Summation: $x+ sum _{ n=1 }^{ infty }{ frac { (-1)^{n-1} (2n-3) (2n-5)cdots 5cdot 3cdot 1}{ 2^{n} n!...












3












$begingroup$


I have to calculate:



$$x+ sum _{ n=1 }^{ infty }{ frac { (-1)^{n-1} (2n-3) (2n-5)cdots 5cdot 3cdot 1}{ 2^{n} n! (2n+1)}x^{2n+1} } $$



We have:



$$(2n-3)!! = frac{(2n-2)!!}{(2n-2)!!}(2n-3)!! = frac{(2n-2)!}{(2n-2)!!} = frac{(2n-2)!}{2^{n-1}(n-1)!}$$



So, the expression in the series is:



$$sum_{ngeq 1} (-1)^{n-1} frac{(2n-2)!}{n!(n-1)! 2^{2n-1}(2n+1)}x^{2n+1} = -2xsum_{ngeq 0} binom{2n}{n}frac{1}{(n+1)(2n+3)}left(-frac{x^2}{4}right)^{n+1}$$



By the ratio test, the series converges when $x^2 < 1$.



We now evaluate the series:



$$S(v) = sum_{ngeq 0}v^{n+1} binom{2n}{n} frac{1}{n+1} = int sum_{ngeq 0} v^n binom{2n}{n}~dv=int frac{1}{sqrt{1-4v}}~dv=- frac12sqrt{1-4v}$$



Letting $v = -x^2/4$ afterwards will give us an expression for the series $S(u)$.



But now I am stuck.






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    (+1) for showing your attempt.
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:31
















3












$begingroup$


I have to calculate:



$$x+ sum _{ n=1 }^{ infty }{ frac { (-1)^{n-1} (2n-3) (2n-5)cdots 5cdot 3cdot 1}{ 2^{n} n! (2n+1)}x^{2n+1} } $$



We have:



$$(2n-3)!! = frac{(2n-2)!!}{(2n-2)!!}(2n-3)!! = frac{(2n-2)!}{(2n-2)!!} = frac{(2n-2)!}{2^{n-1}(n-1)!}$$



So, the expression in the series is:



$$sum_{ngeq 1} (-1)^{n-1} frac{(2n-2)!}{n!(n-1)! 2^{2n-1}(2n+1)}x^{2n+1} = -2xsum_{ngeq 0} binom{2n}{n}frac{1}{(n+1)(2n+3)}left(-frac{x^2}{4}right)^{n+1}$$



By the ratio test, the series converges when $x^2 < 1$.



We now evaluate the series:



$$S(v) = sum_{ngeq 0}v^{n+1} binom{2n}{n} frac{1}{n+1} = int sum_{ngeq 0} v^n binom{2n}{n}~dv=int frac{1}{sqrt{1-4v}}~dv=- frac12sqrt{1-4v}$$



Letting $v = -x^2/4$ afterwards will give us an expression for the series $S(u)$.



But now I am stuck.






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    (+1) for showing your attempt.
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:31














3












3








3


1



$begingroup$


I have to calculate:



$$x+ sum _{ n=1 }^{ infty }{ frac { (-1)^{n-1} (2n-3) (2n-5)cdots 5cdot 3cdot 1}{ 2^{n} n! (2n+1)}x^{2n+1} } $$



We have:



$$(2n-3)!! = frac{(2n-2)!!}{(2n-2)!!}(2n-3)!! = frac{(2n-2)!}{(2n-2)!!} = frac{(2n-2)!}{2^{n-1}(n-1)!}$$



So, the expression in the series is:



$$sum_{ngeq 1} (-1)^{n-1} frac{(2n-2)!}{n!(n-1)! 2^{2n-1}(2n+1)}x^{2n+1} = -2xsum_{ngeq 0} binom{2n}{n}frac{1}{(n+1)(2n+3)}left(-frac{x^2}{4}right)^{n+1}$$



By the ratio test, the series converges when $x^2 < 1$.



We now evaluate the series:



$$S(v) = sum_{ngeq 0}v^{n+1} binom{2n}{n} frac{1}{n+1} = int sum_{ngeq 0} v^n binom{2n}{n}~dv=int frac{1}{sqrt{1-4v}}~dv=- frac12sqrt{1-4v}$$



Letting $v = -x^2/4$ afterwards will give us an expression for the series $S(u)$.



