Infinite series containing positive and negative terms
$begingroup$
Calculation of sum of
$displaystyle 1-frac{1}{5}+frac{1cdot 4}{5cdot 10}-frac{1cdot 4 cdot 7}{5cdot 10 cdot 15}+cdots cdots $
The series is not in arithmetic series or in arithmetic geometric series
I did not understand how to solve such type of problem
help me to solve it plaese
sequences-and-series
asked Dec 5 '18 at 15:45
jackyjacky
842615
$endgroup$
add a comment |
$begingroup$
Calculation of sum of
$displaystyle 1-frac{1}{5}+frac{1cdot 4}{5cdot 10}-frac{1cdot 4 cdot 7}{5cdot 10 cdot 15}+cdots cdots $
The series is not in arithmetic series or in arithmetic geometric series
I did not understand how to solve such type of problem
help me to solve it plaese
sequences-and-series
asked Dec 5 '18 at 15:45
jackyjacky
842615
$endgroup$
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 15:50
add a comment |
$begingroup$
Calculation of sum of
$displaystyle 1-frac{1}{5}+frac{1cdot 4}{5cdot 10}-frac{1cdot 4 cdot 7}{5cdot 10 cdot 15}+cdots cdots $
The series is not in arithmetic series or in arithmetic geometric series
I did not understand how to solve such type of problem
help me to solve it plaese
sequences-and-series
asked Dec 5 '18 at 15:45
jackyjacky
842615
$endgroup$
Calculation of sum of
$displaystyle 1-frac{1}{5}+frac{1cdot 4}{5cdot 10}-frac{1cdot 4 cdot 7}{5cdot 10 cdot 15}+cdots cdots $
The series is not in arithmetic series or in arithmetic geometric series
I did not understand how to solve such type of problem
help me to solve it plaese
sequences-and-series
sequences-and-series
asked Dec 5 '18 at 15:45
jackyjacky
842615
asked Dec 5 '18 at 15:45
jackyjacky
842615
asked Dec 5 '18 at 15:45
jackyjacky
842615
asked Dec 5 '18 at 15:45
jackyjacky
842615
asked Dec 5 '18 at 15:45
jackyjacky
842615
842615
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 15:50
add a comment |
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 15:50
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 15:50
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 15:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 - sum_{n = 0}^{infty}pars{-1}^{n},
{prod_{k = 0}^{n}pars{3k + 1} over prod_{k = 0}^{n}pars{5k + 5}}}
\[5mm] = &
1 - sum_{n = 0}^{infty}pars{-1}^{n},
{3^{n + 1}prod_{k = 0}^{n}pars{k + 1/3} over
5^{n + 1}prod_{k = 0}^{n}pars{k + 1}}
\[5mm] = &
1 + sum_{n = 0}^{infty}pars{-,{3 over 5}}^{n + 1},
{pars{1/3}^{overline{n + 1}} over pars{n + 1}!}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{Gammapars{1/3 + n + 1}/Gammapars{1/3} over pars{n + 1}!},
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{pars{n + 1/3}! over pars{n + 1}!pars{-2/3}!},
pars{-,{3 over 5}}^{n + 1} =
1 + sum_{n = 0}^{infty}{n + 1/3 choose n + 1}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
bracks{{-1/3 choose n + 1}pars{-1}^{n + 1}}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
sum_{n = 0}^{infty}
{-1/3 choose n}pars{3 over 5}^{n} = pars{1 + {3 over 5}}^{-1/3} =
bbx{5^{1/3} over 2} approx 0.8550
end{align}
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
$endgroup$
$begingroup$
cool the passage through binomial (+1)
$endgroup$
– G Cab
Dec 5 '18 at 22:13
$begingroup$
@GCab Thanks. Binomials are always quite fine.
$endgroup$
– Felix Marin
Dec 5 '18 at 22:31
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 - sum_{n = 0}^{infty}pars{-1}^{n},
{prod_{k = 0}^{n}pars{3k + 1} over prod_{k = 0}^{n}pars{5k + 5}}}
\[5mm] = &
1 - sum_{n = 0}^{infty}pars{-1}^{n},
{3^{n + 1}prod_{k = 0}^{n}pars{k + 1/3} over
5^{n + 1}prod_{k = 0}^{n}pars{k + 1}}
\[5mm] = &
1 + sum_{n = 0}^{infty}pars{-,{3 over 5}}^{n + 1},
{pars{1/3}^{overline{n + 1}} over pars{n + 1}!}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{Gammapars{1/3 + n + 1}/Gammapars{1/3} over pars{n + 1}!},
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{pars{n + 1/3}! over pars{n + 1}!pars{-2/3}!},
pars{-,{3 over 5}}^{n + 1} =
1 + sum_{n = 0}^{infty}{n + 1/3 choose n + 1}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
bracks{{-1/3 choose n + 1}pars{-1}^{n + 1}}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
sum_{n = 0}^{infty}
{-1/3 choose n}pars{3 over 5}^{n} = pars{1 + {3 over 5}}^{-1/3} =
bbx{5^{1/3} over 2} approx 0.8550
end{align}
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
$endgroup$
$begingroup$
cool the passage through binomial (+1)
$endgroup$
– G Cab
Dec 5 '18 at 22:13
$begingroup$
@GCab Thanks. Binomials are always quite fine.
