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How to show that $sum_{n=1}^{infty}frac{H_n}{n^2+n}=frac{pi^2}{6}$
$begingroup$
Wolfram Alpha shows that
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=zeta(2)=frac{pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$begin{align}
sum_{n=1}^{infty}frac{H_n}{n^2+n}&=sum_{n=1}^{infty}H_nleft(frac{1}{n}-frac{1}{n+1}right)\
&=H_{1}left(1-frac{1}{2}right)+H_{2}left(frac{1}{2}-frac{1}{3}right)+H_{3}left(frac{1}{3}-frac{1}{4}right)\
&=1-frac{1}{2}+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{6}+frac{1}{3}-frac{1}{4}+frac{1}{6}-frac{1}{8}+frac{1}{9}-frac{1}{12}
end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=int_{0}^{1}frac{1-t^n}{1-t}dt$$
Then,
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=sum_{n=1}^{infty}frac{1}{n^2+n}int_{0}^{1}frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
sequences-and-series
asked Jan 6 at 15:07
LarryLarry
2,27231028
$endgroup$
add a comment |
$begingroup$
Wolfram Alpha shows that
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=zeta(2)=frac{pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$begin{align}
sum_{n=1}^{infty}frac{H_n}{n^2+n}&=sum_{n=1}^{infty}H_nleft(frac{1}{n}-frac{1}{n+1}right)\
&=H_{1}left(1-frac{1}{2}right)+H_{2}left(frac{1}{2}-frac{1}{3}right)+H_{3}left(frac{1}{3}-frac{1}{4}right)\
&=1-frac{1}{2}+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{6}+frac{1}{3}-frac{1}{4}+frac{1}{6}-frac{1}{8}+frac{1}{9}-frac{1}{12}
end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=int_{0}^{1}frac{1-t^n}{1-t}dt$$
Then,
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=sum_{n=1}^{infty}frac{1}{n^2+n}int_{0}^{1}frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
sequences-and-series
asked Jan 6 at 15:07
LarryLarry
2,27231028
$endgroup$
$begingroup$
It should work similarly to this question.
$endgroup$
– Dietrich Burde
Jan 6 at 15:13
add a comment |
$begingroup$
Wolfram Alpha shows that
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=zeta(2)=frac{pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$begin{align}
sum_{n=1}^{infty}frac{H_n}{n^2+n}&=sum_{n=1}^{infty}H_nleft(frac{1}{n}-frac{1}{n+1}right)\
&=H_{1}left(1-frac{1}{2}right)+H_{2}left(frac{1}{2}-frac{1}{3}right)+H_{3}left(frac{1}{3}-frac{1}{4}right)\
&=1-frac{1}{2}+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{6}+frac{1}{3}-frac{1}{4}+frac{1}{6}-frac{1}{8}+frac{1}{9}-frac{1}{12}
end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=int_{0}^{1}frac{1-t^n}{1-t}dt$$
Then,
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=sum_{n=1}^{infty}frac{1}{n^2+n}int_{0}^{1}frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
sequences-and-series
asked Jan 6 at 15:07
LarryLarry
2,27231028
$endgroup$
Wolfram Alpha shows that
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=zeta(2)=frac{pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$begin{align}
sum_{n=1}^{infty}frac{H_n}{n^2+n}&=sum_{n=1}^{infty}H_nleft(frac{1}{n}-frac{1}{n+1}right)\
&=H_{1}left(1-frac{1}{2}right)+H_{2}left(frac{1}{2}-frac{1}{3}right)+H_{3}left(frac{1}{3}-frac{1}{4}right)\
&=1-frac{1}{2}+frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{6}+frac{1}{3}-frac{1}{4}+frac{1}{6}-frac{1}{8}+frac{1}{9}-frac{1}{12}
end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=int_{0}^{1}frac{1-t^n}{1-t}dt$$
Then,
$$sum_{n=1}^{infty}frac{H_n}{n^2+n}=sum_{n=1}^{infty}frac{1}{n^2+n}int_{0}^{1}frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
sequences-and-series
sequences-and-series
asked Jan 6 at 15:07
LarryLarry
2,27231028
asked Jan 6 at 15:07
LarryLarry
2,27231028
asked Jan 6 at 15:07
LarryLarry
2,27231028
asked Jan 6 at 15:07
LarryLarry
2,27231028
asked Jan 6 at 15:07
LarryLarry
2,27231028
2,27231028
$begingroup$
It should work similarly to this question.
$endgroup$
– Dietrich Burde
Jan 6 at 15:13
add a comment |
$begingroup$
It should work similarly to this question.
$endgroup$
– Dietrich Burde
Jan 6 at 15:13
$begingroup$
It should work similarly to this question.
$endgroup$
– Dietrich Burde
Jan 6 at 15:13
$begingroup$
It should work similarly to this question.
