Harmonic series partial sum
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According to Beyer (1987) the following progression is called harmonic series: $$frac {1} {a_{1}},frac {1} {a_{1}+d}, frac {1} {a_{1}+2d},...$$
How it can be calculated the partial sum of the above mentioned sequence?
calculus sequences-and-series
add a comment |
up vote
0
down vote
favorite
According to Beyer (1987) the following progression is called harmonic series: $$frac {1} {a_{1}},frac {1} {a_{1}+d}, frac {1} {a_{1}+2d},...$$
How it can be calculated the partial sum of the above mentioned sequence?
calculus sequences-and-series
I don´t see that there exists a closed form for the partial sum.
– callculus
Nov 17 at 16:03
@callculus what about approximation?
– David
Nov 17 at 16:03
1
Search for "Harmonic numbers". Note however, that your sequence is more general, related to "Digamma function"
– Yuriy S
Nov 17 at 16:06
@YuriyS Thank you for your comment. Can you please provide some material reagrdinf this problem?
– David
Nov 17 at 16:10
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
According to Beyer (1987) the following progression is called harmonic series: $$frac {1} {a_{1}},frac {1} {a_{1}+d}, frac {1} {a_{1}+2d},...$$
How it can be calculated the partial sum of the above mentioned sequence?
calculus sequences-and-series
According to Beyer (1987) the following progression is called harmonic series: $$frac {1} {a_{1}},frac {1} {a_{1}+d}, frac {1} {a_{1}+2d},...$$
How it can be calculated the partial sum of the above mentioned sequence?
calculus sequences-and-series
calculus sequences-and-series
asked Nov 17 at 15:58
David
408
408
I don´t see that there exists a closed form for the partial sum.
– callculus
Nov 17 at 16:03
@callculus what about approximation?
– David
Nov 17 at 16:03
1
Search for "Harmonic numbers". Note however, that your sequence is more general, related to "Digamma function"
– Yuriy S
Nov 17 at 16:06
@YuriyS Thank you for your comment. Can you please provide some material reagrdinf this problem?
– David
Nov 17 at 16:10
add a comment |
I don´t see that there exists a closed form for the partial sum.
– callculus
Nov 17 at 16:03
@callculus what about approximation?
– David
Nov 17 at 16:03
1
Search for "Harmonic numbers". Note however, that your sequence is more general, related to "Digamma function"
– Yuriy S
Nov 17 at 16:06
@YuriyS Thank you for your comment. Can you please provide some material reagrdinf this problem?
– David
Nov 17 at 16:10
I don´t see that there exists a closed form for the partial sum.
– callculus
Nov 17 at 16:03
I don´t see that there exists a closed form for the partial sum.
– callculus
Nov 17 at 16:03
@callculus what about approximation?
– David
Nov 17 at 16:03
@callculus what about approximation?
– David
Nov 17 at 16:03
1
1
Search for "Harmonic numbers". Note however, that your sequence is more general, related to "Digamma function"
– Yuriy S
Nov 17 at 16:06
Search for "Harmonic numbers". Note however, that your sequence is more general, related to "Digamma function"
– Yuriy S
Nov 17 at 16:06
@YuriyS Thank you for your comment. Can you please provide some material reagrdinf this problem?
– David
Nov 17 at 16:10
@YuriyS Thank you for your comment. Can you please provide some material reagrdinf this problem?
– David
Nov 17 at 16:10
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
The sum of terms for $n=0,1,...,m$ is $$sum_{n=0}^m frac{1}{d(a_1/d+n)} =frac{1}{d} left(psi left(m+1+frac{a_1}{d}right) -psi left(frac{a_1}{d}right) right)$$
Where $psi$ is the Digamma function.
Use the definition:
$$psi(z)=-gamma+sum_{n=0}^infty frac{z-1}{(n+1)(n+z)}$$
Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic.
– David
Nov 17 at 16:23
@David, try using the definition I provided, and you will see that the formula is true
– Yuriy S
Nov 17 at 16:25
Thank you. My aim is to find a closed-form expression for the finite series. Is it possible?
– David
Nov 17 at 16:31
@David, I have written it in my answer above
– Yuriy S
Nov 17 at 16:38
The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you.
– David
Nov 17 at 16:52
|
show 3 more comments
up vote
2
down vote
Yuriy S gave the only possible closed form for the summation.
