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Calculating the series $1/8+1/88+1/888+…$
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up vote
7
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I wonder whether this series is calculable or not.
Attempt:
$S=1/8+1/88+1/888+....=dfrac18displaystylesum_k=0^inftydfrac1sum_n=0^k10^n$
where $$displaystylesum_n=0^k10^n=dfrac10^k+1-19$$
then
$S=dfrac98displaystylesum_k=0^inftydfrac110^k+1-1$
I have tried to calculate $displaystylesum_k=0^Kdfrac110^k+1-1$ for finite values but I failed.
What methods can we try?
sequences-and-series
asked Aug 1 at 8:05
user2312512851
1,134521
 |Â
show 3 more comments
up vote
7
down vote
favorite
I wonder whether this series is calculable or not.
Attempt:
$S=1/8+1/88+1/888+....=dfrac18displaystylesum_k=0^inftydfrac1sum_n=0^k10^n$
where $$displaystylesum_n=0^k10^n=dfrac10^k+1-19$$
then
$S=dfrac98displaystylesum_k=0^inftydfrac110^k+1-1$
I have tried to calculate $displaystylesum_k=0^Kdfrac110^k+1-1$ for finite values but I failed.
What methods can we try?
sequences-and-series
asked Aug 1 at 8:05
user2312512851
1,134521
by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test
– user2312512851
Aug 1 at 8:07
I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series?
– user2312512851
Aug 1 at 8:11
WolframAlpha doesn't give an elementary answer, so chances are there isn't one. There are simplifications that WA misses, though, so we can't be certain.
– Arthur
Aug 1 at 8:23
This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function"
– Von Neumann
Aug 1 at 8:23
The sum $$s(z):=sum_n=0^infty,frac1z^n-1text for z>1$$ involves the $q$-polygamma function $psi^(0)_q$ with $q:=frac1z$. WolframAlpha gives $$s(z)=fraclnleft(fraczz-1right)-psi^(0)_frac1z(1)ln(z)text for z>1,.$$ See mathworld.wolfram.com/q-PolygammaFunction.html.
– Batominovski
Aug 1 at 8:28
 |Â
show 3 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I wonder whether this series is calculable or not.
Attempt:
$S=1/8+1/88+1/888+....=dfrac18displaystylesum_k=0^inftydfrac1sum_n=0^k10^n$
where $$displaystylesum_n=0^k10^n=dfrac10^k+1-19$$
then
$S=dfrac98displaystylesum_k=0^inftydfrac110^k+1-1$
I have tried to calculate $displaystylesum_k=0^Kdfrac110^k+1-1$ for finite values but I failed.
What methods can we try?
sequences-and-series
asked Aug 1 at 8:05
user2312512851
1,134521
I wonder whether this series is calculable or not.
Attempt:
$S=1/8+1/88+1/888+....=dfrac18displaystylesum_k=0^inftydfrac1sum_n=0^k10^n$
where $$displaystylesum_n=0^k10^n=dfrac10^k+1-19$$
then
$S=dfrac98displaystylesum_k=0^inftydfrac110^k+1-1$
I have tried to calculate $displaystylesum_k=0^Kdfrac110^k+1-1$ for finite values but I failed.
What methods can we try?
sequences-and-series
asked Aug 1 at 8:05
user2312512851
1,134521
asked Aug 1 at 8:05
user2312512851
1,134521
asked Aug 1 at 8:05
user2312512851
1,134521
asked Aug 1 at 8:05
user2312512851
1,134521
1,134521
by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test
– user2312512851
Aug 1 at 8:07
I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series?
– user2312512851
Aug 1 at 8:11
WolframAlpha doesn't give an elementary answer, so chances are there isn't one. There are simplifications that WA misses, though, so we can't be certain.
– Arthur
Aug 1 at 8:23
This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function"
– Von Neumann
Aug 1 at 8:23
The sum $$s(z):=sum_n=0^infty,frac1z^n-1text for z>1$$ involves the $q$-polygamma function $psi^(0)_q$ with $q:=frac1z$. WolframAlpha gives $$s(z)=fraclnleft(fraczz-1right)-psi^(0)_frac1z(1)ln(z)text for z>1,.$$ See mathworld.wolfram.com/q-PolygammaFunction.html.
– Batominovski
Aug 1 at 8:28
 |Â
show 3 more comments
by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test
– user2312512851
Aug 1 at 8:07
I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series?