$endgroup$
– Dietrich Burde
Jan 6 at 15:13
add a comment |
5 Answers
5
active
oldest
votes
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Beside telescoping, all you need is to reverse the order of the double sum: $$sum_{nge 1}frac{sum_{k=1}^n k^{-1}}{n(n+1)}=sum_{kge 1}k^{-1}sum_{nge k}frac{1}{n(n+1)}=sum_{kge 1}k^{-2}=frac{pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $frac{k^{-1}}{n(n+1)}$ with $n,,k$ integers in the set ${(n,,k)|n,,kge 1,,kle n}$. But I could equivalently describe this set as ${(n,,k)|kge 1,,nge k}$.
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
$endgroup$
$begingroup$
How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit.
$endgroup$
– Larry
Jan 6 at 15:20
$begingroup$
@Larry See edit.
$endgroup$
– J.G.
Jan 6 at 15:23
$begingroup$
$frac{H_n}{n^2(1+frac{1}{n})}=frac{1}{n^2}lim_{n→∞}frac{H_n}{1+H_n}=frac{1}{n^2}$
$endgroup$
– sirous
Jan 6 at 15:24
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@J.G Thanks, I think I understand the reversal now.
$endgroup$
– Larry
Jan 6 at 15:25
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@sirous Your equality is not true: for example $$H_1/(1^2+1)neq 1/1^2$$
$endgroup$
– Nick
Jan 6 at 15:26
|
show 1 more comment
$begingroup$
We have
begin{align}
sum_{n=1}^inftyfrac{H_n}{n^2+n}
&=sum_{n=1}^infty H_nleft(frac1n-frac1{n+1}right)
\&=sum_{n=1}^inftysum_{m=1}^nfrac1mleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1msum_{n=m}^inftyleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1mcdotfrac1m
\&=frac{pi^2}6.
end{align}
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
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add a comment |
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$newcommand{Li}{operatorname{Li}}$Starting with a generating function for the Harmonic Numbers
$$-frac{log(1-x)}{1-x}~=~sum_{n=1}^infty H_nx^ntag1$$
we divide both sides by $x$ and integrate afterwards to get
$$begin{align*}
int-frac{log(1-x)}{x(1-x)}mathrm dx&=intsum_{n=1}^infty H_nx^{n-1}mathrm dx\
Li_2(x)+frac12log^2(1-x)+c&=sum_{n=1}^infty frac{H_n}nx^{n}
end{align*}$$
Plug in $x=0$ to determine the value of $c$ which turns out to be $0$ aswell. Integrating this equation yields to
$$begin{align*}
intLi_2(x)+frac12log^2(1-x)mathrm dx&=intsum_{n=1}^infty frac{H_n}nx^{n}mathrm dx\
xLi_2(x)-x+(x-1)log(1-x)+frac12(x-1)(log^2(1-x)-2log(1-x)+2)+c&=sum_{n=1}^inftyfrac{H_n}{n(n+1)}x^{n}
end{align*}$$
The new constant again equals $0$ due the same argumentation as above. Now we can plug in $x=1$ to compute the value of your given series which overall is given by $Li_2(1)$ and this is well-known to equal $zeta(2)$.
$$therefore~sum_{n=1}^inftyfrac{H_n}{n(n+1)}~=~frac{pi^2}6tag2$$
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
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add a comment |
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Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$int_0^1 x^{n - 1} , dx = frac{1}{n} quad text{and} quad int_0^1 y^n , dy = frac{1}{n + 1},$$
the sum can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = int_0^1 int_0^1 frac{1}{x} sum_{n = 1}^infty H_n (xy)^n , dx dy.tag1$$
Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem.
Making use of the generating function for the Harmonic number, namely
$$-frac{log(1-x)}{1 - x}~=~sum_{n=1}^infty H_n x^n,$$
the infinity sum appearing inside the integral in (1) can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = -int_0^1 int_0^1 frac{ln (1 - xy)}{x(1 - xy)} , dx dy.$$
Enforcing a substitution of $x mapsto x/y$ gives
begin{align}
sum_{n = 1}^infty frac{H_n}{n(n + 1)} &= -int_0^1 int_0^y frac{ln (1 - x)}{x (1 - x)} , dx dy\
&= -int_0^1 int_0^y left [frac{ln (1 - x)}{x} + frac{ln (1 - x)}{1 - x} right ] , dx , dy\
&= int_0^1 operatorname{Li}_2 (y) , dy + frac{1}{2} int_0^1 ln^2 (1 - y) , dy\
&= I_1 + frac{1}{2} I_2.
end{align}
Here $operatorname{Li}_2 (y)$ is the dilogarithm function.
For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have
begin{align}
I_1 &= int_0^1 sum_{n = 1}^infty frac{y^n}{n^2} , dy\
&= sum_{n = 1}^infty frac{1}{n^2} int_0^1 y^n , dy\
&= sum_{n = 1}^infty frac{1}{n^2(n + 1)}\
&= sum_{n = 1}^infty left [frac{1}{n^2} + frac{1}{n + 1} - frac{1}{n} right ]\
&= sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty left (frac{1}{n} - frac{1}{n + 1} right )\
&= frac{pi^2}{6} - 1,
end{align}
where the first of the series is the well-known Basel problem while the second telescopes.
