Series sum with factorial notation Announcing the arrival of Valued Associate #679: Cesar...
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Series sum with factorial notation
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Sum of a simple infinite seriesSum of infinite series with Irrational termsOn a series involving harmonic numbersHow do I calculate the sum of this infinite series? $sumlimits_{n=1}^{infty}frac{1}{4n^{2}-1}$Finding $lim_{nrightarrowinfty} sum^{n}_{k=0}left|frac{2picos(kpi(3-sqrt{5}))}{n}right|$Convergence or not of infinite series: $sum^{infty}_{n=1}frac{n}{1+n^2}$Sum of series $frac{ 4}{10}+frac{4cdot 7}{10cdot 20}+frac{4cdot 7 cdot 10}{10cdot 20 cdot 30}+cdots cdots$Sum of $ frac{1}{1cdot 3}+frac{1}{1cdot 3cdot 5}+frac{1}{1cdot 3cdot 5cdot 7}+ cdots$Infinite series containing positive and negative termsInfinite series sum which have infinite terms
$begingroup$
The sum of series
$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $
what i try:
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$
How do i solve it Help me please
sequences-and-series
asked Mar 23 at 11:59
jackyjacky
1,349816
$endgroup$
add a comment |
$begingroup$
The sum of series
$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $
what i try:
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$
How do i solve it Help me please
sequences-and-series
asked Mar 23 at 11:59
jackyjacky
1,349816
$endgroup$
1
$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49
add a comment |
$begingroup$
The sum of series
$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $
what i try:
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$
How do i solve it Help me please
sequences-and-series
asked Mar 23 at 11:59
jackyjacky
1,349816
$endgroup$
The sum of series
$displaystyle 1+frac{1}{1!}cdot frac{1}{4}+frac{1cdot 3}{2!}cdot frac{1}{4^2}+frac{1cdot 3 cdot 5}{3!}cdot frac{1}{4^3}+cdots $
what i try:
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{1cdot 3cdot cdots (2r-1)}{r!cdot 4^r}$$
$$1+lim_{nrightarrow infty}sum^{n}_{r=1}frac{(2r)!}{r!prod^{n}_{r=1}(2r)!}$$
How do i solve it Help me please
sequences-and-series
sequences-and-series
asked Mar 23 at 11:59
jackyjacky
1,349816
asked Mar 23 at 11:59
jackyjacky
1,349816
asked Mar 23 at 11:59
jackyjacky
1,349816
asked Mar 23 at 11:59
jackyjacky
1,349816
asked Mar 23 at 11:59
jackyjacky
1,349816
1,349816
1
$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49
add a comment |
1
$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49
1
1
$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49
$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The series under consideration is
begin{align*}
color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
&=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
&=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
&=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
&=frac{1}{sqrt{1-frac{1}{2}}}\
&,,color{blue}{=sqrt{2}}
end{align*}
In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
{1cdot 3 over 2!}, {1 over 4^{2}} +
{1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
n!, 4^{n}}
\[5mm] = &
1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
n!, 4^{n}} =
1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
,pars{1 over 2}^{n}
\[5mm] = &
1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
,pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
,pars{1 over 2}^{n}
\[5mm] = &
sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
bracks{1 + pars{-1 over 2}}^{-1/2}
\[5mm] = & bbx{root{2}} approx 1.4142
end{align}
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
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oldest
votes
$begingroup$
The series under consideration is
begin{align*}
color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
&=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
&=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
&=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
&=frac{1}{sqrt{1-frac{1}{2}}}\
&,,color{blue}{=sqrt{2}}
end{align*}
In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
add a comment |
$begingroup$
The series under consideration is
begin{align*}
color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
