Is there a way to sum (part of) the harmonic series to a given total?
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For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.
sequences-and-series summation harmonic-numbers
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add a comment |
$begingroup$
For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.
sequences-and-series summation harmonic-numbers
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No chance for an exact formula. There are very sharp estimates though.
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– A. Pongrácz
Jan 2 at 23:47
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Also, this looks like a duplicate. See here
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– jmerry
Jan 2 at 23:48
add a comment |
$begingroup$
For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.
sequences-and-series summation harmonic-numbers
$endgroup$
For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.
sequences-and-series summation harmonic-numbers
sequences-and-series summation harmonic-numbers
edited Jan 3 at 0:12
Eevee Trainer
7,04321337
7,04321337
asked Jan 2 at 23:41
TristianTristian
1769
1769
$begingroup$
No chance for an exact formula. There are very sharp estimates though.
$endgroup$
– A. Pongrácz
Jan 2 at 23:47
$begingroup$
Also, this looks like a duplicate. See here
$endgroup$
– jmerry
Jan 2 at 23:48
add a comment |
$begingroup$
No chance for an exact formula. There are very sharp estimates though.
$endgroup$
– A. Pongrácz
Jan 2 at 23:47
$begingroup$
Also, this looks like a duplicate. See here
$endgroup$
– jmerry
Jan 2 at 23:48
$begingroup$
No chance for an exact formula. There are very sharp estimates though.
$endgroup$
– A. Pongrácz
Jan 2 at 23:47
$begingroup$
No chance for an exact formula. There are very sharp estimates though.
$endgroup$
– A. Pongrácz
Jan 2 at 23:47
$begingroup$
Also, this looks like a duplicate. See here
$endgroup$
– jmerry
Jan 2 at 23:48
$begingroup$
Also, this looks like a duplicate. See here
$endgroup$
– jmerry
Jan 2 at 23:48
add a comment |
2 Answers
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There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$
$epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.
$gamma$ 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}}} approx 0.5772156649$$
Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.
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add a comment |
$begingroup$
Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.
If you want more accuracy, you could use
$$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
n^2}+Oleft(frac{1}{n^4}right)$$
Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$
$epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$
Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.
$endgroup$
add a comment |
$begingroup$
There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$
$epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$
Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.
$endgroup$
add a comment |
$begingroup$
There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$
$epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$
Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.
$endgroup$
There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$
$epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.
$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.
$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$
Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.
answered Jan 3 at 0:09
Eevee TrainerEevee Trainer
7,04321337
7,04321337
add a comment |
add a comment |
$begingroup$
Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.
If you want more accuracy, you could use
$$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
n^2}+Oleft(frac{1}{n^4}right)$$
Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$
$endgroup$
add a comment |
$begingroup$
Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.
If you want more accuracy, you could use
$$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
n^2}+Oleft(frac{1}{n^4}right)$$
Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$
$endgroup$
add a comment |
$begingroup$
Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.
If you want more accuracy, you could use
$$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
n^2}+Oleft(frac{1}{n^4}right)$$
Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$
$endgroup$
Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.
If you want more accuracy, you could use
$$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
n^2}+Oleft(frac{1}{n^4}right)$$
Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$
answered Jan 3 at 9:42
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
add a comment |
add a comment |
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$begingroup$
No chance for an exact formula. There are very sharp estimates though.
$endgroup$
– A. Pongrácz
Jan 2 at 23:47
$begingroup$
Also, this looks like a duplicate. See here
$endgroup$
– jmerry
Jan 2 at 23:48