$endgroup$
– Felix Marin
Dec 5 '18 at 22:31
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 - sum_{n = 0}^{infty}pars{-1}^{n},
{prod_{k = 0}^{n}pars{3k + 1} over prod_{k = 0}^{n}pars{5k + 5}}}
\[5mm] = &
1 - sum_{n = 0}^{infty}pars{-1}^{n},
{3^{n + 1}prod_{k = 0}^{n}pars{k + 1/3} over
5^{n + 1}prod_{k = 0}^{n}pars{k + 1}}
\[5mm] = &
1 + sum_{n = 0}^{infty}pars{-,{3 over 5}}^{n + 1},
{pars{1/3}^{overline{n + 1}} over pars{n + 1}!}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{Gammapars{1/3 + n + 1}/Gammapars{1/3} over pars{n + 1}!},
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{pars{n + 1/3}! over pars{n + 1}!pars{-2/3}!},
pars{-,{3 over 5}}^{n + 1} =
1 + sum_{n = 0}^{infty}{n + 1/3 choose n + 1}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
bracks{{-1/3 choose n + 1}pars{-1}^{n + 1}}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
sum_{n = 0}^{infty}
{-1/3 choose n}pars{3 over 5}^{n} = pars{1 + {3 over 5}}^{-1/3} =
bbx{5^{1/3} over 2} approx 0.8550
end{align}
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
$endgroup$
$begingroup$
cool the passage through binomial (+1)
$endgroup$
– G Cab
Dec 5 '18 at 22:13
$begingroup$
@GCab Thanks. Binomials are always quite fine.
$endgroup$
– Felix Marin
Dec 5 '18 at 22:31
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 - sum_{n = 0}^{infty}pars{-1}^{n},
{prod_{k = 0}^{n}pars{3k + 1} over prod_{k = 0}^{n}pars{5k + 5}}}
\[5mm] = &
1 - sum_{n = 0}^{infty}pars{-1}^{n},
{3^{n + 1}prod_{k = 0}^{n}pars{k + 1/3} over
5^{n + 1}prod_{k = 0}^{n}pars{k + 1}}
\[5mm] = &
1 + sum_{n = 0}^{infty}pars{-,{3 over 5}}^{n + 1},
{pars{1/3}^{overline{n + 1}} over pars{n + 1}!}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{Gammapars{1/3 + n + 1}/Gammapars{1/3} over pars{n + 1}!},
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{pars{n + 1/3}! over pars{n + 1}!pars{-2/3}!},
pars{-,{3 over 5}}^{n + 1} =
1 + sum_{n = 0}^{infty}{n + 1/3 choose n + 1}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
bracks{{-1/3 choose n + 1}pars{-1}^{n + 1}}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
sum_{n = 0}^{infty}
{-1/3 choose n}pars{3 over 5}^{n} = pars{1 + {3 over 5}}^{-1/3} =
bbx{5^{1/3} over 2} approx 0.8550
end{align}
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{1 - sum_{n = 0}^{infty}pars{-1}^{n},
{prod_{k = 0}^{n}pars{3k + 1} over prod_{k = 0}^{n}pars{5k + 5}}}
\[5mm] = &
1 - sum_{n = 0}^{infty}pars{-1}^{n},
{3^{n + 1}prod_{k = 0}^{n}pars{k + 1/3} over
5^{n + 1}prod_{k = 0}^{n}pars{k + 1}}
\[5mm] = &
1 + sum_{n = 0}^{infty}pars{-,{3 over 5}}^{n + 1},
{pars{1/3}^{overline{n + 1}} over pars{n + 1}!}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{Gammapars{1/3 + n + 1}/Gammapars{1/3} over pars{n + 1}!},
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
{pars{n + 1/3}! over pars{n + 1}!pars{-2/3}!},
pars{-,{3 over 5}}^{n + 1} =
1 + sum_{n = 0}^{infty}{n + 1/3 choose n + 1}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
1 + sum_{n = 0}^{infty}
bracks{{-1/3 choose n + 1}pars{-1}^{n + 1}}
pars{-,{3 over 5}}^{n + 1}
\[5mm] = &
sum_{n = 0}^{infty}
{-1/3 choose n}pars{3 over 5}^{n} = pars{1 + {3 over 5}}^{-1/3} =
bbx{5^{1/3} over 2} approx 0.8550
end{align}
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
answered Dec 5 '18 at 21:57
Felix MarinFelix Marin
68.2k7109144
68.2k7109144
$begingroup$
cool the passage through binomial (+1)
$endgroup$
– G Cab
Dec 5 '18 at 22:13
$begingroup$
@GCab Thanks. Binomials are always quite fine.
$endgroup$
– Felix Marin
Dec 5 '18 at 22:31
add a comment |
$begingroup$
cool the passage through binomial (+1)
$endgroup$
– G Cab
Dec 5 '18 at 22:13
$begingroup$
@GCab Thanks. Binomials are always quite fine.
$endgroup$
– Felix Marin
Dec 5 '18 at 22:31
$begingroup$
cool the passage through binomial (+1)
$endgroup$
– G Cab
Dec 5 '18 at 22:13
$begingroup$
cool the passage through binomial (+1)
$endgroup$
– G Cab
Dec 5 '18 at 22:13
$begingroup$
@GCab Thanks. Binomials are always quite fine.
$endgroup$
– Felix Marin
Dec 5 '18 at 22:31
$begingroup$
@GCab Thanks. Binomials are always quite fine.
$endgroup$
– Felix Marin
Dec 5 '18 at 22:31
add a comment |
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– lab bhattacharjee
Dec 5 '18 at 15:50