Infinite series containing positive and negative terms












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 15:50
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 15:50














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















3












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027224%2finfinite-series-containing-positive-and-negative-terms%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}






share|cite|improve this answer









$endgroup$



$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027224%2finfinite-series-containing-positive-and-negative-terms%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more