How do you calculate this infinite series $sum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1$?
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Evaluate the infinite series $$sum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1$$
On a calculator, for $p=0$ this tends to $1$. For $p=1$ it tends to $frac112$. For all values of $p$ I checked the result was a reciprocal of a natural number. How would you calculate this without a calculator for single values of $p$ and for any $p$ in general (natural numbers).
sequences-and-series
edited Sep 2 at 10:34
Robert Z
85.6k1055123
asked Sep 2 at 9:51
Nimish
807
add a comment |Â
up vote
2
down vote
favorite
Evaluate the infinite series $$sum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1$$
On a calculator, for $p=0$ this tends to $1$. For $p=1$ it tends to $frac112$. For all values of $p$ I checked the result was a reciprocal of a natural number. How would you calculate this without a calculator for single values of $p$ and for any $p$ in general (natural numbers).
sequences-and-series
edited Sep 2 at 10:34
Robert Z
85.6k1055123
asked Sep 2 at 9:51
Nimish
807
Where did you get this series from?
– Henning Makholm
Sep 2 at 10:03
A beginning would be to collect the $n$th powers to get $$frac45 sum_n=0^infty frac12n+1binom2n+1n-p (tfrac425)^n $$
– Henning Makholm
Sep 2 at 10:10
It originated from the trigonometric power formulas and was modified a bit for my use. mathworld.wolfram.com/TrigonometricPowerFormulas.html
– Nimish
Sep 2 at 10:10
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Evaluate the infinite series $$sum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1$$
On a calculator, for $p=0$ this tends to $1$. For $p=1$ it tends to $frac112$. For all values of $p$ I checked the result was a reciprocal of a natural number. How would you calculate this without a calculator for single values of $p$ and for any $p$ in general (natural numbers).
sequences-and-series
edited Sep 2 at 10:34
Robert Z
85.6k1055123
asked Sep 2 at 9:51
Nimish
807
Evaluate the infinite series $$sum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1$$
On a calculator, for $p=0$ this tends to $1$. For $p=1$ it tends to $frac112$. For all values of $p$ I checked the result was a reciprocal of a natural number. How would you calculate this without a calculator for single values of $p$ and for any $p$ in general (natural numbers).
sequences-and-series
sequences-and-series
edited Sep 2 at 10:34
Robert Z
85.6k1055123
asked Sep 2 at 9:51
Nimish
807
edited Sep 2 at 10:34
Robert Z
85.6k1055123
asked Sep 2 at 9:51
Nimish
807
edited Sep 2 at 10:34
Robert Z
85.6k1055123
edited Sep 2 at 10:34
Robert Z
85.6k1055123
edited Sep 2 at 10:34
Robert Z
85.6k1055123
85.6k1055123
asked Sep 2 at 9:51
Nimish
807
asked Sep 2 at 9:51
Nimish
807
asked Sep 2 at 9:51
Nimish
807
807
Where did you get this series from?
– Henning Makholm
Sep 2 at 10:03
A beginning would be to collect the $n$th powers to get $$frac45 sum_n=0^infty frac12n+1binom2n+1n-p (tfrac425)^n $$
– Henning Makholm
Sep 2 at 10:10
It originated from the trigonometric power formulas and was modified a bit for my use. mathworld.wolfram.com/TrigonometricPowerFormulas.html
– Nimish
Sep 2 at 10:10
add a comment |Â
Where did you get this series from?
– Henning Makholm
Sep 2 at 10:03
A beginning would be to collect the $n$th powers to get $$frac45 sum_n=0^infty frac12n+1binom2n+1n-p (tfrac425)^n $$
– Henning Makholm
Sep 2 at 10:10
It originated from the trigonometric power formulas and was modified a bit for my use. mathworld.wolfram.com/TrigonometricPowerFormulas.html
– Nimish
Sep 2 at 10:10
Where did you get this series from?
– Henning Makholm
Sep 2 at 10:03
Where did you get this series from?
– Henning Makholm
Sep 2 at 10:03
A beginning would be to collect the $n$th powers to get $$frac45 sum_n=0^infty frac12n+1binom2n+1n-p (tfrac425)^n $$
– Henning Makholm
Sep 2 at 10:10
A beginning would be to collect the $n$th powers to get $$frac45 sum_n=0^infty frac12n+1binom2n+1n-p (tfrac425)^n $$
– Henning Makholm
Sep 2 at 10:10
It originated from the trigonometric power formulas and was modified a bit for my use. mathworld.wolfram.com/TrigonometricPowerFormulas.html
– Nimish
Sep 2 at 10:10
It originated from the trigonometric power formulas and was modified a bit for my use. mathworld.wolfram.com/TrigonometricPowerFormulas.html
– Nimish
Sep 2 at 10:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Hint. Note that
$$beginalignsum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1&=
2sum_n=0^infty fracbinom2n+1n+p+1(2n+1)(2.5)^2n+1\
&=frac45sum_n=0^infty fracbinom2nn+px^nn+p+1
endalign$$
with $x=4/25$.