– user2312512851
Aug 1 at 8:11
WolframAlpha doesn't give an elementary answer, so chances are there isn't one. There are simplifications that WA misses, though, so we can't be certain.
– Arthur
Aug 1 at 8:23
This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function"
– Von Neumann
Aug 1 at 8:23
The sum $$s(z):=sum_n=0^infty,frac1z^n-1text for z>1$$ involves the $q$-polygamma function $psi^(0)_q$ with $q:=frac1z$. WolframAlpha gives $$s(z)=fraclnleft(fraczz-1right)-psi^(0)_frac1z(1)ln(z)text for z>1,.$$ See mathworld.wolfram.com/q-PolygammaFunction.html.
– Batominovski
Aug 1 at 8:28
by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test
– user2312512851
Aug 1 at 8:07
by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test
– user2312512851
Aug 1 at 8:07
I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series?
– user2312512851
Aug 1 at 8:11
I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series?
– user2312512851
Aug 1 at 8:11
WolframAlpha doesn't give an elementary answer, so chances are there isn't one. There are simplifications that WA misses, though, so we can't be certain.
– Arthur
Aug 1 at 8:23
WolframAlpha doesn't give an elementary answer, so chances are there isn't one. There are simplifications that WA misses, though, so we can't be certain.
– Arthur
Aug 1 at 8:23
This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function"
– Von Neumann
Aug 1 at 8:23
This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function"
– Von Neumann
Aug 1 at 8:23
The sum $$s(z):=sum_n=0^infty,frac1z^n-1text for z>1$$ involves the $q$-polygamma function $psi^(0)_q$ with $q:=frac1z$. WolframAlpha gives $$s(z)=fraclnleft(fraczz-1right)-psi^(0)_frac1z(1)ln(z)text for z>1,.$$ See mathworld.wolfram.com/q-PolygammaFunction.html.
– Batominovski
Aug 1 at 8:28
The sum $$s(z):=sum_n=0^infty,frac1z^n-1text for z>1$$ involves the $q$-polygamma function $psi^(0)_q$ with $q:=frac1z$. WolframAlpha gives $$s(z)=fraclnleft(fraczz-1right)-psi^(0)_frac1z(1)ln(z)text for z>1,.$$ See mathworld.wolfram.com/q-PolygammaFunction.html.
– Batominovski
Aug 1 at 8:28
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
4
down vote
$$displaystylesum_k=0^Kdfrac110^k+1-1=-1-K+fracpsi _10^(0)(K+2)-psi _10^(0)(1)log (10)$$ where appears the generalized PolyGamma function.
It is not surprising that you have problems with it.
Edit
If $K to infty$, the limit is
$$S=-frac9 16
log (10)left(2 psi _10^(0)(1)+log left(frac8110right)right)approx 0.13761477385452509205$$
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
add a comment |Â
up vote
3
down vote
$$frac18+frac188+frac1888+dotsm = frac18left( 1+frac111+frac1111+dotsm right) =
frac98sum_n=1^infty fracx^n1-x^n iff x= frac110$$
Where $sum_n=1^infty fracx^n1-x^n$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we have:
$$
S=frac98sum_n=1^infty fracx^n1-x^n =frac98left(fraclogleft(1-xright)+psi_x^(0)(1)log(x)right)
$$
That gives us:
$$S=frac98left(fraclogleft(frac910right)+psi_frac110^(0)(1)log(frac110)right)=0.137614773854525092047481887706797505400431...$$
Where $psi_x^(y)(z)$ is the generalized PolyGamma function
answered Aug 3 at 10:29
MarioE
486
add a comment |Â
up vote
2
down vote
I will extend the answer soon, but in the meanwhile I throw the rock and I give you the direct answer:
$$frac18 sum_k = 0^+infty frac110^k + 1 = frac18frac-log left(frac109right)+psi _frac110^(0)left(-fraci pi log (10)right)log (10)$$
Where $psi _frac110^(0)$ is the above mentioned PolyGamma generalized function.
answered Aug 1 at 8:24
Von Neumann
15.9k72443
1
The sign is wrong? Shouldn't it be $frac110^k-1$ instead?
– Batominovski
Aug 1 at 8:29
1
I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+.....
– user2312512851
Aug 1 at 14:10
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
$$displaystylesum_k=0^Kdfrac110^k+1-1=-1-K+fracpsi _10^(0)(K+2)-psi _10^(0)(1)log (10)$$ where appears the generalized PolyGamma function.
It is not surprising that you have problems with it.