For the second of the integrals, define
$$I(a) = int_0^1 (1 - y)^a , dy, quad a > - 1.$$
Observe that
$$I''(0) = int_0^1 ln^2 ( 1 - y) , dy,$$
where the derivative is with respect to $a$. Now, as
$$I(a) = operatorname{B} (1, a + 1) = frac{Gamma (a + 1)}{Gamma (a + 2)} = frac{1}{a + 1},$$
where $operatorname{B}(x,y)$ is the Beta function, one readily has
$$I''(a) = frac{2}{(a + 1)^3},$$
yielding $I''(0) = I_2 = 2$.
Thus
$$sum_{n = 1}^infty frac{H_n}{n (n + 1)} = frac{pi^2}{6} - 1 + frac{1}{2} cdot 2 = frac{pi^2}{6}.$$
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
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Using summation by parts we have $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}sum_{nleq N}frac{1}{nleft(n+1right)}-sum_{nleq N-1}frac{1}{n+1}left(sum_{kleq n}frac{1}{kleft(k+1right)}right).$$ Clearly $$sum_{nleq N}frac{1}{nleft(n+1right)}=left(1-frac{1}{N+1}right)$$ then $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}left(1-frac{1}{N+1}right)-H_{N}+sum_{nleq N-1}frac{1}{left(n+1right)^{2}}.$$ Now, since $H_{N}simlogleft(Nright)$ as $Nrightarrow+infty$, the claim follows.
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
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5 Answers
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5 Answers
5
active
oldest
votes
active
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active
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$begingroup$
Beside telescoping, all you need is to reverse the order of the double sum: $$sum_{nge 1}frac{sum_{k=1}^n k^{-1}}{n(n+1)}=sum_{kge 1}k^{-1}sum_{nge k}frac{1}{n(n+1)}=sum_{kge 1}k^{-2}=frac{pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $frac{k^{-1}}{n(n+1)}$ with $n,,k$ integers in the set ${(n,,k)|n,,kge 1,,kle n}$. But I could equivalently describe this set as ${(n,,k)|kge 1,,nge k}$.
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
$endgroup$
$begingroup$
How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit.
$endgroup$
– Larry
Jan 6 at 15:20
$begingroup$
@Larry See edit.
$endgroup$
– J.G.
Jan 6 at 15:23
$begingroup$
$frac{H_n}{n^2(1+frac{1}{n})}=frac{1}{n^2}lim_{n→∞}frac{H_n}{1+H_n}=frac{1}{n^2}$
$endgroup$
– sirous
Jan 6 at 15:24
$begingroup$
@J.G Thanks, I think I understand the reversal now.
$endgroup$
– Larry
Jan 6 at 15:25
$begingroup$
@sirous Your equality is not true: for example $$H_1/(1^2+1)neq 1/1^2$$
$endgroup$
– Nick
Jan 6 at 15:26
|
show 1 more comment
$begingroup$
Beside telescoping, all you need is to reverse the order of the double sum: $$sum_{nge 1}frac{sum_{k=1}^n k^{-1}}{n(n+1)}=sum_{kge 1}k^{-1}sum_{nge k}frac{1}{n(n+1)}=sum_{kge 1}k^{-2}=frac{pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $frac{k^{-1}}{n(n+1)}$ with $n,,k$ integers in the set ${(n,,k)|n,,kge 1,,kle n}$. But I could equivalently describe this set as ${(n,,k)|kge 1,,nge k}$.
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
$endgroup$
$begingroup$
How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit.
$endgroup$
– Larry
Jan 6 at 15:20
$begingroup$
@Larry See edit.
$endgroup$
– J.G.
Jan 6 at 15:23
$begingroup$
$frac{H_n}{n^2(1+frac{1}{n})}=frac{1}{n^2}lim_{n→∞}frac{H_n}{1+H_n}=frac{1}{n^2}$
$endgroup$
– sirous
Jan 6 at 15:24
$begingroup$
@J.G Thanks, I think I understand the reversal now.
$endgroup$
– Larry
Jan 6 at 15:25
$begingroup$
@sirous Your equality is not true: for example $$H_1/(1^2+1)neq 1/1^2$$
$endgroup$
– Nick
Jan 6 at 15:26
|
show 1 more comment
$begingroup$
Beside telescoping, all you need is to reverse the order of the double sum: $$sum_{nge 1}frac{sum_{k=1}^n k^{-1}}{n(n+1)}=sum_{kge 1}k^{-1}sum_{nge k}frac{1}{n(n+1)}=sum_{kge 1}k^{-2}=frac{pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $frac{k^{-1}}{n(n+1)}$ with $n,,k$ integers in the set ${(n,,k)|n,,kge 1,,kle n}$. But I could equivalently describe this set as ${(n,,k)|kge 1,,nge k}$.
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
$endgroup$
Beside telescoping, all you need is to reverse the order of the double sum: $$sum_{nge 1}frac{sum_{k=1}^n k^{-1}}{n(n+1)}=sum_{kge 1}k^{-1}sum_{nge k}frac{1}{n(n+1)}=sum_{kge 1}k^{-2}=frac{pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $frac{k^{-1}}{n(n+1)}$ with $n,,k$ integers in the set ${(n,,k)|n,,kge 1,,kle n}$. But I could equivalently describe this set as ${(n,,k)|kge 1,,nge k}$.
