Calculate $limlimits_{nrightarrow infty }sum_{k=1}^{n}frac{6}{k(k+1)(k+3)}$












6












$begingroup$


$$lim_{nrightarrow infty }sum_{k=1}^{n}frac{6}{k(k+1)(k+3)}$$



I tried to simplify the sum and I got $frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}$ but I can't use this to simplify the terms.Also,I tried to amplify with $k+2$ and I got $$frac{(k+3)-1}{k(k+1)(k+2)(k+3)}=frac{(k+3)}{k(k+1)(k+2)(k+3)}-frac{1}{k(k+1)(k+2)(k+3)}$$ but the terms also don't simplify.






calculus limits







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  • $begingroup$
    It looks like your first simplified version can get you the answer. Just start writing out the first few terms and see what cancels out.
    $endgroup$
    – Peter Foreman
    Jan 27 at 8:53
















6












$begingroup$


$$lim_{nrightarrow infty }sum_{k=1}^{n}frac{6}{k(k+1)(k+3)}$$



I tried to simplify the sum and I got $frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}$ but I can't use this to simplify the terms.Also,I tried to amplify with $k+2$ and I got $$frac{(k+3)-1}{k(k+1)(k+2)(k+3)}=frac{(k+3)}{k(k+1)(k+2)(k+3)}-frac{1}{k(k+1)(k+2)(k+3)}$$ but the terms also don't simplify.






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    It looks like your first simplified version can get you the answer. Just start writing out the first few terms and see what cancels out.
    $endgroup$
    – Peter Foreman
    Jan 27 at 8:53














6












6








6


1



$begingroup$


$$lim_{nrightarrow infty }sum_{k=1}^{n}frac{6}{k(k+1)(k+3)}$$



I tried to simplify the sum and I got $frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}$ but I can't use this to simplify the terms.Also,I tried to amplify with $k+2$ and I got $$frac{(k+3)-1}{k(k+1)(k+2)(k+3)}=frac{(k+3)}{k(k+1)(k+2)(k+3)}-frac{1}{k(k+1)(k+2)(k+3)}$$ but the terms also don't simplify.






calculus limits







share|cite|improve this question











$endgroup$




$$lim_{nrightarrow infty }sum_{k=1}^{n}frac{6}{k(k+1)(k+3)}$$



I tried to simplify the sum and I got $frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}$ but I can't use this to simplify the terms.Also,I tried to amplify with $k+2$ and I got $$frac{(k+3)-1}{k(k+1)(k+2)(k+3)}=frac{(k+3)}{k(k+1)(k+2)(k+3)}-frac{1}{k(k+1)(k+2)(k+3)}$$ but the terms also don't simplify.






calculus limits




calculus limits






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edited Jan 27 at 8:52









DonAntonio

180k1494233




180k1494233










asked Jan 27 at 8:45









DaniVajaDaniVaja

977




977












  • $begingroup$
    It looks like your first simplified version can get you the answer. Just start writing out the first few terms and see what cancels out.
    $endgroup$
    – Peter Foreman
    Jan 27 at 8:53


















  • $begingroup$
    It looks like your first simplified version can get you the answer. Just start writing out the first few terms and see what cancels out.
    $endgroup$
    – Peter Foreman
    Jan 27 at 8:53
















$begingroup$
It looks like your first simplified version can get you the answer. Just start writing out the first few terms and see what cancels out.
$endgroup$
– Peter Foreman
Jan 27 at 8:53




$begingroup$
It looks like your first simplified version can get you the answer. Just start writing out the first few terms and see what cancels out.
$endgroup$
– Peter Foreman
Jan 27 at 8:53










2 Answers
2






active

oldest

votes


















8












$begingroup$

Hint



$$frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}=frac{2}{k}-frac{2}{k+1}-frac{1}{k+1}+frac{1}{k+3}=2left(frac{1}{k}-frac{1}{k+1} right)-left(frac{1}{k+1}-frac{1}{k+3} right)$$






share|cite|improve this answer









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  • $begingroup$
    I did it!Thanks a lot!
    $endgroup$
    – DaniVaja
    Jan 27 at 9:08






  • 1




    $begingroup$
    @DaniVaja. Good to know and you are very welcome ! Remember the small trick since you will need it more than once. Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 27 at 9:10










  • $begingroup$
    I don’t understand what the small trick does though
    $endgroup$
    – Chase Ryan Taylor
    Jan 28 at 2:32



















