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Adventure with infinite series, a curiosity
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It is easily verifiable that
$$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
It is not that difficult to get
$$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$
Question. Is there something similarly "nice" in computing
$$sum_kgeq0binom8k4kfrac12^10k=?$$
Perhaps the same question about
$$sum_kgeq0binom16k8kfrac12^20k=?$$
NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.
real-analysis sequences-and-series
asked 1 hour ago
T. Amdeberhan
16.2k228120
add a comment |Â
up vote
1
down vote
favorite
It is easily verifiable that
$$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
It is not that difficult to get
$$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$
Question. Is there something similarly "nice" in computing
$$sum_kgeq0binom8k4kfrac12^10k=?$$
Perhaps the same question about
$$sum_kgeq0binom16k8kfrac12^20k=?$$
NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.
real-analysis sequences-and-series
asked 1 hour ago
T. Amdeberhan
16.2k228120
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
It is easily verifiable that
$$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
It is not that difficult to get
$$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$
Question. Is there something similarly "nice" in computing
$$sum_kgeq0binom8k4kfrac12^10k=?$$
Perhaps the same question about
$$sum_kgeq0binom16k8kfrac12^20k=?$$
NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.
real-analysis sequences-and-series
asked 1 hour ago
T. Amdeberhan
16.2k228120
It is easily verifiable that
$$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
It is not that difficult to get
$$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$
Question. Is there something similarly "nice" in computing
$$sum_kgeq0binom8k4kfrac12^10k=?$$
Perhaps the same question about
$$sum_kgeq0binom16k8kfrac12^20k=?$$
NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked 1 hour ago
T. Amdeberhan
16.2k228120
asked 1 hour ago
T. Amdeberhan
16.2k228120
asked 1 hour ago
T. Amdeberhan
16.2k228120
asked 1 hour ago
T. Amdeberhan
16.2k228120
asked 1 hour ago
T. Amdeberhan
16.2k228120
16.2k228120
add a comment |Â
add a comment |Â
2 Answers
2
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up vote
5
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It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
and so on.
answered 1 hour ago
Fedor Petrov
45.3k5108213
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up vote
0
down vote
More generally, it seems that
$$
sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
$$
where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).
answered 21 mins ago
AccidentalFourierTransform
541112
Note: this has been confirmed up to $j=7$.
– AccidentalFourierTransform
4 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
and so on.
answered 1 hour ago
Fedor Petrov
45.3k5108213
add a comment |Â
up vote
5
down vote
It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
and so on.
answered 1 hour ago
Fedor Petrov
45.3k5108213
add a comment |Â
up vote
5
down vote
up vote
5
down vote
It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
and so on.
answered 1 hour ago
Fedor Petrov
45.3k5108213
It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
and so on.
answered 1 hour ago
Fedor Petrov
45.3k5108213
answered 1 hour ago
Fedor Petrov
45.3k5108213
answered 1 hour ago
Fedor Petrov
45.3k5108213
answered 1 hour ago
Fedor Petrov
45.3k5108213
45.3k5108213
add a comment |Â
add a comment |Â
up vote
0
down vote
More generally, it seems that
$$
sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
$$
where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).
answered 21 mins ago
AccidentalFourierTransform
541112
Note: this has been confirmed up to $j=7$.
– AccidentalFourierTransform
4 mins ago
add a comment |Â
up vote
0
down vote
More generally, it seems that
$$
sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
$$
where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).
answered 21 mins ago
AccidentalFourierTransform
541112
Note: this has been confirmed up to $j=7$.
– AccidentalFourierTransform
4 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
More generally, it seems that
$$
sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
$$
where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).
answered 21 mins ago
AccidentalFourierTransform
541112
More generally, it seems that
$$
sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
$$
where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).
answered 21 mins ago
AccidentalFourierTransform
541112
answered 21 mins ago
AccidentalFourierTransform
541112
answered 21 mins ago
AccidentalFourierTransform
541112
answered 21 mins ago
AccidentalFourierTransform
541112
541112
Note: this has been confirmed up to $j=7$.
– AccidentalFourierTransform
4 mins ago
add a comment |Â
Note: this has been confirmed up to $j=7$.
– AccidentalFourierTransform
4 mins ago
Note: this has been confirmed up to $j=7$.
– AccidentalFourierTransform
4 mins ago
Note: this has been confirmed up to $j=7$.
– AccidentalFourierTransform
4 mins ago
add a comment |Â
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