For large $n$, for sure, you could use series expansions and get from
$$S_n=sum_{i=0}^n frac{1}{a+i,d}=frac 1d left(psi left(n+1+frac{a}{d}right)-psi left(frac{a}{d}right) right)$$
$$S_n=frac{log(n)-psi left(frac{a}{d}right)}{d}+frac{2 a+d}{2 d^2 n}-frac{6 a^2+6 a d+d^2}{12
d^3 n^2}+frac{2 a^3+3 a^2 d+a d^2}{6 d^4
n^3}+Oleft(frac{1}{n^4}right)$$
Using $a=pi$, $d=e$ and $n=10$, the exact result would be $approx 1.03107$ while the above series expansion would give $approx 1.03113$.
Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 8:56
@David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=alpha=0$ makes the problem more than difficult (at least to me).
– Claude Leibovici
Nov 18 at 9:11
Do you mean both of them?
– David
Nov 18 at 9:12
add a comment |
up vote
1
down vote
The Harmonic Series
The partial sum of the standard Harmonic Series is given by
$$
H_n=sum_{k=1}^nfrac1ktag1
$$
This can be extended to a function that is analytic except at the negative integers
$$
H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)tag2
$$
The Euler-Maclaurin Sum Formula gives the asymptotic expansion
$$
H_n=log(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}+O!left(frac1{n^{12}}right)tag3
$$
where $gamma=0.57721566490153286060651209$ is the Euler-Mascheroni Constant.
The Series in the Question
$$
begin{align}
sum_{k=1}^nfrac1{a+(k-1)d}
&=frac1dsum_{k=1}^nfrac1{frac ad+k-1}\
&=frac1dsum_{k=1}^inftyleft(frac1{frac ad+k-1}-frac1{frac{a-d}d+k-1+n}right)\[3pt]
&=frac1dleft(H!left(n+frac ad-1right)-H!left(frac ad-1right)right)tag4
end{align}
$$
using the extension in $(2)$.
Formula $(2)$ from this answer allows us to compute $H!left(frac ad-1right)$ for rational $a$ and $d$, while formula $(3)$ above allows us to approximate $H!left(n+frac ad-1right)$ for large $n$.
Thank you for your answer.
– David
Nov 18 at 20:34
Please have a look. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 20:35
Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link?
– David
Nov 18 at 20:36
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The sum of terms for $n=0,1,...,m$ is $$sum_{n=0}^m frac{1}{d(a_1/d+n)} =frac{1}{d} left(psi left(m+1+frac{a_1}{d}right) -psi left(frac{a_1}{d}right) right)$$
Where $psi$ is the Digamma function.
Use the definition:
$$psi(z)=-gamma+sum_{n=0}^infty frac{z-1}{(n+1)(n+z)}$$
Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic.
– David
Nov 17 at 16:23
@David, try using the definition I provided, and you will see that the formula is true
– Yuriy S
Nov 17 at 16:25
Thank you. My aim is to find a closed-form expression for the finite series. Is it possible?
– David
Nov 17 at 16:31
@David, I have written it in my answer above
– Yuriy S
Nov 17 at 16:38
The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you.
– David
Nov 17 at 16:52
|
show 3 more comments
up vote
3
down vote
accepted
The sum of terms for $n=0,1,...,m$ is $$sum_{n=0}^m frac{1}{d(a_1/d+n)} =frac{1}{d} left(psi left(m+1+frac{a_1}{d}right) -psi left(frac{a_1}{d}right) right)$$
Where $psi$ is the Digamma function.
Use the definition:
$$psi(z)=-gamma+sum_{n=0}^infty frac{z-1}{(n+1)(n+z)}$$
Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic.
– David
Nov 17 at 16:23
@David, try using the definition I provided, and you will see that the formula is true
– Yuriy S
Nov 17 at 16:25
Thank you. My aim is to find a closed-form expression for the finite series. Is it possible?
– David
Nov 17 at 16:31
@David, I have written it in my answer above
– Yuriy S
Nov 17 at 16:38
The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you.
– David
Nov 17 at 16:52
|
show 3 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The sum of terms for $n=0,1,...,m$ is $$sum_{n=0}^m frac{1}{d(a_1/d+n)} =frac{1}{d} left(psi left(m+1+frac{a_1}{d}right) -psi left(frac{a_1}{d}right) right)$$
Where $psi$ is the Digamma function.
Use the definition:
$$psi(z)=-gamma+sum_{n=0}^infty frac{z-1}{(n+1)(n+z)}$$
The sum of terms for $n=0,1,...,m$ is $$sum_{n=0}^m frac{1}{d(a_1/d+n)} =frac{1}{d} left(psi left(m+1+frac{a_1}{d}right) -psi left(frac{a_1}{d}right) right)$$
Where $psi$ is the Digamma function.
