What can you think of for the harmonic series?
up vote
2
down vote
favorite
Background
When I attempt questions in real analysis, frequently I encounter the following expression like harmonic series:
$$1+frac{1}{2} + frac{1}{3} + dots + frac{1}{n}$$
To my surprise when I arrive at something like this ,the solution book does not give me a closed-form of this expression.
Question
So my question is a simple one. What can you think of for this series? And besides the fact that it does not converge is it related to any mathematics that you know? And most importantly, does it have a closed-form?
real-analysis number-theory harmonic-numbers
add a comment |
up vote
2
down vote
favorite
Background
When I attempt questions in real analysis, frequently I encounter the following expression like harmonic series:
$$1+frac{1}{2} + frac{1}{3} + dots + frac{1}{n}$$
To my surprise when I arrive at something like this ,the solution book does not give me a closed-form of this expression.
Question
So my question is a simple one. What can you think of for this series? And besides the fact that it does not converge is it related to any mathematics that you know? And most importantly, does it have a closed-form?
real-analysis number-theory harmonic-numbers
1
It is divergent .....
– neelkanth
45 mins ago
1
Look out for the Euler-Mascheroni constant $gamma$ if you want to explore the mathematical content behind your question. The sum is asymptotically $ln n + gamma$. The error term can be analysed.
– Mark Bennet
44 mins ago
OP probably meant to say "does not converge," since it's like the most well-known fact about the series.
– Eevee Trainer
44 mins ago
@neelkanth the typo is corrected
– hephaes
29 mins ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Background
When I attempt questions in real analysis, frequently I encounter the following expression like harmonic series:
$$1+frac{1}{2} + frac{1}{3} + dots + frac{1}{n}$$
To my surprise when I arrive at something like this ,the solution book does not give me a closed-form of this expression.
Question
So my question is a simple one. What can you think of for this series? And besides the fact that it does not converge is it related to any mathematics that you know? And most importantly, does it have a closed-form?
real-analysis number-theory harmonic-numbers
Background
When I attempt questions in real analysis, frequently I encounter the following expression like harmonic series:
$$1+frac{1}{2} + frac{1}{3} + dots + frac{1}{n}$$
To my surprise when I arrive at something like this ,the solution book does not give me a closed-form of this expression.
Question
So my question is a simple one. What can you think of for this series? And besides the fact that it does not converge is it related to any mathematics that you know? And most importantly, does it have a closed-form?
real-analysis number-theory harmonic-numbers
real-analysis number-theory harmonic-numbers
edited 29 mins ago
asked 54 mins ago
hephaes
1667
1667
1
It is divergent .....
– neelkanth
45 mins ago
1
Look out for the Euler-Mascheroni constant $gamma$ if you want to explore the mathematical content behind your question. The sum is asymptotically $ln n + gamma$. The error term can be analysed.
– Mark Bennet
44 mins ago
OP probably meant to say "does not converge," since it's like the most well-known fact about the series.
– Eevee Trainer
44 mins ago
@neelkanth the typo is corrected
– hephaes
29 mins ago
add a comment |
1
It is divergent .....
– neelkanth
45 mins ago
1
Look out for the Euler-Mascheroni constant $gamma$ if you want to explore the mathematical content behind your question. The sum is asymptotically $ln n + gamma$. The error term can be analysed.
– Mark Bennet
44 mins ago
OP probably meant to say "does not converge," since it's like the most well-known fact about the series.
– Eevee Trainer
44 mins ago
@neelkanth the typo is corrected
– hephaes
29 mins ago
1
1
It is divergent .....
– neelkanth
45 mins ago
It is divergent .....
– neelkanth
45 mins ago
1
1
Look out for the Euler-Mascheroni constant $gamma$ if you want to explore the mathematical content behind your question. The sum is asymptotically $ln n + gamma$. The error term can be analysed.
– Mark Bennet
44 mins ago
Look out for the Euler-Mascheroni constant $gamma$ if you want to explore the mathematical content behind your question. The sum is asymptotically $ln n + gamma$. The error term can be analysed.
– Mark Bennet
44 mins ago
OP probably meant to say "does not converge," since it's like the most well-known fact about the series.
– Eevee Trainer
44 mins ago
OP probably meant to say "does not converge," since it's like the most well-known fact about the series.
– Eevee Trainer
44 mins ago
@neelkanth the typo is corrected
– hephaes
29 mins ago
@neelkanth the typo is corrected
– hephaes
29 mins ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
The harmonic series diverges, which is somewhat surprising because each term tends to zero as $ntoinfty$. However, the problem is that each term does not tend to zero fast enough.