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
edited Jan 6 at 15:22
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
answered Jan 6 at 15:11
J.G.J.G.
24.9k22539
24.9k22539
$begingroup$
How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit.
$endgroup$
– Larry
Jan 6 at 15:20
$begingroup$
@Larry See edit.
$endgroup$
– J.G.
Jan 6 at 15:23
$begingroup$
$frac{H_n}{n^2(1+frac{1}{n})}=frac{1}{n^2}lim_{n→∞}frac{H_n}{1+H_n}=frac{1}{n^2}$
$endgroup$
– sirous
Jan 6 at 15:24
$begingroup$
@J.G Thanks, I think I understand the reversal now.
$endgroup$
– Larry
Jan 6 at 15:25
$begingroup$
@sirous Your equality is not true: for example $$H_1/(1^2+1)neq 1/1^2$$
$endgroup$
– Nick
Jan 6 at 15:26
|
show 1 more comment
$begingroup$
How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit.
$endgroup$
– Larry
Jan 6 at 15:20
$begingroup$
@Larry See edit.
$endgroup$
– J.G.
Jan 6 at 15:23
$begingroup$
$frac{H_n}{n^2(1+frac{1}{n})}=frac{1}{n^2}lim_{n→∞}frac{H_n}{1+H_n}=frac{1}{n^2}$
$endgroup$
– sirous
Jan 6 at 15:24
$begingroup$
@J.G Thanks, I think I understand the reversal now.
$endgroup$
– Larry
Jan 6 at 15:25
$begingroup$
@sirous Your equality is not true: for example $$H_1/(1^2+1)neq 1/1^2$$
$endgroup$
– Nick
Jan 6 at 15:26
$begingroup$
How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit.
$endgroup$
– Larry
Jan 6 at 15:20
$begingroup$
How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit.
$endgroup$
– Larry
Jan 6 at 15:20
$begingroup$
@Larry See edit.
$endgroup$
– J.G.
Jan 6 at 15:23
$begingroup$
@Larry See edit.
$endgroup$
– J.G.
Jan 6 at 15:23
$begingroup$
$frac{H_n}{n^2(1+frac{1}{n})}=frac{1}{n^2}lim_{n→∞}frac{H_n}{1+H_n}=frac{1}{n^2}$
$endgroup$
– sirous
Jan 6 at 15:24
$begingroup$
$frac{H_n}{n^2(1+frac{1}{n})}=frac{1}{n^2}lim_{n→∞}frac{H_n}{1+H_n}=frac{1}{n^2}$
$endgroup$
– sirous
Jan 6 at 15:24
$begingroup$
@J.G Thanks, I think I understand the reversal now.
$endgroup$
– Larry
Jan 6 at 15:25
$begingroup$
@J.G Thanks, I think I understand the reversal now.
$endgroup$
– Larry
Jan 6 at 15:25
$begingroup$
@sirous Your equality is not true: for example $$H_1/(1^2+1)neq 1/1^2$$
$endgroup$
– Nick
Jan 6 at 15:26
$begingroup$
@sirous Your equality is not true: for example $$H_1/(1^2+1)neq 1/1^2$$
$endgroup$
– Nick
Jan 6 at 15:26
|
show 1 more comment
$begingroup$
We have
begin{align}
sum_{n=1}^inftyfrac{H_n}{n^2+n}
&=sum_{n=1}^infty H_nleft(frac1n-frac1{n+1}right)
\&=sum_{n=1}^inftysum_{m=1}^nfrac1mleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1msum_{n=m}^inftyleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1mcdotfrac1m
\&=frac{pi^2}6.
end{align}
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
$endgroup$
add a comment |
$begingroup$
We have
begin{align}
sum_{n=1}^inftyfrac{H_n}{n^2+n}
&=sum_{n=1}^infty H_nleft(frac1n-frac1{n+1}right)
\&=sum_{n=1}^inftysum_{m=1}^nfrac1mleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1msum_{n=m}^inftyleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1mcdotfrac1m
\&=frac{pi^2}6.
end{align}
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
$endgroup$
add a comment |
$begingroup$
We have
begin{align}
sum_{n=1}^inftyfrac{H_n}{n^2+n}
&=sum_{n=1}^infty H_nleft(frac1n-frac1{n+1}right)
\&=sum_{n=1}^inftysum_{m=1}^nfrac1mleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1msum_{n=m}^inftyleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1mcdotfrac1m
\&=frac{pi^2}6.
end{align}
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
$endgroup$
We have
begin{align}
sum_{n=1}^inftyfrac{H_n}{n^2+n}
&=sum_{n=1}^infty H_nleft(frac1n-frac1{n+1}right)
\&=sum_{n=1}^inftysum_{m=1}^nfrac1mleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1msum_{n=m}^inftyleft(frac1n-frac1{n+1}right)
\&=sum_{m=1}^inftyfrac1mcdotfrac1m
\&=frac{pi^2}6.