Edit
If $K to infty$, the limit is
$$S=-frac9 16
log (10)left(2 psi _10^(0)(1)+log left(frac8110right)right)approx 0.13761477385452509205$$
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
add a comment |Â
up vote
4
down vote
$$displaystylesum_k=0^Kdfrac110^k+1-1=-1-K+fracpsi _10^(0)(K+2)-psi _10^(0)(1)log (10)$$ where appears the generalized PolyGamma function.
It is not surprising that you have problems with it.
Edit
If $K to infty$, the limit is
$$S=-frac9 16
log (10)left(2 psi _10^(0)(1)+log left(frac8110right)right)approx 0.13761477385452509205$$
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$$displaystylesum_k=0^Kdfrac110^k+1-1=-1-K+fracpsi _10^(0)(K+2)-psi _10^(0)(1)log (10)$$ where appears the generalized PolyGamma function.
It is not surprising that you have problems with it.
Edit
If $K to infty$, the limit is
$$S=-frac9 16
log (10)left(2 psi _10^(0)(1)+log left(frac8110right)right)approx 0.13761477385452509205$$
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
$$displaystylesum_k=0^Kdfrac110^k+1-1=-1-K+fracpsi _10^(0)(K+2)-psi _10^(0)(1)log (10)$$ where appears the generalized PolyGamma function.
It is not surprising that you have problems with it.
Edit
If $K to infty$, the limit is
$$S=-frac9 16
log (10)left(2 psi _10^(0)(1)+log left(frac8110right)right)approx 0.13761477385452509205$$
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
edited Aug 3 at 7:26
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
answered Aug 1 at 8:31
Claude Leibovici
111k1054126
111k1054126
add a comment |Â
add a comment |Â
up vote
3
down vote
$$frac18+frac188+frac1888+dotsm = frac18left( 1+frac111+frac1111+dotsm right) =
frac98sum_n=1^infty fracx^n1-x^n iff x= frac110$$
Where $sum_n=1^infty fracx^n1-x^n$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we have:
$$
S=frac98sum_n=1^infty fracx^n1-x^n =frac98left(fraclogleft(1-xright)+psi_x^(0)(1)log(x)right)
$$
That gives us:
$$S=frac98left(fraclogleft(frac910right)+psi_frac110^(0)(1)log(frac110)right)=0.137614773854525092047481887706797505400431...$$
Where $psi_x^(y)(z)$ is the generalized PolyGamma function
answered Aug 3 at 10:29
MarioE
486
add a comment |Â
up vote
3
down vote
$$frac18+frac188+frac1888+dotsm = frac18left( 1+frac111+frac1111+dotsm right) =
frac98sum_n=1^infty fracx^n1-x^n iff x= frac110$$
Where $sum_n=1^infty fracx^n1-x^n$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we have:
$$
S=frac98sum_n=1^infty fracx^n1-x^n =frac98left(fraclogleft(1-xright)+psi_x^(0)(1)log(x)right)
$$
That gives us:
$$S=frac98left(fraclogleft(frac910right)+psi_frac110^(0)(1)log(frac110)right)=0.137614773854525092047481887706797505400431...$$
Where $psi_x^(y)(z)$ is the generalized PolyGamma function
answered Aug 3 at 10:29
MarioE
486
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$frac18+frac188+frac1888+dotsm = frac18left( 1+frac111+frac1111+dotsm right) =
frac98sum_n=1^infty fracx^n1-x^n iff x= frac110$$
Where $sum_n=1^infty fracx^n1-x^n$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we have:
$$
S=frac98sum_n=1^infty fracx^n1-x^n =frac98left(fraclogleft(1-xright)+psi_x^(0)(1)log(x)right)
$$
That gives us:
$$S=frac98left(fraclogleft(frac910right)+psi_frac110^(0)(1)log(frac110)right)=0.137614773854525092047481887706797505400431...$$
Where $psi_x^(y)(z)$ is the generalized PolyGamma function
answered Aug 3 at 10:29
MarioE
486
$$frac18+frac188+frac1888+dotsm = frac18left( 1+frac111+frac1111+dotsm right) =
frac98sum_n=1^infty fracx^n1-x^n iff x= frac110$$
Where $sum_n=1^infty fracx^n1-x^n$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we have:
$$
S=frac98sum_n=1^infty fracx^n1-x^n =frac98left(fraclogleft(1-xright)+psi_x^(0)(1)log(x)right)
$$
That gives us:
$$S=frac98left(fraclogleft(frac910right)+psi_frac110^(0)(1)log(frac110)right)=0.137614773854525092047481887706797505400431...$$
Where $psi_x^(y)(z)$ is the generalized PolyGamma function
answered Aug 3 at 10:29
MarioE
486
answered Aug 3 at 10:29
MarioE
486
answered Aug 3 at 10:29
MarioE
486
answered Aug 3 at 10:29
MarioE
486
486
add a comment |Â
add a comment |Â
up vote
2
down vote
I will extend the answer soon, but in the meanwhile I throw the rock and I give you the direct answer:
$$frac18 sum_k = 0^+infty frac110^k + 1 = frac18frac-log left(frac109right)+psi _frac110^(0)left(-fraci pi log (10)right)log (10)$$
Where $psi _frac110^(0)$ is the above mentioned PolyGamma generalized function.