For $p=0$, recall that, for $|x|<1/4$,
$$sum_n=0^infty fracbinom2nnx^nn+1 = frac1-sqrt1-4x2x=frac21+sqrt1-4x$$
(see the g.f.of the Catalan Numbers). Extend this identity and show that for non-negative integer $p$,
$$sum_n=0^infty fracbinom2nn+px^nn+p+1 = left(frac21+sqrt1-4xright)^2p+1cdot fracx^p2p+1$$
answered Sep 2 at 10:10
Robert Z
85.6k1055123
Is there any way to extend or modify the Catalan Numbers' property to find values for any p?
– Nimish
Sep 2 at 10:39
@Nimish See my edit.
– Robert Z
Sep 2 at 10:51
Sorry about asking this so late but is there a more general extension for the Catalan Numbers' to even include types like this: $sum_n=0^infty fracbinom2nn(x^n)(2n)$? Thanks for the help.
– Nimish
Sep 10 at 4:08
@Nimish We have that $sum_n=1^infty fracbinom2nn(x^n)(2n)=lnleft(frac21+sqrt1-4xright)$.
– Robert Z
Sep 10 at 6:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint. Note that
$$beginalignsum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1&=
2sum_n=0^infty fracbinom2n+1n+p+1(2n+1)(2.5)^2n+1\
&=frac45sum_n=0^infty fracbinom2nn+px^nn+p+1
endalign$$
with $x=4/25$.
For $p=0$, recall that, for $|x|<1/4$,
$$sum_n=0^infty fracbinom2nnx^nn+1 = frac1-sqrt1-4x2x=frac21+sqrt1-4x$$
(see the g.f.of the Catalan Numbers). Extend this identity and show that for non-negative integer $p$,
$$sum_n=0^infty fracbinom2nn+px^nn+p+1 = left(frac21+sqrt1-4xright)^2p+1cdot fracx^p2p+1$$
answered Sep 2 at 10:10
Robert Z
85.6k1055123
Is there any way to extend or modify the Catalan Numbers' property to find values for any p?
– Nimish
Sep 2 at 10:39
@Nimish See my edit.
– Robert Z
Sep 2 at 10:51
Sorry about asking this so late but is there a more general extension for the Catalan Numbers' to even include types like this: $sum_n=0^infty fracbinom2nn(x^n)(2n)$? Thanks for the help.
– Nimish
Sep 10 at 4:08
@Nimish We have that $sum_n=1^infty fracbinom2nn(x^n)(2n)=lnleft(frac21+sqrt1-4xright)$.
– Robert Z
Sep 10 at 6:37
add a comment |Â
up vote
4
down vote
accepted
Hint. Note that
$$beginalignsum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1&=
2sum_n=0^infty fracbinom2n+1n+p+1(2n+1)(2.5)^2n+1\
&=frac45sum_n=0^infty fracbinom2nn+px^nn+p+1
endalign$$
with $x=4/25$.
For $p=0$, recall that, for $|x|<1/4$,
$$sum_n=0^infty fracbinom2nnx^nn+1 = frac1-sqrt1-4x2x=frac21+sqrt1-4x$$
(see the g.f.of the Catalan Numbers). Extend this identity and show that for non-negative integer $p$,
$$sum_n=0^infty fracbinom2nn+px^nn+p+1 = left(frac21+sqrt1-4xright)^2p+1cdot fracx^p2p+1$$
answered Sep 2 at 10:10
Robert Z
85.6k1055123
Is there any way to extend or modify the Catalan Numbers' property to find values for any p?
– Nimish
Sep 2 at 10:39
@Nimish See my edit.
– Robert Z
Sep 2 at 10:51
Sorry about asking this so late but is there a more general extension for the Catalan Numbers' to even include types like this: $sum_n=0^infty fracbinom2nn(x^n)(2n)$? Thanks for the help.
– Nimish
Sep 10 at 4:08
@Nimish We have that $sum_n=1^infty fracbinom2nn(x^n)(2n)=lnleft(frac21+sqrt1-4xright)$.
– Robert Z
Sep 10 at 6:37
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint. Note that
$$beginalignsum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1&=
2sum_n=0^infty fracbinom2n+1n+p+1(2n+1)(2.5)^2n+1\
&=frac45sum_n=0^infty fracbinom2nn+px^nn+p+1
endalign$$
with $x=4/25$.
