How would you prove this infinite sum? [closed]
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$$displaystyle sum_{n=0}^{infty}frac{1}{2^{n+2}}tanleft(frac{pi}{2^{n+3}}right)=frac{2}{pi}-frac{1}{2}$$
sequences-and-series trigonometric-series
edited Jan 13 at 15:32
KM101
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asked Jan 13 at 15:23
ben tenysonben tenyson
364
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closed as off-topic by abiessu, Xander Henderson, RRL, amWhy, max_zorn Jan 13 at 22:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
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add a comment |
$begingroup$
$$displaystyle sum_{n=0}^{infty}frac{1}{2^{n+2}}tanleft(frac{pi}{2^{n+3}}right)=frac{2}{pi}-frac{1}{2}$$
sequences-and-series trigonometric-series
edited Jan 13 at 15:32
KM101
5,9481524
asked Jan 13 at 15:23
ben tenysonben tenyson
364
$endgroup$
closed as off-topic by abiessu, Xander Henderson, RRL, amWhy, max_zorn Jan 13 at 22:13
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." – abiessu, Xander Henderson, RRL, amWhy, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
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See math.stackexchange.com/questions/1778357/…
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– lab bhattacharjee
Jan 13 at 16:23
add a comment |
$begingroup$
$$displaystyle sum_{n=0}^{infty}frac{1}{2^{n+2}}tanleft(frac{pi}{2^{n+3}}right)=frac{2}{pi}-frac{1}{2}$$
sequences-and-series trigonometric-series
edited Jan 13 at 15:32
KM101
5,9481524
asked Jan 13 at 15:23
ben tenysonben tenyson
364
$endgroup$
$$displaystyle sum_{n=0}^{infty}frac{1}{2^{n+2}}tanleft(frac{pi}{2^{n+3}}right)=frac{2}{pi}-frac{1}{2}$$
sequences-and-series trigonometric-series
sequences-and-series trigonometric-series
edited Jan 13 at 15:32
KM101
5,9481524
asked Jan 13 at 15:23
ben tenysonben tenyson
364
edited Jan 13 at 15:32
KM101
5,9481524
asked Jan 13 at 15:23
ben tenysonben tenyson
364
edited Jan 13 at 15:32
KM101
5,9481524
edited Jan 13 at 15:32
KM101
5,9481524
edited Jan 13 at 15:32
KM101
5,9481524
5,9481524
asked Jan 13 at 15:23
ben tenysonben tenyson
364
asked Jan 13 at 15:23
ben tenysonben tenyson
364
asked Jan 13 at 15:23
ben tenysonben tenyson
364
364
closed as off-topic by abiessu, Xander Henderson, RRL, amWhy, max_zorn Jan 13 at 22:13
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." – abiessu, Xander Henderson, RRL, amWhy, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by abiessu, Xander Henderson, RRL, amWhy, max_zorn Jan 13 at 22:13
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." – abiessu, Xander Henderson, RRL, amWhy, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
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See math.stackexchange.com/questions/1778357/…
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– lab bhattacharjee
Jan 13 at 16:23
add a comment |
$begingroup$
See math.stackexchange.com/questions/1778357/…
$endgroup$
– lab bhattacharjee
Jan 13 at 16:23
$begingroup$
See math.stackexchange.com/questions/1778357/…
$endgroup$
– lab bhattacharjee
Jan 13 at 16:23
$begingroup$
See math.stackexchange.com/questions/1778357/…
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– lab bhattacharjee
Jan 13 at 16:23
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1 Answer
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Hint. From the identity
$$
tan{x}=cot{x}-2cot{2 x}
$$ one may write
$$
frac{1}{2^{n+2}}tan{frac{pi}{2^{n+3}}}=frac{1}{2^{n+2}}cot{frac{pi}{2^{n+3}}}-frac{1}{2^{n+1}}cot{frac{pi}{2^{n+2}}}
$$ then one may conclude with a telescopic sum and a classic limit.
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
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1
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Please refrain from answer questions which include absolutely no context whatsoever.
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– amWhy
Jan 13 at 20:47
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint. From the identity
$$
tan{x}=cot{x}-2cot{2 x}
$$ one may write
$$
frac{1}{2^{n+2}}tan{frac{pi}{2^{n+3}}}=frac{1}{2^{n+2}}cot{frac{pi}{2^{n+3}}}-frac{1}{2^{n+1}}cot{frac{pi}{2^{n+2}}}
$$ then one may conclude with a telescopic sum and a classic limit.
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
1
$begingroup$
Please refrain from answer questions which include absolutely no context whatsoever.
$endgroup$
– amWhy
Jan 13 at 20:47
add a comment |
$begingroup$
Hint. From the identity
$$
tan{x}=cot{x}-2cot{2 x}
$$ one may write
$$
frac{1}{2^{n+2}}tan{frac{pi}{2^{n+3}}}=frac{1}{2^{n+2}}cot{frac{pi}{2^{n+3}}}-frac{1}{2^{n+1}}cot{frac{pi}{2^{n+2}}}
$$ then one may conclude with a telescopic sum and a classic limit.
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
1
$begingroup$
Please refrain from answer questions which include absolutely no context whatsoever.
$endgroup$
– amWhy
Jan 13 at 20:47
add a comment |
$begingroup$
Hint. From the identity
$$
tan{x}=cot{x}-2cot{2 x}
$$ one may write
$$
frac{1}{2^{n+2}}tan{frac{pi}{2^{n+3}}}=frac{1}{2^{n+2}}cot{frac{pi}{2^{n+3}}}-frac{1}{2^{n+1}}cot{frac{pi}{2^{n+2}}}
$$ then one may conclude with a telescopic sum and a classic limit.
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
Hint. From the identity
$$
tan{x}=cot{x}-2cot{2 x}
$$ one may write
$$
frac{1}{2^{n+2}}tan{frac{pi}{2^{n+3}}}=frac{1}{2^{n+2}}cot{frac{pi}{2^{n+3}}}-frac{1}{2^{n+1}}cot{frac{pi}{2^{n+2}}}
$$ then one may conclude with a telescopic sum and a classic limit.
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
answered Jan 13 at 15:31
Olivier OloaOlivier Oloa
108k17177294
108k17177294
1
$begingroup$
Please refrain from answer questions which include absolutely no context whatsoever.
$endgroup$
– amWhy
Jan 13 at 20:47
add a comment |
1
$begingroup$
Please refrain from answer questions which include absolutely no context whatsoever.
$endgroup$
– amWhy
Jan 13 at 20:47
1
1
$begingroup$
Please refrain from answer questions which include absolutely no context whatsoever.
$endgroup$
– amWhy
Jan 13 at 20:47
$begingroup$
Please refrain from answer questions which include absolutely no context whatsoever.
$endgroup$
– amWhy
Jan 13 at 20:47
add a comment |
$begingroup$
See math.stackexchange.com/questions/1778357/…
$endgroup$
– lab bhattacharjee
Jan 13 at 16:23