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How to evaluate this infinite sum involving powers and trigonometric terms?
How to turn this sum into an integral?Series Question: $sum_n=1^inftyfrac116n^2-1$How to evaluate the following infinite sum involving the Riemann zeta function?Infinite sum involving ascending powersHow to evaluate this double infinite sum (Catalan number)Help with an infinite sum of exponential terms?Infinite sum involving powers and factorialsHow to evaluate this limit of a finite sumInfinite sum with alternated termsHow to check if the sum of infinite series is convergent?
$begingroup$
I wish to find out the following infinite sum:
$$lim_nto inftysum_k=0^nleft(frac29right)^ksinleft[frac2pi3(2^k)right]$$
I can sum up a GP or an AGP well, and know telescoping series, how can I find this infinite sum? Any help would be appreciated. Thanks in advance!
sequences-and-series limits
edited Mar 22 at 15:35
Paras Khosla
2,883523
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
$endgroup$
add a comment |
$begingroup$
I wish to find out the following infinite sum:
$$lim_nto inftysum_k=0^nleft(frac29right)^ksinleft[frac2pi3(2^k)right]$$
I can sum up a GP or an AGP well, and know telescoping series, how can I find this infinite sum? Any help would be appreciated. Thanks in advance!
sequences-and-series limits
edited Mar 22 at 15:35
Paras Khosla
2,883523
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
$endgroup$
1
$begingroup$
Do you have any reason to think the sum can be expressed in closed form?
$endgroup$
– Robert Israel
Mar 22 at 15:41
$begingroup$
What is the origin of the problem?
$endgroup$
– user
Mar 22 at 15:42
$begingroup$
@RobertIsrael I didnt get you....
$endgroup$
– saisanjeev
Mar 24 at 9:36
add a comment |
$begingroup$
I wish to find out the following infinite sum:
$$lim_nto inftysum_k=0^nleft(frac29right)^ksinleft[frac2pi3(2^k)right]$$
I can sum up a GP or an AGP well, and know telescoping series, how can I find this infinite sum? Any help would be appreciated. Thanks in advance!
sequences-and-series limits
edited Mar 22 at 15:35
Paras Khosla
2,883523
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
$endgroup$
I wish to find out the following infinite sum:
$$lim_nto inftysum_k=0^nleft(frac29right)^ksinleft[frac2pi3(2^k)right]$$
I can sum up a GP or an AGP well, and know telescoping series, how can I find this infinite sum? Any help would be appreciated. Thanks in advance!
sequences-and-series limits
sequences-and-series limits
edited Mar 22 at 15:35
Paras Khosla
2,883523
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
edited Mar 22 at 15:35
Paras Khosla
2,883523
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
edited Mar 22 at 15:35
Paras Khosla
2,883523
edited Mar 22 at 15:35
Paras Khosla
2,883523
edited Mar 22 at 15:35
Paras Khosla
2,883523
2,883523
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
asked Mar 22 at 15:30
saisanjeevsaisanjeev
1,076312
1,076312
1
$begingroup$
Do you have any reason to think the sum can be expressed in closed form?
$endgroup$
– Robert Israel
Mar 22 at 15:41
$begingroup$
What is the origin of the problem?
$endgroup$
– user
Mar 22 at 15:42
$begingroup$
@RobertIsrael I didnt get you....
$endgroup$
– saisanjeev
Mar 24 at 9:36
add a comment |
1
$begingroup$
Do you have any reason to think the sum can be expressed in closed form?
$endgroup$
– Robert Israel
Mar 22 at 15:41
$begingroup$
What is the origin of the problem?
$endgroup$
– user
Mar 22 at 15:42
$begingroup$
@RobertIsrael I didnt get you....
$endgroup$
– saisanjeev
Mar 24 at 9:36
1
1
$begingroup$
Do you have any reason to think the sum can be expressed in closed form?
$endgroup$
– Robert Israel
Mar 22 at 15:41
$begingroup$
Do you have any reason to think the sum can be expressed in closed form?
$endgroup$
– Robert Israel
Mar 22 at 15:41
$begingroup$
What is the origin of the problem?
$endgroup$
– user
Mar 22 at 15:42
$begingroup$
What is the origin of the problem?
$endgroup$
– user
Mar 22 at 15:42
$begingroup$
@RobertIsrael I didnt get you....
$endgroup$
– saisanjeev
Mar 24 at 9:36
$begingroup$
@RobertIsrael I didnt get you....
$endgroup$
– saisanjeev
Mar 24 at 9:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't think that there is a closed form. When $k$ is sufficiently large, the argument of the sine becomes tiny and you can use a linear approximation. Hence, denoting $S$ the infinite sum (which converges),
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb=S_a,b-fraca^n+1b2^n(2-a).$$
A better approximation is obtained by using the next term in the Taylor developments, giving
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb+frac16sum_k=n+1^infty a^k(2^-kb)^3
\=S_a,b-fraca^n+1b2^n(2-a)+fraca^n+1b^36cdot8^n(8-a),$$
and so on.
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Why not use the whole Taylor series? For $|a| <2$, $$ sum_n=0^infty a^n sin(2^-n b) = sum _k=0^infty -2,frac 4^k left( -1 right) ^kb^2 ,k+1 left(a -2cdot 4^k right) left( 2,k+1 right) ! $$
$endgroup$
– Robert Israel
Mar 22 at 16:40
$begingroup$
@RobertIsrael: this was implied by "and so on". ;-)
$endgroup$
– Yves Daoust
Mar 22 at 16:41
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
I don't think that there is a closed form. When $k$ is sufficiently large, the argument of the sine becomes tiny and you can use a linear approximation. Hence, denoting $S$ the infinite sum (which converges),
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb=S_a,b-fraca^n+1b2^n(2-a).$$
A better approximation is obtained by using the next term in the Taylor developments, giving
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb+frac16sum_k=n+1^infty a^k(2^-kb)^3
\=S_a,b-fraca^n+1b2^n(2-a)+fraca^n+1b^36cdot8^n(8-a),$$
and so on.