For $p=0$, recall that, for $|x|<1/4$,
$$sum_n=0^infty fracbinom2nnx^nn+1 = frac1-sqrt1-4x2x=frac21+sqrt1-4x$$
(see the g.f.of the Catalan Numbers). Extend this identity and show that for non-negative integer $p$,
$$sum_n=0^infty fracbinom2nn+px^nn+p+1 = left(frac21+sqrt1-4xright)^2p+1cdot fracx^p2p+1$$
answered Sep 2 at 10:10
Robert Z
85.6k1055123
Hint. Note that
$$beginalignsum_n=0^infty fracbinom2n+1n-p(4^n)(2n+1)(1.25)^2n+1&=
2sum_n=0^infty fracbinom2n+1n+p+1(2n+1)(2.5)^2n+1\
&=frac45sum_n=0^infty fracbinom2nn+px^nn+p+1
endalign$$
with $x=4/25$.
For $p=0$, recall that, for $|x|<1/4$,
$$sum_n=0^infty fracbinom2nnx^nn+1 = frac1-sqrt1-4x2x=frac21+sqrt1-4x$$
(see the g.f.of the Catalan Numbers). Extend this identity and show that for non-negative integer $p$,
$$sum_n=0^infty fracbinom2nn+px^nn+p+1 = left(frac21+sqrt1-4xright)^2p+1cdot fracx^p2p+1$$
answered Sep 2 at 10:10
Robert Z
85.6k1055123
edited Sep 2 at 10:51
answered Sep 2 at 10:10
Robert Z
85.6k1055123
answered Sep 2 at 10:10
Robert Z
85.6k1055123
answered Sep 2 at 10:10
Robert Z
85.6k1055123
85.6k1055123
Is there any way to extend or modify the Catalan Numbers' property to find values for any p?
– Nimish
Sep 2 at 10:39
@Nimish See my edit.
– Robert Z
Sep 2 at 10:51
Sorry about asking this so late but is there a more general extension for the Catalan Numbers' to even include types like this: $sum_n=0^infty fracbinom2nn(x^n)(2n)$? Thanks for the help.
– Nimish
Sep 10 at 4:08
@Nimish We have that $sum_n=1^infty fracbinom2nn(x^n)(2n)=lnleft(frac21+sqrt1-4xright)$.
– Robert Z
Sep 10 at 6:37
add a comment |Â
Is there any way to extend or modify the Catalan Numbers' property to find values for any p?
– Nimish
Sep 2 at 10:39
@Nimish See my edit.
– Robert Z
Sep 2 at 10:51
Sorry about asking this so late but is there a more general extension for the Catalan Numbers' to even include types like this: $sum_n=0^infty fracbinom2nn(x^n)(2n)$? Thanks for the help.
– Nimish
Sep 10 at 4:08
@Nimish We have that $sum_n=1^infty fracbinom2nn(x^n)(2n)=lnleft(frac21+sqrt1-4xright)$.
– Robert Z
Sep 10 at 6:37
Is there any way to extend or modify the Catalan Numbers' property to find values for any p?
– Nimish
Sep 2 at 10:39
Is there any way to extend or modify the Catalan Numbers' property to find values for any p?
– Nimish
Sep 2 at 10:39
@Nimish See my edit.
– Robert Z
Sep 2 at 10:51
@Nimish See my edit.
– Robert Z
Sep 2 at 10:51
Sorry about asking this so late but is there a more general extension for the Catalan Numbers' to even include types like this: $sum_n=0^infty fracbinom2nn(x^n)(2n)$? Thanks for the help.
– Nimish
Sep 10 at 4:08
Sorry about asking this so late but is there a more general extension for the Catalan Numbers' to even include types like this: $sum_n=0^infty fracbinom2nn(x^n)(2n)$? Thanks for the help.
– Nimish
Sep 10 at 4:08
@Nimish We have that $sum_n=1^infty fracbinom2nn(x^n)(2n)=lnleft(frac21+sqrt1-4xright)$.
– Robert Z
Sep 10 at 6:37
@Nimish We have that $sum_n=1^infty fracbinom2nn(x^n)(2n)=lnleft(frac21+sqrt1-4xright)$.
– Robert Z
Sep 10 at 6:37
add a comment |Â
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Where did you get this series from?
– Henning Makholm
Sep 2 at 10:03
A beginning would be to collect the $n$th powers to get $$frac45 sum_n=0^infty frac12n+1binom2n+1n-p (tfrac425)^n $$
– Henning Makholm
Sep 2 at 10:10
It originated from the trigonometric power formulas and was modified a bit for my use. mathworld.wolfram.com/TrigonometricPowerFormulas.html
– Nimish
Sep 2 at 10:10