A function which generalises this series is the Riemann zeta function $$zeta(s)=sum_{n=1}^inftyfrac{1}{n^s}$$
Which converges for $text{Re}(s)>1$. For example,
$$zeta(2)=frac{pi^2}{6}.$$
Euler gave a general formula for $zeta(2n)$ in terms of the Bernoulli numbers, but no closed form is known yet for odd arguments.
It can also be analytically continued to $mathbb{C}setminus{1}$. There is a singularity ar $s=1$.
Zeros of this function lie on the negative even integers (the trivial zeros), whereas the non trivial (purely complex) zeros are all known to lie in the critical strip. In fact it is hypothesised that they all lie on the critical line $s=1/2+it$, which you may know is called the Riemann Hypothesis. If true, this has big implications concerning the distribution of the primes, and many conjectured theorems in Analytic Number Theory.
Coming back to "$zeta(1)$" for a moment, as mentioned in the other answer, it pops up in the definition of Euler's gamma constant,
$$gamma=lim_{ntoinfty}sum_{k=1}^nfrac{1}{k}-log n,$$
where $gamma$ constant with
$$gammaapprox
0.57721566490...$$
It is not yet known whether $gamma$ is irrational, unlike $zeta(2n)$, and also $zeta(3)$ which was proved to be irrational by Apéry.
add a comment |
up vote
4
down vote
If I remember correctly, there is a sort of closed form:
$$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_k$$
$epsilon_k$ in this constant is a constant proportional to $1/2k$, and thus $epsilon_k to 0$ as $k to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}}$$
Granted, I don't think this is really in line with what you want because ... well, it's sort of like a self-referential thing. "Oh, the harmonic series is given by a function, an error constant, and this special constant that comes from the difference from the series and that other function under certain conditions." It's mostly a personal thing so it doesn't mesh quite well with me?
Beyond that I don't really have anything to offer - specifically with relations of the series to mathematics I know - that I wouldn't just be regurgitating from Wikipedia. Weirdly it hasn't popped up much in my coursework thus far.
add a comment |
up vote
0
down vote
It diverges because, if you look at the partial sums, $s_{2n}-s_n=frac1{2n}+frac1{2n-1}+cdots+frac1{n+1}ge ncdot frac1{2n}=frac12,,forall n$. Thus it isn't Cauchy.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
The harmonic series diverges, which is somewhat surprising because each term tends to zero as $ntoinfty$. However, the problem is that each term does not tend to zero fast enough.
A function which generalises this series is the Riemann zeta function $$zeta(s)=sum_{n=1}^inftyfrac{1}{n^s}$$
Which converges for $text{Re}(s)>1$. For example,
$$zeta(2)=frac{pi^2}{6}.$$
Euler gave a general formula for $zeta(2n)$ in terms of the Bernoulli numbers, but no closed form is known yet for odd arguments.
It can also be analytically continued to $mathbb{C}setminus{1}$. There is a singularity ar $s=1$.
Zeros of this function lie on the negative even integers (the trivial zeros), whereas the non trivial (purely complex) zeros are all known to lie in the critical strip. In fact it is hypothesised that they all lie on the critical line $s=1/2+it$, which you may know is called the Riemann Hypothesis. If true, this has big implications concerning the distribution of the primes, and many conjectured theorems in Analytic Number Theory.
Coming back to "$zeta(1)$" for a moment, as mentioned in the other answer, it pops up in the definition of Euler's gamma constant,
$$gamma=lim_{ntoinfty}sum_{k=1}^nfrac{1}{k}-log n,$$
where $gamma$ constant with
$$gammaapprox
0.57721566490...$$
It is not yet known whether $gamma$ is irrational, unlike $zeta(2n)$, and also $zeta(3)$ which was proved to be irrational by Apéry.
add a comment |
up vote
6
down vote
The harmonic series diverges, which is somewhat surprising because each term tends to zero as $ntoinfty$. However, the problem is that each term does not tend to zero fast enough.
A function which generalises this series is the Riemann zeta function $$zeta(s)=sum_{n=1}^inftyfrac{1}{n^s}$$
Which converges for $text{Re}(s)>1$. For example,
$$zeta(2)=frac{pi^2}{6}.$$
Euler gave a general formula for $zeta(2n)$ in terms of the Bernoulli numbers, but no closed form is known yet for odd arguments.
It can also be analytically continued to $mathbb{C}setminus{1}$. There is a singularity ar $s=1$.