Use the definition:
$$psi(z)=-gamma+sum_{n=0}^infty frac{z-1}{(n+1)(n+z)}$$
edited Nov 17 at 16:23
answered Nov 17 at 16:16
Yuriy S
15.3k433115
15.3k433115
Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic.
– David
Nov 17 at 16:23
@David, try using the definition I provided, and you will see that the formula is true
– Yuriy S
Nov 17 at 16:25
Thank you. My aim is to find a closed-form expression for the finite series. Is it possible?
– David
Nov 17 at 16:31
@David, I have written it in my answer above
– Yuriy S
Nov 17 at 16:38
The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you.
– David
Nov 17 at 16:52
|
show 3 more comments
Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic.
– David
Nov 17 at 16:23
@David, try using the definition I provided, and you will see that the formula is true
– Yuriy S
Nov 17 at 16:25
Thank you. My aim is to find a closed-form expression for the finite series. Is it possible?
– David
Nov 17 at 16:31
@David, I have written it in my answer above
– Yuriy S
Nov 17 at 16:38
The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you.
– David
Nov 17 at 16:52
Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic.
– David
Nov 17 at 16:23
Thank you for your answer. Could you please provide some information how it is derived? P.S. sorry, if my comment is weird, i don't have background in this topic.
– David
Nov 17 at 16:23
@David, try using the definition I provided, and you will see that the formula is true
– Yuriy S
Nov 17 at 16:25
@David, try using the definition I provided, and you will see that the formula is true
– Yuriy S
Nov 17 at 16:25
Thank you. My aim is to find a closed-form expression for the finite series. Is it possible?
– David
Nov 17 at 16:31
Thank you. My aim is to find a closed-form expression for the finite series. Is it possible?
– David
Nov 17 at 16:31
@David, I have written it in my answer above
– Yuriy S
Nov 17 at 16:38
@David, I have written it in my answer above
– Yuriy S
Nov 17 at 16:38
The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you.
– David
Nov 17 at 16:52
The Digamma function is expressed in terms of infinite sum, but my aim is to find closed-form expression. Thank you.
– David
Nov 17 at 16:52
|
show 3 more comments
up vote
2
down vote
Yuriy S gave the only possible closed form for the summation.
For large $n$, for sure, you could use series expansions and get from
$$S_n=sum_{i=0}^n frac{1}{a+i,d}=frac 1d left(psi left(n+1+frac{a}{d}right)-psi left(frac{a}{d}right) right)$$
$$S_n=frac{log(n)-psi left(frac{a}{d}right)}{d}+frac{2 a+d}{2 d^2 n}-frac{6 a^2+6 a d+d^2}{12
d^3 n^2}+frac{2 a^3+3 a^2 d+a d^2}{6 d^4
n^3}+Oleft(frac{1}{n^4}right)$$
Using $a=pi$, $d=e$ and $n=10$, the exact result would be $approx 1.03107$ while the above series expansion would give $approx 1.03113$.
Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 8:56
@David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=alpha=0$ makes the problem more than difficult (at least to me).
– Claude Leibovici
Nov 18 at 9:11
Do you mean both of them?
– David
Nov 18 at 9:12
add a comment |
up vote
2
down vote
Yuriy S gave the only possible closed form for the summation.
For large $n$, for sure, you could use series expansions and get from
$$S_n=sum_{i=0}^n frac{1}{a+i,d}=frac 1d left(psi left(n+1+frac{a}{d}right)-psi left(frac{a}{d}right) right)$$
$$S_n=frac{log(n)-psi left(frac{a}{d}right)}{d}+frac{2 a+d}{2 d^2 n}-frac{6 a^2+6 a d+d^2}{12
d^3 n^2}+frac{2 a^3+3 a^2 d+a d^2}{6 d^4
n^3}+Oleft(frac{1}{n^4}right)$$
Using $a=pi$, $d=e$ and $n=10$, the exact result would be $approx 1.03107$ while the above series expansion would give $approx 1.03113$.
Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 8:56
@David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=alpha=0$ makes the problem more than difficult (at least to me).
– Claude Leibovici
Nov 18 at 9:11
Do you mean both of them?
– David
Nov 18 at 9:12
add a comment |
up vote
2
down vote
up vote
2
down vote
Yuriy S gave the only possible closed form for the summation.
