Is there a “greatest function” that converges?
$begingroup$
We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.
But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.
1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?
2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?
I'm just a freshman in high school so I apologize if this is a stupid question.
calculus sequences-and-series convergence
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
$endgroup$
|
show 1 more comment
$begingroup$
We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.
But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.
1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?
2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?
I'm just a freshman in high school so I apologize if this is a stupid question.
calculus sequences-and-series convergence
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
$endgroup$
2
$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
Feb 5 at 21:16
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
Feb 5 at 21:19
9
$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
Feb 5 at 21:21
1
$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
Feb 5 at 21:22
1
$begingroup$
There are lots of other great answers in that linked thread "Is there a slowest rate of divergence of a series?", which I'm sure the OP will find very interesting.
$endgroup$
– Rahul
Feb 6 at 11:36
|
show 1 more comment
$begingroup$
We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.
But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.
1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?
2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?
I'm just a freshman in high school so I apologize if this is a stupid question.
calculus sequences-and-series convergence
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
$endgroup$
We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.
But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.
1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?
2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?
I'm just a freshman in high school so I apologize if this is a stupid question.
calculus sequences-and-series convergence
calculus sequences-and-series convergence
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
asked Feb 5 at 21:11
Electric MoccasinsElectric Moccasins
22126
22126
2
$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
Feb 5 at 21:16
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
Feb 5 at 21:19
9
$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
Feb 5 at 21:21
1
$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
Feb 5 at 21:22
1
$begingroup$
There are lots of other great answers in that linked thread "Is there a slowest rate of divergence of a series?", which I'm sure the OP will find very interesting.
$endgroup$
– Rahul
Feb 6 at 11:36
|
show 1 more comment
2
$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
Feb 5 at 21:16
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
Feb 5 at 21:19
9
$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
Feb 5 at 21:21
1
$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
Feb 5 at 21:22
1
$begingroup$
There are lots of other great answers in that linked thread "Is there a slowest rate of divergence of a series?", which I'm sure the OP will find very interesting.
$endgroup$
– Rahul
Feb 6 at 11:36
2
2
$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
Feb 5 at 21:16
$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
Feb 5 at 21:16
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
Feb 5 at 21:19
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
Feb 5 at 21:19
9
9
$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
Feb 5 at 21:21
$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
Feb 5 at 21:21
1
1
$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
Feb 5 at 21:22
$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
Feb 5 at 21:22
1
1
$begingroup$
There are lots of other great answers in that linked thread "Is there a slowest rate of divergence of a series?", which I'm sure the OP will find very interesting.
$endgroup$
– Rahul
Feb 6 at 11:36
$begingroup$
There are lots of other great answers in that linked thread "Is there a slowest rate of divergence of a series?", which I'm sure the OP will find very interesting.
$endgroup$
– Rahul
Feb 6 at 11:36
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.
2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.
Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.
Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.
Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.
edited Feb 6 at 1:36
user12345
2,0211515
answered Feb 5 at 21:21
MindlackMindlack
4,815210
$endgroup$
add a comment |
$begingroup$
1) $$f(n) = frac{1}{n log(n)^2}$$
2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$
Then
$ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
$endgroup$
1
$begingroup$
If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
$endgroup$
– orlp
Feb 5 at 21:22
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
Feb 6 at 0:54
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
Feb 6 at 1:45
$begingroup$
@Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
$endgroup$
– Electric Moccasins
Feb 6 at 2:32
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
Feb 6 at 5:23
|
show 2 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.
2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.
Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.
Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.
Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.
edited Feb 6 at 1:36
user12345
2,0211515
answered Feb 5 at 21:21
MindlackMindlack
4,815210
$endgroup$
add a comment |
$begingroup$
1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.
2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.
Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.
Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.
Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.
edited Feb 6 at 1:36
user12345
2,0211515
answered Feb 5 at 21:21
MindlackMindlack
4,815210
$endgroup$
add a comment |
$begingroup$
1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.
