{b_n-1}right)^{-1} = left[left(frac{d}{dx}ln x right)|_{x=1}right]^{-1}= 1 $$
Since $b_n to 1$. So, we just need to analyze the convergence of $ln b_n^{n^{1/(p+1)}}$, what I couldn't do! However, in the case $p=0$, the book does the following
$$lim_{n to infty} b_n^n = lim_{n to infty} left(frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}right)^n = lim_{n to infty} frac{(n+1)!}{n!}frac{1}{(n+1)!^{1/(n+1)}} = lim_{n to infty} frac{n+1}{(n+1)!^{1/(n+1)}} = e $$
Where we used ref{1} in the last equality.
real-analysis sequences-and-series limits contest-math
real-analysis sequences-and-series limits contest-math
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
asked Nov 21 '18 at 12:05
Rafael Deiga
662311
662311
1
You can try Cesaro-Stolz on $(log b_{n})/n^{-a}$ where $a=1/(p+1)leq 1$.
– Paramanand Singh
Nov 21 '18 at 12:41
@ParamanandSingh How can I use this? I tried and get again the limit $infty.0$ for $p>0$.
– Rafael Deiga
Nov 21 '18 at 13:34
Well Cesaro-Stolz does complicate the expression and I guess one needs some other kind of simplification here.
– Paramanand Singh
Nov 21 '18 at 14:13
add a comment |
1
You can try Cesaro-Stolz on $(log b_{n})/n^{-a}$ where $a=1/(p+1)leq 1$.
– Paramanand Singh
Nov 21 '18 at 12:41
@ParamanandSingh How can I use this? I tried and get again the limit $infty.0$ for $p>0$.
– Rafael Deiga
Nov 21 '18 at 13:34
Well Cesaro-Stolz does complicate the expression and I guess one needs some other kind of simplification here.
– Paramanand Singh
Nov 21 '18 at 14:13
1
1
You can try Cesaro-Stolz on $(log b_{n})/n^{-a}$ where $a=1/(p+1)leq 1$.
– Paramanand Singh
Nov 21 '18 at 12:41
You can try Cesaro-Stolz on $(log b_{n})/n^{-a}$ where $a=1/(p+1)leq 1$.
– Paramanand Singh
Nov 21 '18 at 12:41
@ParamanandSingh How can I use this? I tried and get again the limit $infty.0$ for $p>0$.
– Rafael Deiga
Nov 21 '18 at 13:34
@ParamanandSingh How can I use this? I tried and get again the limit $infty.0$ for $p>0$.
– Rafael Deiga
Nov 21 '18 at 13:34
Well Cesaro-Stolz does complicate the expression and I guess one needs some other kind of simplification here.
– Paramanand Singh
Nov 21 '18 at 14:13
Well Cesaro-Stolz does complicate the expression and I guess one needs some other kind of simplification here.
– Paramanand Singh
Nov 21 '18 at 14:13
add a comment |
1 Answer
1
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We have that
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=a_{n+1}-a_n$$
with
$$a_n=left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=e^{frac{sum_{k=1}^{n}k^p log k}{n^{p+1}}}sim cn^{frac1{p+1}}$$
indeed
$$frac{sum_{k=1}^{n}k^p log k}{n^{p+1}} =frac1nsum_{k=1}^{n}left(frac knright)^p left(log left(frac knright) +log nright)=$$
$$=int_0^1 x^p log x dx+frac{log n}{n^{p+1}}sum_{k=1}^{n} k^p=frac{log n}{p+1}+I+Oleft(frac{log n}nright)$$
therefore
$$a_{n+1}-a_nsim ccdot left[(n+1)^{frac1{p+1}}-n^{frac1{p+1}}right]$$
which converges for any $p>0$ indeed let $a=frac1{p+1}in (0,1)$
$$(n+1)^a-n^a=n^a(1+1/n)^a-n^asimfrac{a}{n^{1-a}} to 0$$
answered Nov 21 '18 at 14:33
gimusi
1
Before $sim frac{log n}{p+1}$ should be $k^p$ instead of $k^n$?
– Rafael Deiga
Nov 21 '18 at 15:33
@RafaelDeiga Yes of course! I fix that, thanks!
– gimusi
Nov 21 '18 at 15:35
You can't subtract (or add) equivalence relations out of thin air. Be more precise and write down the error terms.
– Gabriel Romon
Nov 21 '18 at 15:46
Yeah, i m a bit lost because i don't know if the errors go to zero when $n$ goes to infinity.
– Rafael Deiga
Nov 21 '18 at 15:48
@GabrielRomon Yes you are right, my aim is to make a simple exposure of the key ideas and let the pleasure to the asker (or readers) to refine the details. I can add more details if requested.