But now I am stuck.






sequences-and-series







share|cite|improve this question











$endgroup$




I have to calculate:



$$x+ sum _{ n=1 }^{ infty }{ frac { (-1)^{n-1} (2n-3) (2n-5)cdots 5cdot 3cdot 1}{ 2^{n} n! (2n+1)}x^{2n+1} } $$



We have:



$$(2n-3)!! = frac{(2n-2)!!}{(2n-2)!!}(2n-3)!! = frac{(2n-2)!}{(2n-2)!!} = frac{(2n-2)!}{2^{n-1}(n-1)!}$$



So, the expression in the series is:



$$sum_{ngeq 1} (-1)^{n-1} frac{(2n-2)!}{n!(n-1)! 2^{2n-1}(2n+1)}x^{2n+1} = -2xsum_{ngeq 0} binom{2n}{n}frac{1}{(n+1)(2n+3)}left(-frac{x^2}{4}right)^{n+1}$$



By the ratio test, the series converges when $x^2 < 1$.



We now evaluate the series:



$$S(v) = sum_{ngeq 0}v^{n+1} binom{2n}{n} frac{1}{n+1} = int sum_{ngeq 0} v^n binom{2n}{n}~dv=int frac{1}{sqrt{1-4v}}~dv=- frac12sqrt{1-4v}$$



Letting $v = -x^2/4$ afterwards will give us an expression for the series $S(u)$.



But now I am stuck.






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








edited Dec 26 '18 at 22:36









Namaste

1




1










asked Dec 26 '18 at 22:27









GeorgeGeorge

263




263












  • $begingroup$
    (+1) for showing your attempt.
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:31


















  • $begingroup$
    (+1) for showing your attempt.
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:31
















$begingroup$
(+1) for showing your attempt.
$endgroup$
– Frpzzd
Dec 26 '18 at 22:31




$begingroup$
(+1) for showing your attempt.
$endgroup$
– Frpzzd
Dec 26 '18 at 22:31










1 Answer
1






active

oldest

votes


















3












$begingroup$

Instead of using $v^{n+1}$, it would have been best to use $v^{2n+1}$. You almost got it; here is a complete solution:



The following is a well-know Taylor series:
$$sum_{n=0}^infty binom{2n}{n}x^{2n+1}=frac{x}{sqrt{1-4x^2}}$$
Integrating both sides of this equation yields the equation
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+2}}{2n+2}=frac{1-sqrt{1-4x^2}}{4}$$
and integrating both sides of this equation gives us
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+3}}{(2n+2)(2n+3)}=frac{4x-2xsqrt{1-4x^2}-arcsin(2x)}{16}$$
Multiplying both sides by $2$ and making an appropriate substitution with $x$ will give you the desired result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well I was a second to late posting my own solution ^^ Nevertheless (+1)
    $endgroup$
    – mrtaurho
    Dec 26 '18 at 22:36










  • $begingroup$
    could you have a more detailed answer, because i get confused and to be ensured for the answer
    $endgroup$
    – George
    Dec 26 '18 at 22:36










  • $begingroup$
    @George Which step don't you understand?
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:37












Your Answer








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%2f3053376%2fsummation-x-sum-n-1-infty-frac-1n-1-2n-3-2n-5-cdots%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












$begingroup$

Instead of using $v^{n+1}$, it would have been best to use $v^{2n+1}$. You almost got it; here is a complete solution:



The following is a well-know Taylor series:
$$sum_{n=0}^infty binom{2n}{n}x^{2n+1}=frac{x}{sqrt{1-4x^2}}$$
Integrating both sides of this equation yields the equation
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+2}}{2n+2}=frac{1-sqrt{1-4x^2}}{4}$$
and integrating both sides of this equation gives us
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+3}}{(2n+2)(2n+3)}=frac{4x-2xsqrt{1-4x^2}-arcsin(2x)}{16}$$
Multiplying both sides by $2$ and making an appropriate substitution with $x$ will give you the desired result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well I was a second to late posting my own solution ^^ Nevertheless (+1)
    $endgroup$
    – mrtaurho
    Dec 26 '18 at 22:36










  • $begingroup$
    could you have a more detailed answer, because i get confused and to be ensured for the answer
    $endgroup$
    – George
    Dec 26 '18 at 22:36










  • $begingroup$
    @George Which step don't you understand?
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:37
















3












$begingroup$

Instead of using $v^{n+1}$, it would have been best to use $v^{2n+1}$. You almost got it; here is a complete solution:



The following is a well-know Taylor series:
$$sum_{n=0}^infty binom{2n}{n}x^{2n+1}=frac{x}{sqrt{1-4x^2}}$$
Integrating both sides of this equation yields the equation
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+2}}{2n+2}=frac{1-sqrt{1-4x^2}}{4}$$
and integrating both sides of this equation gives us
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+3}}{(2n+2)(2n+3)}=frac{4x-2xsqrt{1-4x^2}-arcsin(2x)}{16}$$
Multiplying both sides by $2$ and making an appropriate substitution with $x$ will give you the desired result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well I was a second to late posting my own solution ^^ Nevertheless (+1)
    $endgroup$
    – mrtaurho
    Dec 26 '18 at 22:36










  • $begingroup$
    could you have a more detailed answer, because i get confused and to be ensured for the answer
    $endgroup$
    – George
    Dec 26 '18 at 22:36










  • $begingroup$
    @George Which step don't you understand?
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:37














3












3








3





$begingroup$

Instead of using $v^{n+1}$, it would have been best to use $v^{2n+1}$. You almost got it; here is a complete solution:



The following is a well-know Taylor series:
$$sum_{n=0}^infty binom{2n}{n}x^{2n+1}=frac{x}{sqrt{1-4x^2}}$$
Integrating both sides of this equation yields the equation
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+2}}{2n+2}=frac{1-sqrt{1-4x^2}}{4}$$
and integrating both sides of this equation gives us
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+3}}{(2n+2)(2n+3)}=frac{4x-2xsqrt{1-4x^2}-arcsin(2x)}{16}$$
Multiplying both sides by $2$ and making an appropriate substitution with $x$ will give you the desired result.






share|cite|improve this answer









$endgroup$



Instead of using $v^{n+1}$, it would have been best to use $v^{2n+1}$. You almost got it; here is a complete solution:



The following is a well-know Taylor series:
$$sum_{n=0}^infty binom{2n}{n}x^{2n+1}=frac{x}{sqrt{1-4x^2}}$$
Integrating both sides of this equation yields the equation
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+2}}{2n+2}=frac{1-sqrt{1-4x^2}}{4}$$
and integrating both sides of this equation gives us
$$sum_{n=0}^infty binom{2n}{n}frac{x^{2n+3}}{(2n+2)(2n+3)}=frac{4x-2xsqrt{1-4x^2}-arcsin(2x)}{16}$$
Multiplying both sides by $2$ and making an appropriate substitution with $x$ will give you the desired result.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 22:35









FrpzzdFrpzzd

23k841112




23k841112












  • $begingroup$
    Well I was a second to late posting my own solution ^^ Nevertheless (+1)
    $endgroup$
    – mrtaurho
    Dec 26 '18 at 22:36










  • $begingroup$
    could you have a more detailed answer, because i get confused and to be ensured for the answer
    $endgroup$
    – George
    Dec 26 '18 at 22:36










  • $begingroup$
    @George Which step don't you understand?
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:37


















  • $begingroup$
    Well I was a second to late posting my own solution ^^ Nevertheless (+1)
    $endgroup$
    – mrtaurho
    Dec 26 '18 at 22:36










  • $begingroup$
    could you have a more detailed answer, because i get confused and to be ensured for the answer
    $endgroup$
    – George
    Dec 26 '18 at 22:36










  • $begingroup$
    @George Which step don't you understand?
    $endgroup$
    – Frpzzd
    Dec 26 '18 at 22:37
















$begingroup$
Well I was a second to late posting my own solution ^^ Nevertheless (+1)
$endgroup$
– mrtaurho
Dec 26 '18 at 22:36




$begingroup$
Well I was a second to late posting my own solution ^^ Nevertheless (+1)
$endgroup$
– mrtaurho
Dec 26 '18 at 22:36












$begingroup$
could you have a more detailed answer, because i get confused and to be ensured for the answer
$endgroup$
– George
Dec 26 '18 at 22:36




$begingroup$
could you have a more detailed answer, because i get confused and to be ensured for the answer
$endgroup$
– George
Dec 26 '18 at 22:36












$begingroup$
@George Which step don't you understand?
$endgroup$
– Frpzzd
Dec 26 '18 at 22:37




$begingroup$
@George Which step don't you understand?
$endgroup$
– Frpzzd
Dec 26 '18 at 22:37


















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%2f3053376%2fsummation-x-sum-n-1-infty-frac-1n-1-2n-3-2n-5-cdots%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

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more