end{align}
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
answered Jan 6 at 15:21
SmileyCraftSmileyCraft
3,506517
3,506517
add a comment |
add a comment |
$begingroup$
$newcommand{Li}{operatorname{Li}}$Starting with a generating function for the Harmonic Numbers
$$-frac{log(1-x)}{1-x}~=~sum_{n=1}^infty H_nx^ntag1$$
we divide both sides by $x$ and integrate afterwards to get
$$begin{align*}
int-frac{log(1-x)}{x(1-x)}mathrm dx&=intsum_{n=1}^infty H_nx^{n-1}mathrm dx\
Li_2(x)+frac12log^2(1-x)+c&=sum_{n=1}^infty frac{H_n}nx^{n}
end{align*}$$
Plug in $x=0$ to determine the value of $c$ which turns out to be $0$ aswell. Integrating this equation yields to
$$begin{align*}
intLi_2(x)+frac12log^2(1-x)mathrm dx&=intsum_{n=1}^infty frac{H_n}nx^{n}mathrm dx\
xLi_2(x)-x+(x-1)log(1-x)+frac12(x-1)(log^2(1-x)-2log(1-x)+2)+c&=sum_{n=1}^inftyfrac{H_n}{n(n+1)}x^{n}
end{align*}$$
The new constant again equals $0$ due the same argumentation as above. Now we can plug in $x=1$ to compute the value of your given series which overall is given by $Li_2(1)$ and this is well-known to equal $zeta(2)$.
$$therefore~sum_{n=1}^inftyfrac{H_n}{n(n+1)}~=~frac{pi^2}6tag2$$
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
$endgroup$
add a comment |
$begingroup$
$newcommand{Li}{operatorname{Li}}$Starting with a generating function for the Harmonic Numbers
$$-frac{log(1-x)}{1-x}~=~sum_{n=1}^infty H_nx^ntag1$$
we divide both sides by $x$ and integrate afterwards to get
$$begin{align*}
int-frac{log(1-x)}{x(1-x)}mathrm dx&=intsum_{n=1}^infty H_nx^{n-1}mathrm dx\
Li_2(x)+frac12log^2(1-x)+c&=sum_{n=1}^infty frac{H_n}nx^{n}
end{align*}$$
Plug in $x=0$ to determine the value of $c$ which turns out to be $0$ aswell. Integrating this equation yields to
$$begin{align*}
intLi_2(x)+frac12log^2(1-x)mathrm dx&=intsum_{n=1}^infty frac{H_n}nx^{n}mathrm dx\
xLi_2(x)-x+(x-1)log(1-x)+frac12(x-1)(log^2(1-x)-2log(1-x)+2)+c&=sum_{n=1}^inftyfrac{H_n}{n(n+1)}x^{n}
end{align*}$$
The new constant again equals $0$ due the same argumentation as above. Now we can plug in $x=1$ to compute the value of your given series which overall is given by $Li_2(1)$ and this is well-known to equal $zeta(2)$.
$$therefore~sum_{n=1}^inftyfrac{H_n}{n(n+1)}~=~frac{pi^2}6tag2$$
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
$endgroup$
add a comment |
$begingroup$
$newcommand{Li}{operatorname{Li}}$Starting with a generating function for the Harmonic Numbers
$$-frac{log(1-x)}{1-x}~=~sum_{n=1}^infty H_nx^ntag1$$
we divide both sides by $x$ and integrate afterwards to get
$$begin{align*}
int-frac{log(1-x)}{x(1-x)}mathrm dx&=intsum_{n=1}^infty H_nx^{n-1}mathrm dx\
Li_2(x)+frac12log^2(1-x)+c&=sum_{n=1}^infty frac{H_n}nx^{n}
end{align*}$$
Plug in $x=0$ to determine the value of $c$ which turns out to be $0$ aswell. Integrating this equation yields to
$$begin{align*}
intLi_2(x)+frac12log^2(1-x)mathrm dx&=intsum_{n=1}^infty frac{H_n}nx^{n}mathrm dx\
xLi_2(x)-x+(x-1)log(1-x)+frac12(x-1)(log^2(1-x)-2log(1-x)+2)+c&=sum_{n=1}^inftyfrac{H_n}{n(n+1)}x^{n}
end{align*}$$
The new constant again equals $0$ due the same argumentation as above. Now we can plug in $x=1$ to compute the value of your given series which overall is given by $Li_2(1)$ and this is well-known to equal $zeta(2)$.
$$therefore~sum_{n=1}^inftyfrac{H_n}{n(n+1)}~=~frac{pi^2}6tag2$$
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
$endgroup$
$newcommand{Li}{operatorname{Li}}$Starting with a generating function for the Harmonic Numbers
$$-frac{log(1-x)}{1-x}~=~sum_{n=1}^infty H_nx^ntag1$$
we divide both sides by $x$ and integrate afterwards to get
$$begin{align*}
int-frac{log(1-x)}{x(1-x)}mathrm dx&=intsum_{n=1}^infty H_nx^{n-1}mathrm dx\
Li_2(x)+frac12log^2(1-x)+c&=sum_{n=1}^infty frac{H_n}nx^{n}
end{align*}$$
Plug in $x=0$ to determine the value of $c$ which turns out to be $0$ aswell. Integrating this equation yields to
$$begin{align*}
intLi_2(x)+frac12log^2(1-x)mathrm dx&=intsum_{n=1}^infty frac{H_n}nx^{n}mathrm dx\
xLi_2(x)-x+(x-1)log(1-x)+frac12(x-1)(log^2(1-x)-2log(1-x)+2)+c&=sum_{n=1}^inftyfrac{H_n}{n(n+1)}x^{n}
end{align*}$$
The new constant again equals $0$ due the same argumentation as above. Now we can plug in $x=1$ to compute the value of your given series which overall is given by $Li_2(1)$ and this is well-known to equal $zeta(2)$.