– gimusi
Nov 21 '18 at 15:49
|
show 5 more comments
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1 Answer
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1 Answer
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We have that
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=a_{n+1}-a_n$$
with
$$a_n=left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=e^{frac{sum_{k=1}^{n}k^p log k}{n^{p+1}}}sim cn^{frac1{p+1}}$$
indeed
$$frac{sum_{k=1}^{n}k^p log k}{n^{p+1}} =frac1nsum_{k=1}^{n}left(frac knright)^p left(log left(frac knright) +log nright)=$$
$$=int_0^1 x^p log x dx+frac{log n}{n^{p+1}}sum_{k=1}^{n} k^p=frac{log n}{p+1}+I+Oleft(frac{log n}nright)$$
therefore
$$a_{n+1}-a_nsim ccdot left[(n+1)^{frac1{p+1}}-n^{frac1{p+1}}right]$$
which converges for any $p>0$ indeed let $a=frac1{p+1}in (0,1)$
$$(n+1)^a-n^a=n^a(1+1/n)^a-n^asimfrac{a}{n^{1-a}} to 0$$
answered Nov 21 '18 at 14:33
gimusi
1
Before $sim frac{log n}{p+1}$ should be $k^p$ instead of $k^n$?
– Rafael Deiga
Nov 21 '18 at 15:33
@RafaelDeiga Yes of course! I fix that, thanks!
– gimusi
Nov 21 '18 at 15:35
You can't subtract (or add) equivalence relations out of thin air. Be more precise and write down the error terms.
– Gabriel Romon
Nov 21 '18 at 15:46
Yeah, i m a bit lost because i don't know if the errors go to zero when $n$ goes to infinity.
– Rafael Deiga
Nov 21 '18 at 15:48
@GabrielRomon Yes you are right, my aim is to make a simple exposure of the key ideas and let the pleasure to the asker (or readers) to refine the details. I can add more details if requested.
– gimusi
Nov 21 '18 at 15:49
|
show 5 more comments
We have that
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=a_{n+1}-a_n$$
with
$$a_n=left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=e^{frac{sum_{k=1}^{n}k^p log k}{n^{p+1}}}sim cn^{frac1{p+1}}$$
indeed
$$frac{sum_{k=1}^{n}k^p log k}{n^{p+1}} =frac1nsum_{k=1}^{n}left(frac knright)^p left(log left(frac knright) +log nright)=$$
$$=int_0^1 x^p log x dx+frac{log n}{n^{p+1}}sum_{k=1}^{n} k^p=frac{log n}{p+1}+I+Oleft(frac{log n}nright)$$
therefore
$$a_{n+1}-a_nsim ccdot left[(n+1)^{frac1{p+1}}-n^{frac1{p+1}}right]$$
which converges for any $p>0$ indeed let $a=frac1{p+1}in (0,1)$
$$(n+1)^a-n^a=n^a(1+1/n)^a-n^asimfrac{a}{n^{1-a}} to 0$$
answered Nov 21 '18 at 14:33
gimusi
1
Before $sim frac{log n}{p+1}$ should be $k^p$ instead of $k^n$?
– Rafael Deiga
Nov 21 '18 at 15:33
@RafaelDeiga Yes of course! I fix that, thanks!
– gimusi
Nov 21 '18 at 15:35
You can't subtract (or add) equivalence relations out of thin air. Be more precise and write down the error terms.
– Gabriel Romon
Nov 21 '18 at 15:46
Yeah, i m a bit lost because i don't know if the errors go to zero when $n$ goes to infinity.
– Rafael Deiga
Nov 21 '18 at 15:48
@GabrielRomon Yes you are right, my aim is to make a simple exposure of the key ideas and let the pleasure to the asker (or readers) to refine the details. I can add more details if requested.
– gimusi
Nov 21 '18 at 15:49
|
show 5 more comments
We have that
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=a_{n+1}-a_n$$
with
$$a_n=left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=e^{frac{sum_{k=1}^{n}k^p log k}{n^{p+1}}}sim cn^{frac1{p+1}}$$
indeed
$$frac{sum_{k=1}^{n}k^p log k}{n^{p+1}} =frac1nsum_{k=1}^{n}left(frac knright)^p left(log left(frac knright) +log nright)=$$
$$=int_0^1 x^p log x dx+frac{log n}{n^{p+1}}sum_{k=1}^{n} k^p=frac{log n}{p+1}+I+Oleft(frac{log n}nright)$$
therefore
$$a_{n+1}-a_nsim ccdot left[(n+1)^{frac1{p+1}}-n^{frac1{p+1}}right]$$
which converges for any $p>0$ indeed let $a=frac1{p+1}in (0,1)$
$$(n+1)^a-n^a=n^a(1+1/n)^a-n^asimfrac{a}{n^{1-a}} to 0$$
answered Nov 21 '18 at 14:33
gimusi
1
We have that
$$x_n = left[1^{1^p}2^{2^p}cdots (n+1)^{(n+1)^p}right]^{1/(n+1)^{p+1}}-left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=a_{n+1}-a_n$$
with
$$a_n=left[1^{1^p}2^{2^p}cdots n^{n^p}right]^{1/n^{p+1}}=e^{frac{sum_{k=1}^{n}k^p log k}{n^{p+1}}}sim cn^{frac1{p+1}}$$
indeed
$$frac{sum_{k=1}^{n}k^p log k}{n^{p+1}} =frac1nsum_{k=1}^{n}left(frac knright)^p left(log left(frac knright) +log nright)=$$
$$=int_0^1 x^p log x dx+frac{log n}{n^{p+1}}sum_{k=1}^{n} k^p=frac{log n}{p+1}+I+Oleft(frac{log n}nright)$$
therefore
$$a_{n+1}-a_nsim ccdot left[(n+1)^{frac1{p+1}}-n^{frac1{p+1}}right]$$
which converges for any $p>0$ indeed let $a=frac1{p+1}in (0,1)$
$$(n+1)^a-n^a=n^a(1+1/n)^a-n^asimfrac{a}{n^{1-a}} to 0$$
answered Nov 21 '18 at 14:33
gimusi
1
edited Nov 21 '18 at 15:56
answered Nov 21 '18 at 14:33
gimusi
1
answered Nov 21 '18 at 14:33
gimusi
1
answered Nov 21 '18 at 14:33
gimusi
1
1
Before $sim frac{log n}{p+1}$ should be $k^p$ instead of $k^n$?
