Mixing
$1+ {1over 11}+ {1over 111}+ {1over 1111}+…=?$ [closed]
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What is the sum of the series $$1+ {1over 11}+ {1over 111}+ {1over 1111}+....$$.The partial sum is a monotonically increasing and bounded above sequence, so sum must exits in real.
sequences-and-series
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
$endgroup$
closed as off-topic by Saad, Did, José Carlos Santos, Arnaud D., clathratus Jan 12 at 21:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, José Carlos Santos, Arnaud D., clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is the sum of the series $$1+ {1over 11}+ {1over 111}+ {1over 1111}+....$$.The partial sum is a monotonically increasing and bounded above sequence, so sum must exits in real.
sequences-and-series
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
$endgroup$
closed as off-topic by Saad, Did, José Carlos Santos, Arnaud D., clathratus Jan 12 at 21:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, José Carlos Santos, Arnaud D., clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Not sure if this helps, but you could also write the sum as $frac{9}{9}+frac{9}{99}+frac{9}{999}+...$, which becomes $sum_{i=1}^infty frac{9}{10^i-1}$.
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– Noble Mushtak
Jan 12 at 15:16
2
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Did you come up with this question? I'm not sure it convergence to anything special.
$endgroup$
– Yanko
Jan 12 at 15:17
1
$begingroup$
Wolfram Alpha gave me a closed-form answer to this sum in terms of the $q$-digamma function, but I'm not sure how they derived this answer: wolframalpha.com/input/…
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– Noble Mushtak
Jan 12 at 15:18
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@NobleMushtak wow never heard of this function.
$endgroup$
– Yanko
Jan 12 at 15:19
1
$begingroup$
In fact, it's a corollary of Eq. (4) here with $a=10$.
$endgroup$
– J.G.
Jan 12 at 15:34
add a comment |
$begingroup$
What is the sum of the series $$1+ {1over 11}+ {1over 111}+ {1over 1111}+....$$.The partial sum is a monotonically increasing and bounded above sequence, so sum must exits in real.
sequences-and-series
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
$endgroup$
What is the sum of the series $$1+ {1over 11}+ {1over 111}+ {1over 1111}+....$$.The partial sum is a monotonically increasing and bounded above sequence, so sum must exits in real.
sequences-and-series
sequences-and-series
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
asked Jan 12 at 15:13
Supriyo HalderSupriyo Halder
635113
635113
closed as off-topic by Saad, Did, José Carlos Santos, Arnaud D., clathratus Jan 12 at 21:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, José Carlos Santos, Arnaud D., clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Did, José Carlos Santos, Arnaud D., clathratus Jan 12 at 21:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, José Carlos Santos, Arnaud D., clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Not sure if this helps, but you could also write the sum as $frac{9}{9}+frac{9}{99}+frac{9}{999}+...$, which becomes $sum_{i=1}^infty frac{9}{10^i-1}$.
$endgroup$
– Noble Mushtak
Jan 12 at 15:16
2
$begingroup$
Did you come up with this question? I'm not sure it convergence to anything special.
$endgroup$
– Yanko
Jan 12 at 15:17
1
$begingroup$
Wolfram Alpha gave me a closed-form answer to this sum in terms of the $q$-digamma function, but I'm not sure how they derived this answer: wolframalpha.com/input/…
$endgroup$
– Noble Mushtak
Jan 12 at 15:18
$begingroup$
@NobleMushtak wow never heard of this function.
$endgroup$
– Yanko
Jan 12 at 15:19
1
$begingroup$
In fact, it's a corollary of Eq. (4) here with $a=10$.
$endgroup$
– J.G.
Jan 12 at 15:34
add a comment |
3
$begingroup$
Not sure if this helps, but you could also write the sum as $frac{9}{9}+frac{9}{99}+frac{9}{999}+...$, which becomes $sum_{i=1}^infty frac{9}{10^i-1}$.
$endgroup$
– Noble Mushtak
Jan 12 at 15:16
2
$begingroup$
Did you come up with this question? I'm not sure it convergence to anything special.
$endgroup$
– Yanko
Jan 12 at 15:17
1
$begingroup$
Wolfram Alpha gave me a closed-form answer to this sum in terms of the $q$-digamma function, but I'm not sure how they derived this answer: wolframalpha.com/input/…
$endgroup$
– Noble Mushtak
Jan 12 at 15:18
$begingroup$
@NobleMushtak wow never heard of this function.
$endgroup$
– Yanko
Jan 12 at 15:19
1
$begingroup$
In fact, it's a corollary of Eq. (4) here with $a=10$.
