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

Le Mesnil-Réaume

Image
Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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

web3.py web3.isConnected() returns false always

Image
1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
Read more