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 yesterday
Electric MoccasinsElectric Moccasins
17614
$endgroup$
add a 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 yesterday
Electric MoccasinsElectric Moccasins
17614
$endgroup$
1
$begingroup$
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
yesterday
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
yesterday
8
$begingroup$
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
yesterday
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
yesterday
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
14 hours ago
add a 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 yesterday
Electric MoccasinsElectric Moccasins
17614
$endgroup$
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 yesterday
Electric MoccasinsElectric Moccasins
17614
asked yesterday
Electric MoccasinsElectric Moccasins
17614
asked yesterday
Electric MoccasinsElectric Moccasins
17614
asked yesterday
Electric MoccasinsElectric Moccasins
17614
asked yesterday
Electric MoccasinsElectric Moccasins
17614
17614
1
$begingroup$
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
yesterday
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
yesterday
8
$begingroup$
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
yesterday
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
yesterday
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
14 hours ago
add a comment |
1
$begingroup$
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
yesterday
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
yesterday
8
$begingroup$
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
yesterday
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
yesterday
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
14 hours ago
1
1
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
yesterday
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
yesterday
8
8
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
$begingroup$
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
yesterday
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
14 hours ago
$begingroup$
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
14 hours ago
add a 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 yesterday
user12345
2,0231515
answered yesterday
MindlackMindlack
3,73518
$endgroup$
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 yesterday
Robert IsraelRobert Israel
322k23212465
$endgroup$
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
yesterday
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
yesterday
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
yesterday
$begingroup$
@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
23 hours ago
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
20 hours ago
|
show 2 more comments
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2 Answers
2
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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 yesterday
user12345
2,0231515
answered yesterday
MindlackMindlack
3,73518
$endgroup$
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 yesterday
user12345
2,0231515
answered yesterday
MindlackMindlack
3,73518
$endgroup$
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 yesterday
user12345
2,0231515
answered yesterday
MindlackMindlack
3,73518
$endgroup$
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 yesterday
user12345
2,0231515
answered yesterday
MindlackMindlack
3,73518
edited yesterday
user12345
2,0231515
edited yesterday
user12345
2,0231515
edited yesterday
user12345
2,0231515
2,0231515
answered yesterday
MindlackMindlack
3,73518
answered yesterday
MindlackMindlack
3,73518
answered yesterday
MindlackMindlack
3,73518
3,73518
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 yesterday
Robert IsraelRobert Israel
322k23212465
$endgroup$
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
yesterday
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
yesterday
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
yesterday
$begingroup$
@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
23 hours ago
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
20 hours ago
|
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 yesterday
Robert IsraelRobert Israel
322k23212465
$endgroup$
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
yesterday
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
yesterday
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
yesterday
$begingroup$
@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
23 hours ago
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
20 hours ago
|
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 yesterday
Robert IsraelRobert Israel
322k23212465
$endgroup$
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 yesterday
Robert IsraelRobert Israel
322k23212465
answered yesterday
Robert IsraelRobert Israel
322k23212465
answered yesterday
Robert IsraelRobert Israel
322k23212465
answered yesterday
Robert IsraelRobert Israel
322k23212465
322k23212465
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
yesterday
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
yesterday
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
yesterday
$begingroup$
@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".
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– Electric Moccasins
23 hours ago
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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
20 hours ago
|
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.
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– orlp
yesterday
$begingroup$
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
yesterday
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
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– Solomonoff's Secret
yesterday
$begingroup$
@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
23 hours ago
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
20 hours ago
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
yesterday
$begingroup$
Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
$endgroup$
– Electric Moccasins
yesterday
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
yesterday
$begingroup$
@ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
$endgroup$
– Solomonoff's Secret
yesterday
$begingroup$
@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
23 hours ago
$begingroup$
@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
23 hours ago
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
20 hours ago
$begingroup$
$$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
$endgroup$
– Robert Israel
20 hours ago
|
show 2 more comments
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1
$begingroup$
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
yesterday
$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
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– Jakobian
yesterday
8
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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
yesterday
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 $lim_{nrightarrowinfty} C_n/c_n = infty$.
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– Martin R
yesterday
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.
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– Rahul
14 hours ago