Mixing
Harmonic number and the golden ratio
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I just don't understand how $(1)$ can have this simple closed form.
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=phi^{-6}tag1$$
Where $phi=frac{1+sqrt{5}}{2}$, is the golden ratio and $H_n$ harmonic number.
How can we show that $(1)=phi^{-6}?$
sequences-and-series harmonic-numbers golden-ratio
asked Jan 12 at 22:09
user583851user583851
1
$endgroup$
add a comment |
$begingroup$
I just don't understand how $(1)$ can have this simple closed form.
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=phi^{-6}tag1$$
Where $phi=frac{1+sqrt{5}}{2}$, is the golden ratio and $H_n$ harmonic number.
How can we show that $(1)=phi^{-6}?$
sequences-and-series harmonic-numbers golden-ratio
asked Jan 12 at 22:09
user583851user583851
1
$endgroup$
1
$begingroup$
This is just a small observation: $H_k - H_{k+1} = - frac{1}{k+1}$.
$endgroup$
– Parker Glynn-Adey
Jan 12 at 22:28
2
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To downvoters. Please take a look to the profile of the proposer. Do you really think that this question is some kind of homework?
$endgroup$
– Robert Z
Jan 13 at 15:09
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@RobertZ Clearly it is not but nevertheless $-$ as nearly all posts of this user $-$ there is no context or own attempt given.
$endgroup$
– mrtaurho
Jan 14 at 11:03
add a comment |
$begingroup$
I just don't understand how $(1)$ can have this simple closed form.
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=phi^{-6}tag1$$
Where $phi=frac{1+sqrt{5}}{2}$, is the golden ratio and $H_n$ harmonic number.
How can we show that $(1)=phi^{-6}?$
sequences-and-series harmonic-numbers golden-ratio
asked Jan 12 at 22:09
user583851user583851
1
$endgroup$
I just don't understand how $(1)$ can have this simple closed form.
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=phi^{-6}tag1$$
Where $phi=frac{1+sqrt{5}}{2}$, is the golden ratio and $H_n$ harmonic number.
How can we show that $(1)=phi^{-6}?$
sequences-and-series harmonic-numbers golden-ratio
sequences-and-series harmonic-numbers golden-ratio
asked Jan 12 at 22:09
user583851user583851
1
asked Jan 12 at 22:09
user583851user583851
1
edited Jan 12 at 22:24
user583851
asked Jan 12 at 22:09
user583851user583851
1
asked Jan 12 at 22:09
user583851user583851
1
asked Jan 12 at 22:09
user583851user583851
1
1
1
$begingroup$
This is just a small observation: $H_k - H_{k+1} = - frac{1}{k+1}$.
$endgroup$
– Parker Glynn-Adey
Jan 12 at 22:28
2
$begingroup$
To downvoters. Please take a look to the profile of the proposer. Do you really think that this question is some kind of homework?
$endgroup$
– Robert Z
Jan 13 at 15:09
$begingroup$
@RobertZ Clearly it is not but nevertheless $-$ as nearly all posts of this user $-$ there is no context or own attempt given.
$endgroup$
– mrtaurho
Jan 14 at 11:03
add a comment |
1
$begingroup$
This is just a small observation: $H_k - H_{k+1} = - frac{1}{k+1}$.
$endgroup$
– Parker Glynn-Adey
Jan 12 at 22:28
2
$begingroup$
To downvoters. Please take a look to the profile of the proposer. Do you really think that this question is some kind of homework?
$endgroup$
– Robert Z
Jan 13 at 15:09
$begingroup$
@RobertZ Clearly it is not but nevertheless $-$ as nearly all posts of this user $-$ there is no context or own attempt given.
$endgroup$
– mrtaurho
Jan 14 at 11:03
1
1
$begingroup$
This is just a small observation: $H_k - H_{k+1} = - frac{1}{k+1}$.
$endgroup$
– Parker Glynn-Adey
Jan 12 at 22:28
$begingroup$
This is just a small observation: $H_k - H_{k+1} = - frac{1}{k+1}$.
$endgroup$
– Parker Glynn-Adey
Jan 12 at 22:28
2
2
$begingroup$
To downvoters. Please take a look to the profile of the proposer. Do you really think that this question is some kind of homework?
$endgroup$
– Robert Z
Jan 13 at 15:09
$begingroup$
To downvoters. Please take a look to the profile of the proposer. Do you really think that this question is some kind of homework?
$endgroup$
– Robert Z
Jan 13 at 15:09
$begingroup$
@RobertZ Clearly it is not but nevertheless $-$ as nearly all posts of this user $-$ there is no context or own attempt given.
$endgroup$
– mrtaurho
Jan 14 at 11:03
$begingroup$
@RobertZ Clearly it is not but nevertheless $-$ as nearly all posts of this user $-$ there is no context or own attempt given.
$endgroup$
– mrtaurho
Jan 14 at 11:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that after letting $x=-1/16$, we have that
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=-sum_{k=1}^{infty}C_kx^k=1-frac{1-sqrt{1-4x}}{2x}=9-4 sqrt{5}=phi^{-6}$$
where we used the generating function of the Catalan numbers $C_k=frac{1}{k+1}binom{2k}{k}$.
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
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add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
Note that after letting $x=-1/16$, we have that
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=-sum_{k=1}^{infty}C_kx^k=1-frac{1-sqrt{1-4x}}{2x}=9-4 sqrt{5}=phi^{-6}$$
where we used the generating function of the Catalan numbers $C_k=frac{1}{k+1}binom{2k}{k}$.
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
$endgroup$
add a comment |
$begingroup$
Note that after letting $x=-1/16$, we have that
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=-sum_{k=1}^{infty}C_kx^k=1-frac{1-sqrt{1-4x}}{2x}=9-4 sqrt{5}=phi^{-6}$$
where we used the generating function of the Catalan numbers $C_k=frac{1}{k+1}binom{2k}{k}$.
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
$endgroup$
add a comment |
$begingroup$
Note that after letting $x=-1/16$, we have that
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=-sum_{k=1}^{infty}C_kx^k=1-frac{1-sqrt{1-4x}}{2x}=9-4 sqrt{5}=phi^{-6}$$
where we used the generating function of the Catalan numbers $C_k=frac{1}{k+1}binom{2k}{k}$.
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
$endgroup$
Note that after letting $x=-1/16$, we have that
$$sum_{k=1}^{infty}frac{{2k choose k}}{(-16)^k}[H_k-H_{k+1}]=-sum_{k=1}^{infty}C_kx^k=1-frac{1-sqrt{1-4x}}{2x}=9-4 sqrt{5}=phi^{-6}$$
where we used the generating function of the Catalan numbers $C_k=frac{1}{k+1}binom{2k}{k}$.
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
edited Jan 12 at 22:34
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
answered Jan 12 at 22:27
Robert ZRobert Z
97.3k1066137
97.3k1066137
add a comment |
add a comment |
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1
$begingroup$
This is just a small observation: $H_k - H_{k+1} = - frac{1}{k+1}$.
$endgroup$
– Parker Glynn-Adey
Jan 12 at 22:28
2
$begingroup$
To downvoters. Please take a look to the profile of the proposer. Do you really think that this question is some kind of homework?
$endgroup$
– Robert Z
Jan 13 at 15:09
$begingroup$
@RobertZ Clearly it is not but nevertheless $-$ as nearly all posts of this user $-$ there is no context or own attempt given.
$endgroup$
– mrtaurho
Jan 14 at 11:03