$endgroup$
– J.G.
Jan 12 at 15:34
3
3
$begingroup$
Not sure if this helps, but you could also write the sum as $frac{9}{9}+frac{9}{99}+frac{9}{999}+...$, which becomes $sum_{i=1}^infty frac{9}{10^i-1}$.
$endgroup$
– Noble Mushtak
Jan 12 at 15:16
$begingroup$
Not sure if this helps, but you could also write the sum as $frac{9}{9}+frac{9}{99}+frac{9}{999}+...$, which becomes $sum_{i=1}^infty frac{9}{10^i-1}$.
$endgroup$
– Noble Mushtak
Jan 12 at 15:16
2
2
$begingroup$
Did you come up with this question? I'm not sure it convergence to anything special.
$endgroup$
– Yanko
Jan 12 at 15:17
$begingroup$
Did you come up with this question? I'm not sure it convergence to anything special.
$endgroup$
– Yanko
Jan 12 at 15:17
1
1
$begingroup$
Wolfram Alpha gave me a closed-form answer to this sum in terms of the $q$-digamma function, but I'm not sure how they derived this answer: wolframalpha.com/input/…
$endgroup$
– Noble Mushtak
Jan 12 at 15:18
$begingroup$
Wolfram Alpha gave me a closed-form answer to this sum in terms of the $q$-digamma function, but I'm not sure how they derived this answer: wolframalpha.com/input/…
$endgroup$
– Noble Mushtak
Jan 12 at 15:18
$begingroup$
@NobleMushtak wow never heard of this function.
$endgroup$
– Yanko
Jan 12 at 15:19
$begingroup$
@NobleMushtak wow never heard of this function.
$endgroup$
– Yanko
Jan 12 at 15:19
1
1
$begingroup$
In fact, it's a corollary of Eq. (4) here with $a=10$.
$endgroup$
– J.G.
Jan 12 at 15:34
$begingroup$
In fact, it's a corollary of Eq. (4) here with $a=10$.
$endgroup$
– J.G.
Jan 12 at 15:34
add a comment |
1 Answer
1
active
oldest
votes
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This answer follows on from Noble Mushtak's comments regarding the simplification of the sum, and the closed form solution on Wolfram Alpha.
The q-digamma function can be written as
$$psi_q(z)=-ln(1-q)+ln qsum_{n=0}^inftyfrac{q^{n+z}}{1-q^{n+z}}$$
So the sum $$sum_{n=1}^inftyfrac{9}{10^n-1}=9sum_{n=0}^inftyfrac{10^{-n-1}}{1-10^{-n-1}}$$ So if we let $q=frac1{10}$, then this is $$9sum_{n=0}^inftyfrac{q^{n+1}}{1-q^{n+1}}=frac{9left(psi_{frac1{10}}(1)+lnfrac9{10}right)}{lnfrac1{10}}=frac{9left(lnfrac{10}9-psi_{frac1{10}}(1)right)}{ln{10}}$$
As given by Wolfram Alpha.
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
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2
$begingroup$
"Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
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– Did
Jan 12 at 16:25
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@Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question.
$endgroup$
– John Doe
Jan 12 at 16:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This answer follows on from Noble Mushtak's comments regarding the simplification of the sum, and the closed form solution on Wolfram Alpha.
The q-digamma function can be written as
$$psi_q(z)=-ln(1-q)+ln qsum_{n=0}^inftyfrac{q^{n+z}}{1-q^{n+z}}$$
So the sum $$sum_{n=1}^inftyfrac{9}{10^n-1}=9sum_{n=0}^inftyfrac{10^{-n-1}}{1-10^{-n-1}}$$ So if we let $q=frac1{10}$, then this is $$9sum_{n=0}^inftyfrac{q^{n+1}}{1-q^{n+1}}=frac{9left(psi_{frac1{10}}(1)+lnfrac9{10}right)}{lnfrac1{10}}=frac{9left(lnfrac{10}9-psi_{frac1{10}}(1)right)}{ln{10}}$$
As given by Wolfram Alpha.
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
$endgroup$
2
$begingroup$
"Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
$endgroup$
– Did
Jan 12 at 16:25
$begingroup$
@Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question.
$endgroup$
– John Doe
Jan 12 at 16:44
add a comment |
$begingroup$
This answer follows on from Noble Mushtak's comments regarding the simplification of the sum, and the closed form solution on Wolfram Alpha.