2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.
Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.
Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.
Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.
edited Feb 6 at 1:36
user12345
2,0211515
answered Feb 5 at 21:21
MindlackMindlack
4,815210
$endgroup$
1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.
2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.
Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.
Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.
Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.
edited Feb 6 at 1:36
user12345
2,0211515
answered Feb 5 at 21:21
MindlackMindlack
4,815210
edited Feb 6 at 1:36
user12345
2,0211515
edited Feb 6 at 1:36
user12345
2,0211515
edited Feb 6 at 1:36
user12345
2,0211515
2,0211515
answered Feb 5 at 21:21
MindlackMindlack
4,815210
answered Feb 5 at 21:21
MindlackMindlack
4,815210
answered Feb 5 at 21:21
MindlackMindlack
4,815210
4,815210
add a comment |
add a comment |
$begingroup$
1) $$f(n) = frac{1}{n log(n)^2}$$
2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$
Then
$ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
$endgroup$
1
$begingroup$
If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
$endgroup$
– orlp
Feb 5 at 21:22
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
Feb 6 at 0:54
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
Feb 6 at 1:45
$begingroup$
@Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
$endgroup$
– Electric Moccasins
Feb 6 at 2:32
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
Feb 6 at 5:23
|
show 2 more comments
$begingroup$
1) $$f(n) = frac{1}{n log(n)^2}$$
2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$
Then
$ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
$endgroup$
1
$begingroup$
If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
$endgroup$
– orlp
Feb 5 at 21:22
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
Feb 6 at 0:54
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
Feb 6 at 1:45
$begingroup$
@Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
$endgroup$
– Electric Moccasins
Feb 6 at 2:32
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
Feb 6 at 5:23
|
show 2 more comments
$begingroup$
1) $$f(n) = frac{1}{n log(n)^2}$$
2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$
Then
$ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
$endgroup$
1) $$f(n) = frac{1}{n log(n)^2}$$
2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$
Then
$ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
answered Feb 5 at 21:21
Robert IsraelRobert Israel
326k23215469
326k23215469
1
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If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
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– orlp
Feb 5 at 21:22
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Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
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– Electric Moccasins
Feb 6 at 0:54
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@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
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– Solomonoff's Secret
Feb 6 at 1:45
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@Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
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– Electric Moccasins
Feb 6 at 2:32
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$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
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– Robert Israel
Feb 6 at 5:23
|
show 2 more comments
1
$begingroup$
If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
$endgroup$
– orlp
Feb 5 at 21:22
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
Feb 6 at 0:54
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
Feb 6 at 1:45
$begingroup$
@Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
$endgroup$
– Electric Moccasins
Feb 6 at 2:32
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
Feb 6 at 5:23
1
1
$begingroup$
If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
$endgroup$
– orlp
Feb 5 at 21:22
$begingroup$
If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
$endgroup$
– orlp
Feb 5 at 21:22
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
Feb 6 at 0:54
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
Feb 6 at 0:54
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
Feb 6 at 1:45
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
Feb 6 at 1:45
$begingroup$
@Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
$endgroup$
– Electric Moccasins
Feb 6 at 2:32
$begingroup$
@Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
$endgroup$
– Electric Moccasins
Feb 6 at 2:32
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
Feb 6 at 5:23
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
Feb 6 at 5:23
|
show 2 more comments
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2
$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
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– Electric Moccasins
Feb 5 at 21:16
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How about $f(x) = frac{1}{xln^2(x)}$ for your first question
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– Jakobian
Feb 5 at 21:19
9
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It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
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– Henning Makholm
Feb 5 at 21:21
1
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From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
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– Martin R
Feb 5 at 21:22
1
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There are lots of other great answers in that linked thread "Is there a slowest rate of divergence of a series?", which I'm sure the OP will find very interesting.
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– Rahul
Feb 6 at 11:36