For large $n$, for sure, you could use series expansions and get from
$$S_n=sum_{i=0}^n frac{1}{a+i,d}=frac 1d left(psi left(n+1+frac{a}{d}right)-psi left(frac{a}{d}right) right)$$
$$S_n=frac{log(n)-psi left(frac{a}{d}right)}{d}+frac{2 a+d}{2 d^2 n}-frac{6 a^2+6 a d+d^2}{12
d^3 n^2}+frac{2 a^3+3 a^2 d+a d^2}{6 d^4
n^3}+Oleft(frac{1}{n^4}right)$$
Using $a=pi$, $d=e$ and $n=10$, the exact result would be $approx 1.03107$ while the above series expansion would give $approx 1.03113$.
Yuriy S gave the only possible closed form for the summation.
For large $n$, for sure, you could use series expansions and get from
$$S_n=sum_{i=0}^n frac{1}{a+i,d}=frac 1d left(psi left(n+1+frac{a}{d}right)-psi left(frac{a}{d}right) right)$$
$$S_n=frac{log(n)-psi left(frac{a}{d}right)}{d}+frac{2 a+d}{2 d^2 n}-frac{6 a^2+6 a d+d^2}{12
d^3 n^2}+frac{2 a^3+3 a^2 d+a d^2}{6 d^4
n^3}+Oleft(frac{1}{n^4}right)$$
Using $a=pi$, $d=e$ and $n=10$, the exact result would be $approx 1.03107$ while the above series expansion would give $approx 1.03113$.
answered Nov 18 at 4:40
Claude Leibovici
116k1156131
116k1156131
Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 8:56
@David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=alpha=0$ makes the problem more than difficult (at least to me).
– Claude Leibovici
Nov 18 at 9:11
Do you mean both of them?
– David
Nov 18 at 9:12
add a comment |
Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 8:56
@David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=alpha=0$ makes the problem more than difficult (at least to me).
– Claude Leibovici
Nov 18 at 9:11
Do you mean both of them?
– David
Nov 18 at 9:12
Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 8:56
Please have a look. Thank you for your support. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 8:56
@David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=alpha=0$ makes the problem more than difficult (at least to me).
– Claude Leibovici
Nov 18 at 9:11
@David. No idea about this other one. I am quite skeptical about even an approximation. Even the case $a=alpha=0$ makes the problem more than difficult (at least to me).
– Claude Leibovici
Nov 18 at 9:11
Do you mean both of them?
– David
Nov 18 at 9:12
Do you mean both of them?
– David
Nov 18 at 9:12
add a comment |
up vote
1
down vote
The Harmonic Series
The partial sum of the standard Harmonic Series is given by
$$
H_n=sum_{k=1}^nfrac1ktag1
$$
This can be extended to a function that is analytic except at the negative integers
$$
H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)tag2
$$
The Euler-Maclaurin Sum Formula gives the asymptotic expansion
$$
H_n=log(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}+O!left(frac1{n^{12}}right)tag3
$$
where $gamma=0.57721566490153286060651209$ is the Euler-Mascheroni Constant.
The Series in the Question
$$
begin{align}
sum_{k=1}^nfrac1{a+(k-1)d}
&=frac1dsum_{k=1}^nfrac1{frac ad+k-1}\
&=frac1dsum_{k=1}^inftyleft(frac1{frac ad+k-1}-frac1{frac{a-d}d+k-1+n}right)\[3pt]
&=frac1dleft(H!left(n+frac ad-1right)-H!left(frac ad-1right)right)tag4
end{align}
$$
using the extension in $(2)$.
Formula $(2)$ from this answer allows us to compute $H!left(frac ad-1right)$ for rational $a$ and $d$, while formula $(3)$ above allows us to approximate $H!left(n+frac ad-1right)$ for large $n$.
Thank you for your answer.
– David
Nov 18 at 20:34
Please have a look. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 20:35
Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link?
– David
Nov 18 at 20:36
add a comment |
up vote
1
down vote
The Harmonic Series
The partial sum of the standard Harmonic Series is given by
$$
H_n=sum_{k=1}^nfrac1ktag1
$$
This can be extended to a function that is analytic except at the negative integers
$$
H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)tag2
$$
The Euler-Maclaurin Sum Formula gives the asymptotic expansion
$$
H_n=log(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}+O!left(frac1{n^{12}}right)tag3
$$
where $gamma=0.57721566490153286060651209$ is the Euler-Mascheroni Constant.