0












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{k = 1}^{infty}{6 over kpars{k + 1}pars{k+ 3}}} =
6sum_{k = 1}^{infty}2int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
{1 over pars{k + u_{1} + 2u_{2}}^{3}}
\[5mm] = &
12int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
underbrace{sum_{k = 1}^{infty}{1 over pars{k + u_{1} + 2u_{2}}^{3}}}
_{ds{-Psi''pars{1 + u_{1} + 2u_{2}}/2}}qquad
pars{~Psi: Digamma Function~}
\[5mm] = &
-3int_{0}^{1}dd u_{1}bracks{Psi, 'pars{1 + 3u_{1}} -
Psi, 'pars{1 + u_{1}}}
\[5mm] = &
-3bracks{{1 over 3},Psipars{1 + 3u_{1}} - Psipars{1 + u_{1}}}_{ 0}^{ 1} =
-
Psipars{4} + 3Psipars{2} + Psipars{1} - 3Psipars{1}
\[5mm] = &
3 underbrace{bracks{Psipars{2} - Psipars{1}}}_{ds{= 1}} -
underbrace{bracks{Psipars{4} - Psipars{1}}}
_{ds{= {11 over 6}}} = bbx{7 over 6}
end{align}




Note that $quadleft{begin{array}{rclcl}
ds{Psipars{2}} & ds{=} & ds{Psipars{1} + {1 over 1}} &&
\[1mm]
ds{Psipars{3}} & ds{=} & ds{Psipars{2} + {1 over 2}}
& ds{=} & ds{Psipars{1} + {3 over 2}}
\[1mm]
ds{Psipars{4}} & ds{=} & ds{Psipars{3} + {1 over 3}}
& ds{=} & ds{Psipars{1} + {11 over 6}}
end{array}right.$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ChaseRyanTaylor Yes. That's quite true. Indeed, I was testing Feynman Parametrization.
    $endgroup$
    – Felix Marin
    Jan 28 at 2:20











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

Hint



$$frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}=frac{2}{k}-frac{2}{k+1}-frac{1}{k+1}+frac{1}{k+3}=2left(frac{1}{k}-frac{1}{k+1} right)-left(frac{1}{k+1}-frac{1}{k+3} right)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did it!Thanks a lot!
    $endgroup$
    – DaniVaja
    Jan 27 at 9:08






  • 1




    $begingroup$
    @DaniVaja. Good to know and you are very welcome ! Remember the small trick since you will need it more than once. Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 27 at 9:10










  • $begingroup$
    I don’t understand what the small trick does though
    $endgroup$
    – Chase Ryan Taylor
    Jan 28 at 2:32
















8












$begingroup$

Hint



$$frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}=frac{2}{k}-frac{2}{k+1}-frac{1}{k+1}+frac{1}{k+3}=2left(frac{1}{k}-frac{1}{k+1} right)-left(frac{1}{k+1}-frac{1}{k+3} right)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did it!Thanks a lot!
    $endgroup$
    – DaniVaja
    Jan 27 at 9:08






  • 1




    $begingroup$
    @DaniVaja. Good to know and you are very welcome ! Remember the small trick since you will need it more than once. Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 27 at 9:10










  • $begingroup$
    I don’t understand what the small trick does though
    $endgroup$
    – Chase Ryan Taylor
    Jan 28 at 2:32














8












8








8





$begingroup$

Hint



$$frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}=frac{2}{k}-frac{2}{k+1}-frac{1}{k+1}+frac{1}{k+3}=2left(frac{1}{k}-frac{1}{k+1} right)-left(frac{1}{k+1}-frac{1}{k+3} right)$$






share|cite|improve this answer









$endgroup$



Hint



$$frac{2}{k}-frac{3}{k+1}+frac{1}{k+3}=frac{2}{k}-frac{2}{k+1}-frac{1}{k+1}+frac{1}{k+3}=2left(frac{1}{k}-frac{1}{k+1} right)-left(frac{1}{k+1}-frac{1}{k+3} right)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 27 at 8:58









Claude LeiboviciClaude Leibovici

125k1158135




125k1158135












  • $begingroup$
    I did it!Thanks a lot!
    $endgroup$
    – DaniVaja
    Jan 27 at 9:08






  • 1




    $begingroup$
    @DaniVaja. Good to know and you are very welcome ! Remember the small trick since you will need it more than once. Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 27 at 9:10










  • $begingroup$
    I don’t understand what the small trick does though
    $endgroup$
    – Chase Ryan Taylor
    Jan 28 at 2:32


















  • $begingroup$
    I did it!Thanks a lot!
    $endgroup$
    – DaniVaja
    Jan 27 at 9:08






  • 1




    $begingroup$
    @DaniVaja. Good to know and you are very welcome ! Remember the small trick since you will need it more than once. Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 27 at 9:10










  • $begingroup$
    I don’t understand what the small trick does though
    $endgroup$
    – Chase Ryan Taylor
    Jan 28 at 2:32
