Zeros of this function lie on the negative even integers (the trivial zeros), whereas the non trivial (purely complex) zeros are all known to lie in the critical strip. In fact it is hypothesised that they all lie on the critical line $s=1/2+it$, which you may know is called the Riemann Hypothesis. If true, this has big implications concerning the distribution of the primes, and many conjectured theorems in Analytic Number Theory.
Coming back to "$zeta(1)$" for a moment, as mentioned in the other answer, it pops up in the definition of Euler's gamma constant,
$$gamma=lim_{ntoinfty}sum_{k=1}^nfrac{1}{k}-log n,$$
where $gamma$ constant with
$$gammaapprox
0.57721566490...$$
It is not yet known whether $gamma$ is irrational, unlike $zeta(2n)$, and also $zeta(3)$ which was proved to be irrational by Apéry.
add a comment |
up vote
6
down vote
up vote
6
down vote
The harmonic series diverges, which is somewhat surprising because each term tends to zero as $ntoinfty$. However, the problem is that each term does not tend to zero fast enough.
A function which generalises this series is the Riemann zeta function $$zeta(s)=sum_{n=1}^inftyfrac{1}{n^s}$$
Which converges for $text{Re}(s)>1$. For example,
$$zeta(2)=frac{pi^2}{6}.$$
Euler gave a general formula for $zeta(2n)$ in terms of the Bernoulli numbers, but no closed form is known yet for odd arguments.
It can also be analytically continued to $mathbb{C}setminus{1}$. There is a singularity ar $s=1$.
Zeros of this function lie on the negative even integers (the trivial zeros), whereas the non trivial (purely complex) zeros are all known to lie in the critical strip. In fact it is hypothesised that they all lie on the critical line $s=1/2+it$, which you may know is called the Riemann Hypothesis. If true, this has big implications concerning the distribution of the primes, and many conjectured theorems in Analytic Number Theory.
Coming back to "$zeta(1)$" for a moment, as mentioned in the other answer, it pops up in the definition of Euler's gamma constant,
$$gamma=lim_{ntoinfty}sum_{k=1}^nfrac{1}{k}-log n,$$
where $gamma$ constant with
$$gammaapprox
0.57721566490...$$
It is not yet known whether $gamma$ is irrational, unlike $zeta(2n)$, and also $zeta(3)$ which was proved to be irrational by Apéry.
The harmonic series diverges, which is somewhat surprising because each term tends to zero as $ntoinfty$. However, the problem is that each term does not tend to zero fast enough.
A function which generalises this series is the Riemann zeta function $$zeta(s)=sum_{n=1}^inftyfrac{1}{n^s}$$
Which converges for $text{Re}(s)>1$. For example,
$$zeta(2)=frac{pi^2}{6}.$$
Euler gave a general formula for $zeta(2n)$ in terms of the Bernoulli numbers, but no closed form is known yet for odd arguments.
It can also be analytically continued to $mathbb{C}setminus{1}$. There is a singularity ar $s=1$.
Zeros of this function lie on the negative even integers (the trivial zeros), whereas the non trivial (purely complex) zeros are all known to lie in the critical strip. In fact it is hypothesised that they all lie on the critical line $s=1/2+it$, which you may know is called the Riemann Hypothesis. If true, this has big implications concerning the distribution of the primes, and many conjectured theorems in Analytic Number Theory.
Coming back to "$zeta(1)$" for a moment, as mentioned in the other answer, it pops up in the definition of Euler's gamma constant,
$$gamma=lim_{ntoinfty}sum_{k=1}^nfrac{1}{k}-log n,$$
where $gamma$ constant with
$$gammaapprox
0.57721566490...$$
It is not yet known whether $gamma$ is irrational, unlike $zeta(2n)$, and also $zeta(3)$ which was proved to be irrational by Apéry.
edited 17 mins ago
answered 45 mins ago
![](https://i.stack.imgur.com/Hykis.png?s=32&g=1)
![](https://i.stack.imgur.com/Hykis.png?s=32&g=1)
Antinous
5,63542051
5,63542051
add a comment |
add a comment |
up vote
4
down vote
If I remember correctly, there is a sort of closed form:
$$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_k$$
$epsilon_k$ in this constant is a constant proportional to $1/2k$, and thus $epsilon_k to 0$ as $k to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}}$$
Granted, I don't think this is really in line with what you want because ... well, it's sort of like a self-referential thing. "Oh, the harmonic series is given by a function, an error constant, and this special constant that comes from the difference from the series and that other function under certain conditions." It's mostly a personal thing so it doesn't mesh quite well with me?