– Rafael Deiga
Nov 21 '18 at 15:33
@RafaelDeiga Yes of course! I fix that, thanks!
– gimusi
Nov 21 '18 at 15:35
You can't subtract (or add) equivalence relations out of thin air. Be more precise and write down the error terms.
– Gabriel Romon
Nov 21 '18 at 15:46
Yeah, i m a bit lost because i don't know if the errors go to zero when $n$ goes to infinity.
– Rafael Deiga
Nov 21 '18 at 15:48
@GabrielRomon Yes you are right, my aim is to make a simple exposure of the key ideas and let the pleasure to the asker (or readers) to refine the details. I can add more details if requested.
– gimusi
Nov 21 '18 at 15:49
|
show 5 more comments
Before $sim frac{log n}{p+1}$ should be $k^p$ instead of $k^n$?
– Rafael Deiga
Nov 21 '18 at 15:33
@RafaelDeiga Yes of course! I fix that, thanks!
– gimusi
Nov 21 '18 at 15:35
You can't subtract (or add) equivalence relations out of thin air. Be more precise and write down the error terms.
– Gabriel Romon
Nov 21 '18 at 15:46
Yeah, i m a bit lost because i don't know if the errors go to zero when $n$ goes to infinity.
– Rafael Deiga
Nov 21 '18 at 15:48
@GabrielRomon Yes you are right, my aim is to make a simple exposure of the key ideas and let the pleasure to the asker (or readers) to refine the details. I can add more details if requested.
– gimusi
Nov 21 '18 at 15:49
Before $sim frac{log n}{p+1}$ should be $k^p$ instead of $k^n$?
– Rafael Deiga
Nov 21 '18 at 15:33
Before $sim frac{log n}{p+1}$ should be $k^p$ instead of $k^n$?
– Rafael Deiga
Nov 21 '18 at 15:33
@RafaelDeiga Yes of course! I fix that, thanks!
– gimusi
Nov 21 '18 at 15:35
@RafaelDeiga Yes of course! I fix that, thanks!
– gimusi
Nov 21 '18 at 15:35
You can't subtract (or add) equivalence relations out of thin air. Be more precise and write down the error terms.
– Gabriel Romon
Nov 21 '18 at 15:46
You can't subtract (or add) equivalence relations out of thin air. Be more precise and write down the error terms.
– Gabriel Romon
Nov 21 '18 at 15:46
Yeah, i m a bit lost because i don't know if the errors go to zero when $n$ goes to infinity.
– Rafael Deiga
Nov 21 '18 at 15:48
Yeah, i m a bit lost because i don't know if the errors go to zero when $n$ goes to infinity.
– Rafael Deiga
Nov 21 '18 at 15:48
@GabrielRomon Yes you are right, my aim is to make a simple exposure of the key ideas and let the pleasure to the asker (or readers) to refine the details. I can add more details if requested.
– gimusi
Nov 21 '18 at 15:49
@GabrielRomon Yes you are right, my aim is to make a simple exposure of the key ideas and let the pleasure to the asker (or readers) to refine the details. I can add more details if requested.
– gimusi
Nov 21 '18 at 15:49
|
show 5 more comments
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1
You can try Cesaro-Stolz on $(log b_{n})/n^{-a}$ where $a=1/(p+1)leq 1$.
– Paramanand Singh
Nov 21 '18 at 12:41
@ParamanandSingh How can I use this? I tried and get again the limit $infty.0$ for $p>0$.
– Rafael Deiga
Nov 21 '18 at 13:34
Well Cesaro-Stolz does complicate the expression and I guess one needs some other kind of simplification here.
– Paramanand Singh
Nov 21 '18 at 14:13