$begingroup$
I did it!Thanks a lot!
$endgroup$
– DaniVaja
Jan 27 at 9:08




$begingroup$
I did it!Thanks a lot!
$endgroup$
– DaniVaja
Jan 27 at 9:08




1




1




$begingroup$
@DaniVaja. Good to know and you are very welcome ! Remember the small trick since you will need it more than once. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 9:10




$begingroup$
@DaniVaja. Good to know and you are very welcome ! Remember the small trick since you will need it more than once. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 9:10












$begingroup$
I don’t understand what the small trick does though
$endgroup$
– Chase Ryan Taylor
Jan 28 at 2:32




$begingroup$
I don’t understand what the small trick does though
$endgroup$
– Chase Ryan Taylor
Jan 28 at 2:32











0












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{k = 1}^{infty}{6 over kpars{k + 1}pars{k+ 3}}} =
6sum_{k = 1}^{infty}2int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
{1 over pars{k + u_{1} + 2u_{2}}^{3}}
\[5mm] = &
12int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
underbrace{sum_{k = 1}^{infty}{1 over pars{k + u_{1} + 2u_{2}}^{3}}}
_{ds{-Psi''pars{1 + u_{1} + 2u_{2}}/2}}qquad
pars{~Psi: Digamma Function~}
\[5mm] = &
-3int_{0}^{1}dd u_{1}bracks{Psi, 'pars{1 + 3u_{1}} -
Psi, 'pars{1 + u_{1}}}
\[5mm] = &
-3bracks{{1 over 3},Psipars{1 + 3u_{1}} - Psipars{1 + u_{1}}}_{ 0}^{ 1} =
-
Psipars{4} + 3Psipars{2} + Psipars{1} - 3Psipars{1}
\[5mm] = &
3 underbrace{bracks{Psipars{2} - Psipars{1}}}_{ds{= 1}} -
underbrace{bracks{Psipars{4} - Psipars{1}}}
_{ds{= {11 over 6}}} = bbx{7 over 6}
end{align}




Note that $quadleft{begin{array}{rclcl}
ds{Psipars{2}} & ds{=} & ds{Psipars{1} + {1 over 1}} &&
\[1mm]
ds{Psipars{3}} & ds{=} & ds{Psipars{2} + {1 over 2}}
& ds{=} & ds{Psipars{1} + {3 over 2}}
\[1mm]
ds{Psipars{4}} & ds{=} & ds{Psipars{3} + {1 over 3}}
& ds{=} & ds{Psipars{1} + {11 over 6}}
end{array}right.$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ChaseRyanTaylor Yes. That's quite true. Indeed, I was testing Feynman Parametrization.
    $endgroup$
    – Felix Marin
    Jan 28 at 2:20
















0












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{k = 1}^{infty}{6 over kpars{k + 1}pars{k+ 3}}} =
6sum_{k = 1}^{infty}2int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
{1 over pars{k + u_{1} + 2u_{2}}^{3}}
\[5mm] = &
12int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
underbrace{sum_{k = 1}^{infty}{1 over pars{k + u_{1} + 2u_{2}}^{3}}}
_{ds{-Psi''pars{1 + u_{1} + 2u_{2}}/2}}qquad
pars{~Psi: Digamma Function~}
\[5mm] = &
-3int_{0}^{1}dd u_{1}bracks{Psi, 'pars{1 + 3u_{1}} -
Psi, 'pars{1 + u_{1}}}
\[5mm] = &
-3bracks{{1 over 3},Psipars{1 + 3u_{1}} - Psipars{1 + u_{1}}}_{ 0}^{ 1} =
-
Psipars{4} + 3Psipars{2} + Psipars{1} - 3Psipars{1}
\[5mm] = &
3 underbrace{bracks{Psipars{2} - Psipars{1}}}_{ds{= 1}} -
underbrace{bracks{Psipars{4} - Psipars{1}}}
_{ds{= {11 over 6}}} = bbx{7 over 6}
end{align}




Note that $quadleft{begin{array}{rclcl}
ds{Psipars{2}} & ds{=} & ds{Psipars{1} + {1 over 1}} &&
\[1mm]
ds{Psipars{3}} & ds{=} & ds{Psipars{2} + {1 over 2}}
& ds{=} & ds{Psipars{1} + {3 over 2}}
\[1mm]
ds{Psipars{4}} & ds{=} & ds{Psipars{3} + {1 over 3}}
& ds{=} & ds{Psipars{1} + {11 over 6}}
end{array}right.$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ChaseRyanTaylor Yes. That's quite true. Indeed, I was testing Feynman Parametrization.
    $endgroup$
    – Felix Marin
    Jan 28 at 2:20