answered Aug 1 at 8:24
Von Neumann
15.9k72443
1
The sign is wrong? Shouldn't it be $frac110^k-1$ instead?
– Batominovski
Aug 1 at 8:29
1
I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+.....
– user2312512851
Aug 1 at 14:10
add a comment |Â
up vote
2
down vote
I will extend the answer soon, but in the meanwhile I throw the rock and I give you the direct answer:
$$frac18 sum_k = 0^+infty frac110^k + 1 = frac18frac-log left(frac109right)+psi _frac110^(0)left(-fraci pi log (10)right)log (10)$$
Where $psi _frac110^(0)$ is the above mentioned PolyGamma generalized function.
answered Aug 1 at 8:24
Von Neumann
15.9k72443
1
The sign is wrong? Shouldn't it be $frac110^k-1$ instead?
– Batominovski
Aug 1 at 8:29
1
I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+.....
– user2312512851
Aug 1 at 14:10
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I will extend the answer soon, but in the meanwhile I throw the rock and I give you the direct answer:
$$frac18 sum_k = 0^+infty frac110^k + 1 = frac18frac-log left(frac109right)+psi _frac110^(0)left(-fraci pi log (10)right)log (10)$$
Where $psi _frac110^(0)$ is the above mentioned PolyGamma generalized function.
answered Aug 1 at 8:24
Von Neumann
15.9k72443
I will extend the answer soon, but in the meanwhile I throw the rock and I give you the direct answer:
$$frac18 sum_k = 0^+infty frac110^k + 1 = frac18frac-log left(frac109right)+psi _frac110^(0)left(-fraci pi log (10)right)log (10)$$
Where $psi _frac110^(0)$ is the above mentioned PolyGamma generalized function.
answered Aug 1 at 8:24
Von Neumann
15.9k72443
answered Aug 1 at 8:24
Von Neumann
15.9k72443
answered Aug 1 at 8:24
Von Neumann
15.9k72443
answered Aug 1 at 8:24
Von Neumann
15.9k72443
15.9k72443
1
The sign is wrong? Shouldn't it be $frac110^k-1$ instead?
– Batominovski
Aug 1 at 8:29
1
I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+.....
– user2312512851
Aug 1 at 14:10
add a comment |Â
1
The sign is wrong? Shouldn't it be $frac110^k-1$ instead?
– Batominovski
Aug 1 at 8:29
1
I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+.....
– user2312512851
Aug 1 at 14:10
1
1
The sign is wrong? Shouldn't it be $frac110^k-1$ instead?
– Batominovski
Aug 1 at 8:29
The sign is wrong? Shouldn't it be $frac110^k-1$ instead?
– Batominovski
Aug 1 at 8:29
1
1
I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+.....
– user2312512851
Aug 1 at 14:10
I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+.....
– user2312512851
Aug 1 at 14:10
add a comment |Â
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by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test
– user2312512851
Aug 1 at 8:07
I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series?
– user2312512851
Aug 1 at 8:11
WolframAlpha doesn't give an elementary answer, so chances are there isn't one. There are simplifications that WA misses, though, so we can't be certain.
– Arthur
Aug 1 at 8:23
This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function"
– Von Neumann
Aug 1 at 8:23
The sum $$s(z):=sum_n=0^infty,frac1z^n-1text for z>1$$ involves the $q$-polygamma function $psi^(0)_q$ with $q:=frac1z$. WolframAlpha gives $$s(z)=fraclnleft(fraczz-1right)-psi^(0)_frac1z(1)ln(z)text for z>1,.$$ See mathworld.wolfram.com/q-PolygammaFunction.html.
– Batominovski
Aug 1 at 8:28