The q-digamma function can be written as
$$psi_q(z)=-ln(1-q)+ln qsum_{n=0}^inftyfrac{q^{n+z}}{1-q^{n+z}}$$
So the sum $$sum_{n=1}^inftyfrac{9}{10^n-1}=9sum_{n=0}^inftyfrac{10^{-n-1}}{1-10^{-n-1}}$$ So if we let $q=frac1{10}$, then this is $$9sum_{n=0}^inftyfrac{q^{n+1}}{1-q^{n+1}}=frac{9left(psi_{frac1{10}}(1)+lnfrac9{10}right)}{lnfrac1{10}}=frac{9left(lnfrac{10}9-psi_{frac1{10}}(1)right)}{ln{10}}$$
As given by Wolfram Alpha.
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
$endgroup$
2
$begingroup$
"Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
$endgroup$
– Did
Jan 12 at 16:25
$begingroup$
@Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question.
$endgroup$
– John Doe
Jan 12 at 16:44
add a comment |
$begingroup$
This answer follows on from Noble Mushtak's comments regarding the simplification of the sum, and the closed form solution on Wolfram Alpha.
The q-digamma function can be written as
$$psi_q(z)=-ln(1-q)+ln qsum_{n=0}^inftyfrac{q^{n+z}}{1-q^{n+z}}$$
So the sum $$sum_{n=1}^inftyfrac{9}{10^n-1}=9sum_{n=0}^inftyfrac{10^{-n-1}}{1-10^{-n-1}}$$ So if we let $q=frac1{10}$, then this is $$9sum_{n=0}^inftyfrac{q^{n+1}}{1-q^{n+1}}=frac{9left(psi_{frac1{10}}(1)+lnfrac9{10}right)}{lnfrac1{10}}=frac{9left(lnfrac{10}9-psi_{frac1{10}}(1)right)}{ln{10}}$$
As given by Wolfram Alpha.
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
$endgroup$
This answer follows on from Noble Mushtak's comments regarding the simplification of the sum, and the closed form solution on Wolfram Alpha.
The q-digamma function can be written as
$$psi_q(z)=-ln(1-q)+ln qsum_{n=0}^inftyfrac{q^{n+z}}{1-q^{n+z}}$$
So the sum $$sum_{n=1}^inftyfrac{9}{10^n-1}=9sum_{n=0}^inftyfrac{10^{-n-1}}{1-10^{-n-1}}$$ So if we let $q=frac1{10}$, then this is $$9sum_{n=0}^inftyfrac{q^{n+1}}{1-q^{n+1}}=frac{9left(psi_{frac1{10}}(1)+lnfrac9{10}right)}{lnfrac1{10}}=frac{9left(lnfrac{10}9-psi_{frac1{10}}(1)right)}{ln{10}}$$
As given by Wolfram Alpha.
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
answered Jan 12 at 15:38
John DoeJohn Doe
11.1k11238
11.1k11238
2
$begingroup$
"Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
$endgroup$
– Did
Jan 12 at 16:25
$begingroup$
@Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question.
$endgroup$
– John Doe
Jan 12 at 16:44
add a comment |
2
$begingroup$
"Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
$endgroup$
– Did
Jan 12 at 16:25
$begingroup$
@Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question.
$endgroup$
– John Doe
Jan 12 at 16:44
2
2
$begingroup$
"Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
$endgroup$
– Did
Jan 12 at 16:25
$begingroup$
"Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
$endgroup$
– Did
Jan 12 at 16:25
$begingroup$
@Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question.
$endgroup$
– John Doe
Jan 12 at 16:44
$begingroup$
@Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question.
$endgroup$
– John Doe
Jan 12 at 16:44
add a comment |
3
$begingroup$
Not sure if this helps, but you could also write the sum as $frac{9}{9}+frac{9}{99}+frac{9}{999}+...$, which becomes $sum_{i=1}^infty frac{9}{10^i-1}$.
$endgroup$
– Noble Mushtak
Jan 12 at 15:16
2
$begingroup$
Did you come up with this question? I'm not sure it convergence to anything special.
$endgroup$
– Yanko
Jan 12 at 15:17
1
$begingroup$
Wolfram Alpha gave me a closed-form answer to this sum in terms of the $q$-digamma function, but I'm not sure how they derived this answer: wolframalpha.com/input/…
$endgroup$
– Noble Mushtak
Jan 12 at 15:18
$begingroup$
@NobleMushtak wow never heard of this function.
$endgroup$
– Yanko
Jan 12 at 15:19
1
$begingroup$
In fact, it's a corollary of Eq. (4) here with $a=10$.
$endgroup$
– J.G.
Jan 12 at 15:34