$$therefore~sum_{n=1}^inftyfrac{H_n}{n(n+1)}~=~frac{pi^2}6tag2$$
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
edited Jan 6 at 15:32
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
answered Jan 6 at 15:27
mrtaurhomrtaurho
4,09121234
4,09121234
add a comment |
add a comment |
$begingroup$
Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$int_0^1 x^{n - 1} , dx = frac{1}{n} quad text{and} quad int_0^1 y^n , dy = frac{1}{n + 1},$$
the sum can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = int_0^1 int_0^1 frac{1}{x} sum_{n = 1}^infty H_n (xy)^n , dx dy.tag1$$
Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem.
Making use of the generating function for the Harmonic number, namely
$$-frac{log(1-x)}{1 - x}~=~sum_{n=1}^infty H_n x^n,$$
the infinity sum appearing inside the integral in (1) can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = -int_0^1 int_0^1 frac{ln (1 - xy)}{x(1 - xy)} , dx dy.$$
Enforcing a substitution of $x mapsto x/y$ gives
begin{align}
sum_{n = 1}^infty frac{H_n}{n(n + 1)} &= -int_0^1 int_0^y frac{ln (1 - x)}{x (1 - x)} , dx dy\
&= -int_0^1 int_0^y left [frac{ln (1 - x)}{x} + frac{ln (1 - x)}{1 - x} right ] , dx , dy\
&= int_0^1 operatorname{Li}_2 (y) , dy + frac{1}{2} int_0^1 ln^2 (1 - y) , dy\
&= I_1 + frac{1}{2} I_2.
end{align}
Here $operatorname{Li}_2 (y)$ is the dilogarithm function.
For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have
begin{align}
I_1 &= int_0^1 sum_{n = 1}^infty frac{y^n}{n^2} , dy\
&= sum_{n = 1}^infty frac{1}{n^2} int_0^1 y^n , dy\
&= sum_{n = 1}^infty frac{1}{n^2(n + 1)}\
&= sum_{n = 1}^infty left [frac{1}{n^2} + frac{1}{n + 1} - frac{1}{n} right ]\
&= sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty left (frac{1}{n} - frac{1}{n + 1} right )\
&= frac{pi^2}{6} - 1,
end{align}
where the first of the series is the well-known Basel problem while the second telescopes.
For the second of the integrals, define
$$I(a) = int_0^1 (1 - y)^a , dy, quad a > - 1.$$
Observe that
$$I''(0) = int_0^1 ln^2 ( 1 - y) , dy,$$
where the derivative is with respect to $a$. Now, as
$$I(a) = operatorname{B} (1, a + 1) = frac{Gamma (a + 1)}{Gamma (a + 2)} = frac{1}{a + 1},$$
where $operatorname{B}(x,y)$ is the Beta function, one readily has
$$I''(a) = frac{2}{(a + 1)^3},$$
yielding $I''(0) = I_2 = 2$.
Thus
$$sum_{n = 1}^infty frac{H_n}{n (n + 1)} = frac{pi^2}{6} - 1 + frac{1}{2} cdot 2 = frac{pi^2}{6}.$$
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
$endgroup$
add a comment |
$begingroup$
Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$int_0^1 x^{n - 1} , dx = frac{1}{n} quad text{and} quad int_0^1 y^n , dy = frac{1}{n + 1},$$
the sum can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = int_0^1 int_0^1 frac{1}{x} sum_{n = 1}^infty H_n (xy)^n , dx dy.tag1$$
Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem.
Making use of the generating function for the Harmonic number, namely
$$-frac{log(1-x)}{1 - x}~=~sum_{n=1}^infty H_n x^n,$$
the infinity sum appearing inside the integral in (1) can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = -int_0^1 int_0^1 frac{ln (1 - xy)}{x(1 - xy)} , dx dy.$$
Enforcing a substitution of $x mapsto x/y$ gives
begin{align}
sum_{n = 1}^infty frac{H_n}{n(n + 1)} &= -int_0^1 int_0^y frac{ln (1 - x)}{x (1 - x)} , dx dy\
&= -int_0^1 int_0^y left [frac{ln (1 - x)}{x} + frac{ln (1 - x)}{1 - x} right ] , dx , dy\
&= int_0^1 operatorname{Li}_2 (y) , dy + frac{1}{2} int_0^1 ln^2 (1 - y) , dy\
&= I_1 + frac{1}{2} I_2.
end{align}
Here $operatorname{Li}_2 (y)$ is the dilogarithm function.