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Why not use the whole Taylor series? For $|a| <2$, $$ sum_n=0^infty a^n sin(2^-n b) = sum _k=0^infty -2,frac 4^k left( -1 right) ^kb^2 ,k+1 left(a -2cdot 4^k right) left( 2,k+1 right) ! $$
$endgroup$
– Robert Israel
Mar 22 at 16:40
$begingroup$
@RobertIsrael: this was implied by "and so on". ;-)
$endgroup$
– Yves Daoust
Mar 22 at 16:41
add a comment |
$begingroup$
I don't think that there is a closed form. When $k$ is sufficiently large, the argument of the sine becomes tiny and you can use a linear approximation. Hence, denoting $S$ the infinite sum (which converges),
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb=S_a,b-fraca^n+1b2^n(2-a).$$
A better approximation is obtained by using the next term in the Taylor developments, giving
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb+frac16sum_k=n+1^infty a^k(2^-kb)^3
\=S_a,b-fraca^n+1b2^n(2-a)+fraca^n+1b^36cdot8^n(8-a),$$
and so on.
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Why not use the whole Taylor series? For $|a| <2$, $$ sum_n=0^infty a^n sin(2^-n b) = sum _k=0^infty -2,frac 4^k left( -1 right) ^kb^2 ,k+1 left(a -2cdot 4^k right) left( 2,k+1 right) ! $$
$endgroup$
– Robert Israel
Mar 22 at 16:40
$begingroup$
@RobertIsrael: this was implied by "and so on". ;-)
$endgroup$
– Yves Daoust
Mar 22 at 16:41
add a comment |
$begingroup$
I don't think that there is a closed form. When $k$ is sufficiently large, the argument of the sine becomes tiny and you can use a linear approximation. Hence, denoting $S$ the infinite sum (which converges),
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb=S_a,b-fraca^n+1b2^n(2-a).$$
A better approximation is obtained by using the next term in the Taylor developments, giving
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb+frac16sum_k=n+1^infty a^k(2^-kb)^3
\=S_a,b-fraca^n+1b2^n(2-a)+fraca^n+1b^36cdot8^n(8-a),$$
and so on.
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
$endgroup$
I don't think that there is a closed form. When $k$ is sufficiently large, the argument of the sine becomes tiny and you can use a linear approximation. Hence, denoting $S$ the infinite sum (which converges),
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb=S_a,b-fraca^n+1b2^n(2-a).$$
A better approximation is obtained by using the next term in the Taylor developments, giving
$$sum_k=0^na^ksin(2^-kb)approx S_a,b-sum_k=n+1^infty a^k2^-kb+frac16sum_k=n+1^infty a^k(2^-kb)^3
\=S_a,b-fraca^n+1b2^n(2-a)+fraca^n+1b^36cdot8^n(8-a),$$
and so on.
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
answered Mar 22 at 15:48
Yves DaoustYves Daoust
132k676230
132k676230
$begingroup$
Why not use the whole Taylor series? For $|a| <2$, $$ sum_n=0^infty a^n sin(2^-n b) = sum _k=0^infty -2,frac 4^k left( -1 right) ^kb^2 ,k+1 left(a -2cdot 4^k right) left( 2,k+1 right) ! $$
$endgroup$
– Robert Israel
Mar 22 at 16:40
$begingroup$
@RobertIsrael: this was implied by "and so on". ;-)
$endgroup$
– Yves Daoust
Mar 22 at 16:41
add a comment |
$begingroup$
Why not use the whole Taylor series? For $|a| <2$, $$ sum_n=0^infty a^n sin(2^-n b) = sum _k=0^infty -2,frac 4^k left( -1 right) ^kb^2 ,k+1 left(a -2cdot 4^k right) left( 2,k+1 right) ! $$
$endgroup$
– Robert Israel
Mar 22 at 16:40
$begingroup$
@RobertIsrael: this was implied by "and so on". ;-)
$endgroup$
– Yves Daoust
Mar 22 at 16:41
$begingroup$
Why not use the whole Taylor series? For $|a| <2$, $$ sum_n=0^infty a^n sin(2^-n b) = sum _k=0^infty -2,frac 4^k left( -1 right) ^kb^2 ,k+1 left(a -2cdot 4^k right) left( 2,k+1 right) ! $$
$endgroup$
– Robert Israel
Mar 22 at 16:40
$begingroup$
Why not use the whole Taylor series? For $|a| <2$, $$ sum_n=0^infty a^n sin(2^-n b) = sum _k=0^infty -2,frac 4^k left( -1 right) ^kb^2 ,k+1 left(a -2cdot 4^k right) left( 2,k+1 right) ! $$
$endgroup$
– Robert Israel
Mar 22 at 16:40
$begingroup$
@RobertIsrael: this was implied by "and so on". ;-)
$endgroup$
– Yves Daoust
Mar 22 at 16:41
$begingroup$
@RobertIsrael: this was implied by "and so on". ;-)
$endgroup$
– Yves Daoust
Mar 22 at 16:41
add a comment |
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$begingroup$
Do you have any reason to think the sum can be expressed in closed form?
$endgroup$
– Robert Israel
Mar 22 at 15:41
$begingroup$
What is the origin of the problem?
$endgroup$
– user
Mar 22 at 15:42
$begingroup$
@RobertIsrael I didnt get you....
$endgroup$
– saisanjeev
Mar 24 at 9:36