Beyond that I don't really have anything to offer - specifically with relations of the series to mathematics I know - that I wouldn't just be regurgitating from Wikipedia. Weirdly it hasn't popped up much in my coursework thus far.
add a comment |
up vote
4
down vote
If I remember correctly, there is a sort of closed form:
$$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_k$$
$epsilon_k$ in this constant is a constant proportional to $1/2k$, and thus $epsilon_k to 0$ as $k to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}}$$
Granted, I don't think this is really in line with what you want because ... well, it's sort of like a self-referential thing. "Oh, the harmonic series is given by a function, an error constant, and this special constant that comes from the difference from the series and that other function under certain conditions." It's mostly a personal thing so it doesn't mesh quite well with me?
Beyond that I don't really have anything to offer - specifically with relations of the series to mathematics I know - that I wouldn't just be regurgitating from Wikipedia. Weirdly it hasn't popped up much in my coursework thus far.
add a comment |
up vote
4
down vote
up vote
4
down vote
If I remember correctly, there is a sort of closed form:
$$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_k$$
$epsilon_k$ in this constant is a constant proportional to $1/2k$, and thus $epsilon_k to 0$ as $k to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}}$$
Granted, I don't think this is really in line with what you want because ... well, it's sort of like a self-referential thing. "Oh, the harmonic series is given by a function, an error constant, and this special constant that comes from the difference from the series and that other function under certain conditions." It's mostly a personal thing so it doesn't mesh quite well with me?
Beyond that I don't really have anything to offer - specifically with relations of the series to mathematics I know - that I wouldn't just be regurgitating from Wikipedia. Weirdly it hasn't popped up much in my coursework thus far.
If I remember correctly, there is a sort of closed form:
$$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_k$$
$epsilon_k$ in this constant is a constant proportional to $1/2k$, and thus $epsilon_k to 0$ as $k to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}}$$
Granted, I don't think this is really in line with what you want because ... well, it's sort of like a self-referential thing. "Oh, the harmonic series is given by a function, an error constant, and this special constant that comes from the difference from the series and that other function under certain conditions." It's mostly a personal thing so it doesn't mesh quite well with me?
Beyond that I don't really have anything to offer - specifically with relations of the series to mathematics I know - that I wouldn't just be regurgitating from Wikipedia. Weirdly it hasn't popped up much in my coursework thus far.
answered 44 mins ago
![](https://lh5.googleusercontent.com/-4tiEjr4jwCI/AAAAAAAAAAI/AAAAAAAAALo/TVwmGutzfYA/photo.jpg?sz=32)
![](https://lh5.googleusercontent.com/-4tiEjr4jwCI/AAAAAAAAAAI/AAAAAAAAALo/TVwmGutzfYA/photo.jpg?sz=32)
Eevee Trainer
1,879216
1,879216
add a comment |
add a comment |
up vote
0
down vote
It diverges because, if you look at the partial sums, $s_{2n}-s_n=frac1{2n}+frac1{2n-1}+cdots+frac1{n+1}ge ncdot frac1{2n}=frac12,,forall n$. Thus it isn't Cauchy.
add a comment |
up vote
0
down vote
It diverges because, if you look at the partial sums, $s_{2n}-s_n=frac1{2n}+frac1{2n-1}+cdots+frac1{n+1}ge ncdot frac1{2n}=frac12,,forall n$. Thus it isn't Cauchy.
add a comment |
up vote
0
down vote
up vote
0
down vote
It diverges because, if you look at the partial sums, $s_{2n}-s_n=frac1{2n}+frac1{2n-1}+cdots+frac1{n+1}ge ncdot frac1{2n}=frac12,,forall n$. Thus it isn't Cauchy.
It diverges because, if you look at the partial sums, $s_{2n}-s_n=frac1{2n}+frac1{2n-1}+cdots+frac1{n+1}ge ncdot frac1{2n}=frac12,,forall n$. Thus it isn't Cauchy.
edited 22 mins ago
answered 30 mins ago
Chris Custer
9,5303624
9,5303624
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028150%2fwhat-can-you-think-of-for-the-harmonic-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
It is divergent .....
– neelkanth
45 mins ago
1
Look out for the Euler-Mascheroni constant $gamma$ if you want to explore the mathematical content behind your question. The sum is asymptotically $ln n + gamma$. The error term can be analysed.
– Mark Bennet
44 mins ago
OP probably meant to say "does not converge," since it's like the most well-known fact about the series.
– Eevee Trainer
44 mins ago
@neelkanth the typo is corrected
– hephaes
29 mins ago