For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have
begin{align}
I_1 &= int_0^1 sum_{n = 1}^infty frac{y^n}{n^2} , dy\
&= sum_{n = 1}^infty frac{1}{n^2} int_0^1 y^n , dy\
&= sum_{n = 1}^infty frac{1}{n^2(n + 1)}\
&= sum_{n = 1}^infty left [frac{1}{n^2} + frac{1}{n + 1} - frac{1}{n} right ]\
&= sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty left (frac{1}{n} - frac{1}{n + 1} right )\
&= frac{pi^2}{6} - 1,
end{align}
where the first of the series is the well-known Basel problem while the second telescopes.
For the second of the integrals, define
$$I(a) = int_0^1 (1 - y)^a , dy, quad a > - 1.$$
Observe that
$$I''(0) = int_0^1 ln^2 ( 1 - y) , dy,$$
where the derivative is with respect to $a$. Now, as
$$I(a) = operatorname{B} (1, a + 1) = frac{Gamma (a + 1)}{Gamma (a + 2)} = frac{1}{a + 1},$$
where $operatorname{B}(x,y)$ is the Beta function, one readily has
$$I''(a) = frac{2}{(a + 1)^3},$$
yielding $I''(0) = I_2 = 2$.
Thus
$$sum_{n = 1}^infty frac{H_n}{n (n + 1)} = frac{pi^2}{6} - 1 + frac{1}{2} cdot 2 = frac{pi^2}{6}.$$
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
$endgroup$
add a comment |
$begingroup$
Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$int_0^1 x^{n - 1} , dx = frac{1}{n} quad text{and} quad int_0^1 y^n , dy = frac{1}{n + 1},$$
the sum can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = int_0^1 int_0^1 frac{1}{x} sum_{n = 1}^infty H_n (xy)^n , dx dy.tag1$$
Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem.
Making use of the generating function for the Harmonic number, namely
$$-frac{log(1-x)}{1 - x}~=~sum_{n=1}^infty H_n x^n,$$
the infinity sum appearing inside the integral in (1) can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = -int_0^1 int_0^1 frac{ln (1 - xy)}{x(1 - xy)} , dx dy.$$
Enforcing a substitution of $x mapsto x/y$ gives
begin{align}
sum_{n = 1}^infty frac{H_n}{n(n + 1)} &= -int_0^1 int_0^y frac{ln (1 - x)}{x (1 - x)} , dx dy\
&= -int_0^1 int_0^y left [frac{ln (1 - x)}{x} + frac{ln (1 - x)}{1 - x} right ] , dx , dy\
&= int_0^1 operatorname{Li}_2 (y) , dy + frac{1}{2} int_0^1 ln^2 (1 - y) , dy\
&= I_1 + frac{1}{2} I_2.
end{align}
Here $operatorname{Li}_2 (y)$ is the dilogarithm function.
For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have
begin{align}
I_1 &= int_0^1 sum_{n = 1}^infty frac{y^n}{n^2} , dy\
&= sum_{n = 1}^infty frac{1}{n^2} int_0^1 y^n , dy\
&= sum_{n = 1}^infty frac{1}{n^2(n + 1)}\
&= sum_{n = 1}^infty left [frac{1}{n^2} + frac{1}{n + 1} - frac{1}{n} right ]\
&= sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty left (frac{1}{n} - frac{1}{n + 1} right )\
&= frac{pi^2}{6} - 1,
end{align}
where the first of the series is the well-known Basel problem while the second telescopes.
For the second of the integrals, define
$$I(a) = int_0^1 (1 - y)^a , dy, quad a > - 1.$$
Observe that
$$I''(0) = int_0^1 ln^2 ( 1 - y) , dy,$$
where the derivative is with respect to $a$. Now, as
$$I(a) = operatorname{B} (1, a + 1) = frac{Gamma (a + 1)}{Gamma (a + 2)} = frac{1}{a + 1},$$
where $operatorname{B}(x,y)$ is the Beta function, one readily has
$$I''(a) = frac{2}{(a + 1)^3},$$
yielding $I''(0) = I_2 = 2$.
Thus
$$sum_{n = 1}^infty frac{H_n}{n (n + 1)} = frac{pi^2}{6} - 1 + frac{1}{2} cdot 2 = frac{pi^2}{6}.$$
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
$endgroup$
Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$int_0^1 x^{n - 1} , dx = frac{1}{n} quad text{and} quad int_0^1 y^n , dy = frac{1}{n + 1},$$
the sum can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = int_0^1 int_0^1 frac{1}{x} sum_{n = 1}^infty H_n (xy)^n , dx dy.tag1$$
Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem.