0












0








0





$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
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newcommand{mrm}[1]{mathrm{#1}}
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newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{k = 1}^{infty}{6 over kpars{k + 1}pars{k+ 3}}} =
6sum_{k = 1}^{infty}2int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
{1 over pars{k + u_{1} + 2u_{2}}^{3}}
\[5mm] = &
12int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
underbrace{sum_{k = 1}^{infty}{1 over pars{k + u_{1} + 2u_{2}}^{3}}}
_{ds{-Psi''pars{1 + u_{1} + 2u_{2}}/2}}qquad
pars{~Psi: Digamma Function~}
\[5mm] = &
-3int_{0}^{1}dd u_{1}bracks{Psi, 'pars{1 + 3u_{1}} -
Psi, 'pars{1 + u_{1}}}
\[5mm] = &
-3bracks{{1 over 3},Psipars{1 + 3u_{1}} - Psipars{1 + u_{1}}}_{ 0}^{ 1} =
-
Psipars{4} + 3Psipars{2} + Psipars{1} - 3Psipars{1}
\[5mm] = &
3 underbrace{bracks{Psipars{2} - Psipars{1}}}_{ds{= 1}} -
underbrace{bracks{Psipars{4} - Psipars{1}}}
_{ds{= {11 over 6}}} = bbx{7 over 6}
end{align}




Note that $quadleft{begin{array}{rclcl}
ds{Psipars{2}} & ds{=} & ds{Psipars{1} + {1 over 1}} &&
\[1mm]
ds{Psipars{3}} & ds{=} & ds{Psipars{2} + {1 over 2}}
& ds{=} & ds{Psipars{1} + {3 over 2}}
\[1mm]
ds{Psipars{4}} & ds{=} & ds{Psipars{3} + {1 over 3}}
& ds{=} & ds{Psipars{1} + {11 over 6}}
end{array}right.$







share|cite|improve this answer











$endgroup$



$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{k = 1}^{infty}{6 over kpars{k + 1}pars{k+ 3}}} =
6sum_{k = 1}^{infty}2int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
{1 over pars{k + u_{1} + 2u_{2}}^{3}}
\[5mm] = &
12int_{0}^{1}dd u_{1}int_{0}^{u_{1}}dd u_{2},
underbrace{sum_{k = 1}^{infty}{1 over pars{k + u_{1} + 2u_{2}}^{3}}}
_{ds{-Psi''pars{1 + u_{1} + 2u_{2}}/2}}qquad
pars{~Psi: Digamma Function~}
\[5mm] = &
-3int_{0}^{1}dd u_{1}bracks{Psi, 'pars{1 + 3u_{1}} -
Psi, 'pars{1 + u_{1}}}
\[5mm] = &
-3bracks{{1 over 3},Psipars{1 + 3u_{1}} - Psipars{1 + u_{1}}}_{ 0}^{ 1} =
-
Psipars{4} + 3Psipars{2} + Psipars{1} - 3Psipars{1}
\[5mm] = &
3 underbrace{bracks{Psipars{2} - Psipars{1}}}_{ds{= 1}} -
underbrace{bracks{Psipars{4} - Psipars{1}}}
_{ds{= {11 over 6}}} = bbx{7 over 6}
end{align}




Note that $quadleft{begin{array}{rclcl}
ds{Psipars{2}} & ds{=} & ds{Psipars{1} + {1 over 1}} &&
\[1mm]
ds{Psipars{3}} & ds{=} & ds{Psipars{2} + {1 over 2}}
& ds{=} & ds{Psipars{1} + {3 over 2}}
\[1mm]
ds{Psipars{4}} & ds{=} & ds{Psipars{3} + {1 over 3}}
& ds{=} & ds{Psipars{1} + {11 over 6}}
end{array}right.$








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edited Jan 28 at 2:12

























answered Jan 28 at 2:03









Felix MarinFelix Marin

68.8k7109146




68.8k7109146












  • $begingroup$
    @ChaseRyanTaylor Yes. That's quite true. Indeed, I was testing Feynman Parametrization.
    $endgroup$
    – Felix Marin
    Jan 28 at 2:20


















  • $begingroup$
    @ChaseRyanTaylor Yes. That's quite true. Indeed, I was testing Feynman Parametrization.
    $endgroup$
    – Felix Marin
    Jan 28 at 2:20
















$begingroup$
@ChaseRyanTaylor Yes. That's quite true. Indeed, I was testing Feynman Parametrization.
$endgroup$
– Felix Marin
Jan 28 at 2:20




$begingroup$
@ChaseRyanTaylor Yes. That's quite true. Indeed, I was testing Feynman Parametrization.
$endgroup$
– Felix Marin
Jan 28 at 2:20


















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