&=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
&=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
&=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
&=frac{1}{sqrt{1-frac{1}{2}}}\
&,,color{blue}{=sqrt{2}}
end{align*}
In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
add a comment |
$begingroup$
The series under consideration is
begin{align*}
color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
&=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
&=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
&=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
&=frac{1}{sqrt{1-frac{1}{2}}}\
&,,color{blue}{=sqrt{2}}
end{align*}
In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
The series under consideration is
begin{align*}
color{blue}{1+sum_{r=1}^inftyfrac{(2r-1)!!}{r!4^r}}&=1+sum_{r=1}^inftyfrac{(2r)!}{r!4^r(2r)!!}\
&=1+sum_{r=1}^inftyfrac{(2r)!}{r!r!}left(frac{1}{8}right)^r\
&=sum_{r=0}^inftybinom{2r}{r}left(frac{1}{8}right)^r\
&=left.frac{1}{sqrt{1-4z}}right|_{z=1/8}tag{1}\
&=frac{1}{sqrt{1-frac{1}{2}}}\
&,,color{blue}{=sqrt{2}}
end{align*}
In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
edited Mar 23 at 16:13
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
answered Mar 23 at 15:59
Markus ScheuerMarkus Scheuer
64.4k460152
64.4k460152
add a comment |
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
{1cdot 3 over 2!}, {1 over 4^{2}} +
{1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
n!, 4^{n}}
\[5mm] = &
1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
n!, 4^{n}} =
1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
,pars{1 over 2}^{n}
\[5mm] = &
1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
,pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
,pars{1 over 2}^{n}
\[5mm] = &
sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
bracks{1 + pars{-1 over 2}}^{-1/2}
\[5mm] = & bbx{root{2}} approx 1.4142
end{align}
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
{1cdot 3 over 2!}, {1 over 4^{2}} +
{1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
n!, 4^{n}}
\[5mm] = &
1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
n!, 4^{n}} =
1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
,pars{1 over 2}^{n}
\[5mm] = &
1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
,pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
,pars{1 over 2}^{n}
\[5mm] = &
sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
bracks{1 + pars{-1 over 2}}^{-1/2}
\[5mm] = & bbx{root{2}} approx 1.4142
end{align}
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
{1cdot 3 over 2!}, {1 over 4^{2}} +
{1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
n!, 4^{n}}
\[5mm] = &
1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
n!, 4^{n}} =
1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
,pars{1 over 2}^{n}
\[5mm] = &
1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
,pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
,pars{1 over 2}^{n}
\[5mm] = &
sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
bracks{1 + pars{-1 over 2}}^{-1/2}
\[5mm] = & bbx{root{2}} approx 1.4142
end{align}
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 + {1 over 1!},{1 over 4} +
{1cdot 3 over 2!}, {1 over 4^{2}} +
{1cdot 3 cdot 5 over 3!},{1 over 4^{3}} + cdots} =
1 + sum_{n = 1}^{infty}{prod_{k = 0}^{n - 1}pars{2k + 1} over
n!, 4^{n}}
\[5mm] = &
1 + sum_{n = 1}^{infty}{2^{n}prod_{k = 0}^{n - 1}pars{k + 1/2} over
n!, 4^{n}} =
1 + sum_{n = 1}^{infty}{pars{1/2}^{overline{n}} over n!}
,pars{1 over 2}^{n}
\[5mm] = &
1 + sum_{n = 1}^{infty}{Gammapars{1/2 + n}/Gammapars{1/2} over n!}
,pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{pars{n - 1/2}! over n!pars{-1/2}!}
,pars{1 over 2}^{n}
\[5mm] = &
sum_{n = 0}^{infty}{n - 1/2 choose n},pars{1 over 2}^{n} =
sum_{n = 0}^{infty}{-1/2 choose n},pars{-,{1 over 2}}^{n} =
bracks{1 + pars{-1 over 2}}^{-1/2}
\[5mm] = & bbx{root{2}} approx 1.4142
end{align}
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
answered Mar 27 at 20:24
Felix MarinFelix Marin
68.9k7110147
68.9k7110147
add a comment |
add a comment |
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$begingroup$
what is the question...?
$endgroup$
– Jneven
Mar 23 at 12:49