A limit question without ideas [on hold]
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I did not calculate it, I hope you can give me some ideas.
$$lim_{nrightarrow infty}left( frac{1}{3-1}+frac{1}{3^2-1}+...+frac{1}{3^n-1} right). $$
calculus
asked Nov 15 at 11:35
daimengjie
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put on hold as off-topic by amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo Nov 15 at 13:59
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-5
down vote
favorite
1
I did not calculate it, I hope you can give me some ideas.
$$lim_{nrightarrow infty}left( frac{1}{3-1}+frac{1}{3^2-1}+...+frac{1}{3^n-1} right). $$
calculus
asked Nov 15 at 11:35
daimengjie
9
New contributor
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo Nov 15 at 13:59
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
You should show some effort that you did yourself. What did you try to do? Where did you get stuck?
– cholo14
Nov 15 at 11:37
1
Wolframalpha told me the answer is related to the q-digamma function, which is not elementary.
– Tianlalu
Nov 15 at 11:41
You will find an answer, of a kind, here at Wolfram Alpha. But it doesn't give the derivation.
– TonyK
Nov 15 at 11:41
$$sum_{ngeq 1}frac{1}{3^n-1}=sum_{ngeq 1}sum_{kgeq 1} 3^{-kn} = sum_{mgeq 1}frac{d(m)}{3^m}$$ is a trascendental number (Tachiya) which can be effectively approximated through the Mellin transform (Riedel). It is $0.6821535ldots$
– Jack D'Aurizio
Nov 15 at 18:48
add a comment |
up vote
-5
down vote
favorite
1
up vote
-5
down vote
favorite
1
1
I did not calculate it, I hope you can give me some ideas.
$$lim_{nrightarrow infty}left( frac{1}{3-1}+frac{1}{3^2-1}+...+frac{1}{3^n-1} right). $$
calculus
asked Nov 15 at 11:35
daimengjie
9
New contributor
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I did not calculate it, I hope you can give me some ideas.
$$lim_{nrightarrow infty}left( frac{1}{3-1}+frac{1}{3^2-1}+...+frac{1}{3^n-1} right). $$
calculus
calculus
asked Nov 15 at 11:35
daimengjie
9
New contributor
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 11:35
daimengjie
9
New contributor
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 11:35
daimengjie
9
New contributor
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 11:35
daimengjie
9
asked Nov 15 at 11:35
daimengjie
9
9
New contributor
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo Nov 15 at 13:59
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo Nov 15 at 13:59
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Nosrati, Théophile, rschwieb, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
You should show some effort that you did yourself. What did you try to do? Where did you get stuck?
– cholo14
Nov 15 at 11:37
1
Wolframalpha told me the answer is related to the q-digamma function, which is not elementary.
– Tianlalu
Nov 15 at 11:41
You will find an answer, of a kind, here at Wolfram Alpha. But it doesn't give the derivation.
– TonyK
Nov 15 at 11:41
$$sum_{ngeq 1}frac{1}{3^n-1}=sum_{ngeq 1}sum_{kgeq 1} 3^{-kn} = sum_{mgeq 1}frac{d(m)}{3^m}$$ is a trascendental number (Tachiya) which can be effectively approximated through the Mellin transform (Riedel). It is $0.6821535ldots$
– Jack D'Aurizio
Nov 15 at 18:48
add a comment |
You should show some effort that you did yourself. What did you try to do? Where did you get stuck?
– cholo14
Nov 15 at 11:37
1
Wolframalpha told me the answer is related to the q-digamma function, which is not elementary.
– Tianlalu
Nov 15 at 11:41
You will find an answer, of a kind, here at Wolfram Alpha. But it doesn't give the derivation.
– TonyK
Nov 15 at 11:41
$$sum_{ngeq 1}frac{1}{3^n-1}=sum_{ngeq 1}sum_{kgeq 1} 3^{-kn} = sum_{mgeq 1}frac{d(m)}{3^m}$$ is a trascendental number (Tachiya) which can be effectively approximated through the Mellin transform (Riedel). It is $0.6821535ldots$
– Jack D'Aurizio
Nov 15 at 18:48
You should show some effort that you did yourself. What did you try to do? Where did you get stuck?
– cholo14
Nov 15 at 11:37
You should show some effort that you did yourself. What did you try to do? Where did you get stuck?
– cholo14
Nov 15 at 11:37
1
1
Wolframalpha told me the answer is related to the q-digamma function, which is not elementary.
– Tianlalu
Nov 15 at 11:41
Wolframalpha told me the answer is related to the q-digamma function, which is not elementary.
– Tianlalu
Nov 15 at 11:41
You will find an answer, of a kind, here at Wolfram Alpha. But it doesn't give the derivation.
– TonyK
Nov 15 at 11:41
You will find an answer, of a kind, here at Wolfram Alpha. But it doesn't give the derivation.
– TonyK
Nov 15 at 11:41
$$sum_{ngeq 1}frac{1}{3^n-1}=sum_{ngeq 1}sum_{kgeq 1} 3^{-kn} = sum_{mgeq 1}frac{d(m)}{3^m}$$ is a trascendental number (Tachiya) which can be effectively approximated through the Mellin transform (Riedel). It is $0.6821535ldots$
– Jack D'Aurizio
Nov 15 at 18:48
$$sum_{ngeq 1}frac{1}{3^n-1}=sum_{ngeq 1}sum_{kgeq 1} 3^{-kn} = sum_{mgeq 1}frac{d(m)}{3^m}$$ is a trascendental number (Tachiya) which can be effectively approximated through the Mellin transform (Riedel). It is $0.6821535ldots$
– Jack D'Aurizio
Nov 15 at 18:48
add a comment |
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You should show some effort that you did yourself. What did you try to do? Where did you get stuck?
– cholo14
Nov 15 at 11:37
1
Wolframalpha told me the answer is related to the q-digamma function, which is not elementary.
– Tianlalu
Nov 15 at 11:41
You will find an answer, of a kind, here at Wolfram Alpha. But it doesn't give the derivation.
– TonyK
Nov 15 at 11:41
$$sum_{ngeq 1}frac{1}{3^n-1}=sum_{ngeq 1}sum_{kgeq 1} 3^{-kn} = sum_{mgeq 1}frac{d(m)}{3^m}$$ is a trascendental number (Tachiya) which can be effectively approximated through the Mellin transform (Riedel). It is $0.6821535ldots$
– Jack D'Aurizio
Nov 15 at 18:48