The Series in the Question
$$
begin{align}
sum_{k=1}^nfrac1{a+(k-1)d}
&=frac1dsum_{k=1}^nfrac1{frac ad+k-1}\
&=frac1dsum_{k=1}^inftyleft(frac1{frac ad+k-1}-frac1{frac{a-d}d+k-1+n}right)\[3pt]
&=frac1dleft(H!left(n+frac ad-1right)-H!left(frac ad-1right)right)tag4
end{align}
$$
using the extension in $(2)$.
Formula $(2)$ from this answer allows us to compute $H!left(frac ad-1right)$ for rational $a$ and $d$, while formula $(3)$ above allows us to approximate $H!left(n+frac ad-1right)$ for large $n$.
Thank you for your answer.
– David
Nov 18 at 20:34
Please have a look. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 20:35
Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link?
– David
Nov 18 at 20:36
add a comment |
up vote
1
down vote
up vote
1
down vote
The Harmonic Series
The partial sum of the standard Harmonic Series is given by
$$
H_n=sum_{k=1}^nfrac1ktag1
$$
This can be extended to a function that is analytic except at the negative integers
$$
H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)tag2
$$
The Euler-Maclaurin Sum Formula gives the asymptotic expansion
$$
H_n=log(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}+O!left(frac1{n^{12}}right)tag3
$$
where $gamma=0.57721566490153286060651209$ is the Euler-Mascheroni Constant.
The Series in the Question
$$
begin{align}
sum_{k=1}^nfrac1{a+(k-1)d}
&=frac1dsum_{k=1}^nfrac1{frac ad+k-1}\
&=frac1dsum_{k=1}^inftyleft(frac1{frac ad+k-1}-frac1{frac{a-d}d+k-1+n}right)\[3pt]
&=frac1dleft(H!left(n+frac ad-1right)-H!left(frac ad-1right)right)tag4
end{align}
$$
using the extension in $(2)$.
Formula $(2)$ from this answer allows us to compute $H!left(frac ad-1right)$ for rational $a$ and $d$, while formula $(3)$ above allows us to approximate $H!left(n+frac ad-1right)$ for large $n$.
The Harmonic Series
The partial sum of the standard Harmonic Series is given by
$$
H_n=sum_{k=1}^nfrac1ktag1
$$
This can be extended to a function that is analytic except at the negative integers
$$
H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)tag2
$$
The Euler-Maclaurin Sum Formula gives the asymptotic expansion
$$
H_n=log(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}+O!left(frac1{n^{12}}right)tag3
$$
where $gamma=0.57721566490153286060651209$ is the Euler-Mascheroni Constant.
The Series in the Question
$$
begin{align}
sum_{k=1}^nfrac1{a+(k-1)d}
&=frac1dsum_{k=1}^nfrac1{frac ad+k-1}\
&=frac1dsum_{k=1}^inftyleft(frac1{frac ad+k-1}-frac1{frac{a-d}d+k-1+n}right)\[3pt]
&=frac1dleft(H!left(n+frac ad-1right)-H!left(frac ad-1right)right)tag4
end{align}
$$
using the extension in $(2)$.
Formula $(2)$ from this answer allows us to compute $H!left(frac ad-1right)$ for rational $a$ and $d$, while formula $(3)$ above allows us to approximate $H!left(n+frac ad-1right)$ for large $n$.
edited Nov 18 at 17:42
answered Nov 18 at 15:15
robjohn♦
263k27301622
263k27301622
Thank you for your answer.
– David
Nov 18 at 20:34
Please have a look. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 20:35
Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link?
– David
Nov 18 at 20:36
add a comment |
Thank you for your answer.
– David
Nov 18 at 20:34
Please have a look. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 20:35
Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link?
– David
Nov 18 at 20:36
Thank you for your answer.
– David
Nov 18 at 20:34
Thank you for your answer.
– David
Nov 18 at 20:34
Please have a look. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 20:35
Please have a look. math.stackexchange.com/questions/3003278/…
– David
Nov 18 at 20:35
Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link?
– David
Nov 18 at 20:36
Could you, please, support in order to calculate or approximate the sums represented in the above mentioned link?
– David
Nov 18 at 20:36
add a comment |
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I don´t see that there exists a closed form for the partial sum.
– callculus
Nov 17 at 16:03
@callculus what about approximation?
– David
Nov 17 at 16:03
1
Search for "Harmonic numbers". Note however, that your sequence is more general, related to "Digamma function"
– Yuriy S
Nov 17 at 16:06
@YuriyS Thank you for your comment. Can you please provide some material reagrdinf this problem?
– David
Nov 17 at 16:10