Making use of the generating function for the Harmonic number, namely
$$-frac{log(1-x)}{1 - x}~=~sum_{n=1}^infty H_n x^n,$$
the infinity sum appearing inside the integral in (1) can be written as
$$sum_{n = 1}^infty frac{H_n}{n(n + 1)} = -int_0^1 int_0^1 frac{ln (1 - xy)}{x(1 - xy)} , dx dy.$$
Enforcing a substitution of $x mapsto x/y$ gives
begin{align}
sum_{n = 1}^infty frac{H_n}{n(n + 1)} &= -int_0^1 int_0^y frac{ln (1 - x)}{x (1 - x)} , dx dy\
&= -int_0^1 int_0^y left [frac{ln (1 - x)}{x} + frac{ln (1 - x)}{1 - x} right ] , dx , dy\
&= int_0^1 operatorname{Li}_2 (y) , dy + frac{1}{2} int_0^1 ln^2 (1 - y) , dy\
&= I_1 + frac{1}{2} I_2.
end{align}
Here $operatorname{Li}_2 (y)$ is the dilogarithm function.
For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have
begin{align}
I_1 &= int_0^1 sum_{n = 1}^infty frac{y^n}{n^2} , dy\
&= sum_{n = 1}^infty frac{1}{n^2} int_0^1 y^n , dy\
&= sum_{n = 1}^infty frac{1}{n^2(n + 1)}\
&= sum_{n = 1}^infty left [frac{1}{n^2} + frac{1}{n + 1} - frac{1}{n} right ]\
&= sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty left (frac{1}{n} - frac{1}{n + 1} right )\
&= frac{pi^2}{6} - 1,
end{align}
where the first of the series is the well-known Basel problem while the second telescopes.
For the second of the integrals, define
$$I(a) = int_0^1 (1 - y)^a , dy, quad a > - 1.$$
Observe that
$$I''(0) = int_0^1 ln^2 ( 1 - y) , dy,$$
where the derivative is with respect to $a$. Now, as
$$I(a) = operatorname{B} (1, a + 1) = frac{Gamma (a + 1)}{Gamma (a + 2)} = frac{1}{a + 1},$$
where $operatorname{B}(x,y)$ is the Beta function, one readily has
$$I''(a) = frac{2}{(a + 1)^3},$$
yielding $I''(0) = I_2 = 2$.
Thus
$$sum_{n = 1}^infty frac{H_n}{n (n + 1)} = frac{pi^2}{6} - 1 + frac{1}{2} cdot 2 = frac{pi^2}{6}.$$
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
edited Jan 8 at 13:04
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
answered Jan 8 at 3:23
omegadotomegadot
5,1922727
5,1922727
add a comment |
add a comment |
$begingroup$
Using summation by parts we have $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}sum_{nleq N}frac{1}{nleft(n+1right)}-sum_{nleq N-1}frac{1}{n+1}left(sum_{kleq n}frac{1}{kleft(k+1right)}right).$$ Clearly $$sum_{nleq N}frac{1}{nleft(n+1right)}=left(1-frac{1}{N+1}right)$$ then $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}left(1-frac{1}{N+1}right)-H_{N}+sum_{nleq N-1}frac{1}{left(n+1right)^{2}}.$$ Now, since $H_{N}simlogleft(Nright)$ as $Nrightarrow+infty$, the claim follows.
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
$endgroup$
add a comment |
$begingroup$
Using summation by parts we have $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}sum_{nleq N}frac{1}{nleft(n+1right)}-sum_{nleq N-1}frac{1}{n+1}left(sum_{kleq n}frac{1}{kleft(k+1right)}right).$$ Clearly $$sum_{nleq N}frac{1}{nleft(n+1right)}=left(1-frac{1}{N+1}right)$$ then $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}left(1-frac{1}{N+1}right)-H_{N}+sum_{nleq N-1}frac{1}{left(n+1right)^{2}}.$$ Now, since $H_{N}simlogleft(Nright)$ as $Nrightarrow+infty$, the claim follows.
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
$endgroup$
add a comment |
$begingroup$
Using summation by parts we have $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}sum_{nleq N}frac{1}{nleft(n+1right)}-sum_{nleq N-1}frac{1}{n+1}left(sum_{kleq n}frac{1}{kleft(k+1right)}right).$$ Clearly $$sum_{nleq N}frac{1}{nleft(n+1right)}=left(1-frac{1}{N+1}right)$$ then $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}left(1-frac{1}{N+1}right)-H_{N}+sum_{nleq N-1}frac{1}{left(n+1right)^{2}}.$$ Now, since $H_{N}simlogleft(Nright)$ as $Nrightarrow+infty$, the claim follows.
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
$endgroup$
Using summation by parts we have $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}sum_{nleq N}frac{1}{nleft(n+1right)}-sum_{nleq N-1}frac{1}{n+1}left(sum_{kleq n}frac{1}{kleft(k+1right)}right).$$ Clearly $$sum_{nleq N}frac{1}{nleft(n+1right)}=left(1-frac{1}{N+1}right)$$ then $$sum_{nleq N}frac{H_{n}}{nleft(n+1right)}=H_{N}left(1-frac{1}{N+1}right)-H_{N}+sum_{nleq N-1}frac{1}{left(n+1right)^{2}}.$$ Now, since $H_{N}simlogleft(Nright)$ as $Nrightarrow+infty$, the claim follows.
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
answered Jan 8 at 13:14
Marco CantariniMarco Cantarini
29.1k23373
29.1k23373
add a comment |
add a comment |
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$begingroup$
It should work similarly to this question.
$endgroup$
– Dietrich Burde
Jan 6 at 15:13