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












3














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.






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
















3














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









New contributor




George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















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














3












3








3


2





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









New contributor




George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited Dec 26 '18 at 22:36









amWhy

191k28224439




191k28224439






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asked Dec 26 '18 at 22:27









George

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161




New contributor




George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












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


















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
















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




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










1 Answer
1






active

oldest

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2














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





















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










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










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











Your Answer





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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2














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





















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










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










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
















2














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





















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










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










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














2












2








2






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












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









Frpzzd

21.9k839107




21.9k839107












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










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










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


















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










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










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
















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




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












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




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












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




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










George is a new contributor. Be nice, and check out our Code of Conduct.










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