The sum of the series $frac{15}{16}+frac{15}{16} times frac{21}{24}+frac{15}{16} times frac{21}{24} times...

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This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




Suppose $S=frac{15}{16}+frac{15}{16} times frac{21}{24}+frac{15}{16} times frac{21}{24} times frac{27}{32}+dots dots$

Does it converge? If so find the sum.



What I attempted:- On inspection of the successive terms, it easy to deduce that the $n^{th}$ term of the series is $t_{n}=left(frac{3}{4}right)^n quadfrac{5.7.9.dots dots (2n+3)}{4.6.8. dots dots (2n+2)}$



Thus $frac{t_{n+1}}{t_n}=frac{3}{4} times frac{2n+5}{2n+4}$. As $n to infty$ this ratio tends to $frac{3}{4}<1$. Hence by Ratio test it turns out to be convergent.



A similar type of question has already been asked here. One of the commenters has provided a nice method to evaluate the sum of such series using recurrence relation and finally using the asymptotic form of Catalan Number.



To proceed exactly in the similar way, I wrote $t_n$ as follows:-
$t_n=frac{1}{3} left(frac{3}{8}right)^n quadfrac{1.3.5.7.9.dots dots (2n+3)}{(n+1)!}=frac{1}{3} left(frac{3}{8}right)^n frac{n+2}{2^{n+2}} binom{2n+4}{n+2}approx left(frac{3}{4}right)^{n-1}frac{sqrt{n+2}}{sqrt{pi}} quad (mbox{For large $n$})$.



I have used the recurrence relation $S_n=S_{n-1}+T_n$, along with the initial condition $S_1=frac{15}{16}$, in order to get a solution like this $$S_n=7.5+frac{T_n^2}{T_n-T_{n-1}}$$.



I am getting trouble in evaluating the limit of the second term as $n to infty$.



I haven't cross checked all the steps. Hope I would be pointed in case of any mistake.










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marked as duplicate by lab bhattacharjee sequences-and-series
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Jan 2 at 4:20


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    See math.stackexchange.com/questions/3054867/… and math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 4:20










  • $begingroup$
    @labbhattacharjee I do not think that the question is a duplicate of the cited one.
    $endgroup$
    – user
    Jan 7 at 15:07
















3












$begingroup$



This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




Suppose $S=frac{15}{16}+frac{15}{16} times frac{21}{24}+frac{15}{16} times frac{21}{24} times frac{27}{32}+dots dots$

Does it converge? If so find the sum.



What I attempted:- On inspection of the successive terms, it easy to deduce that the $n^{th}$ term of the series is $t_{n}=left(frac{3}{4}right)^n quadfrac{5.7.9.dots dots (2n+3)}{4.6.8. dots dots (2n+2)}$



Thus $frac{t_{n+1}}{t_n}=frac{3}{4} times frac{2n+5}{2n+4}$. As $n to infty$ this ratio tends to $frac{3}{4}<1$. Hence by Ratio test it turns out to be convergent.



A similar type of question has already been asked here. One of the commenters has provided a nice method to evaluate the sum of such series using recurrence relation and finally using the asymptotic form of Catalan Number.



To proceed exactly in the similar way, I wrote $t_n$ as follows:-
$t_n=frac{1}{3} left(frac{3}{8}right)^n quadfrac{1.3.5.7.9.dots dots (2n+3)}{(n+1)!}=frac{1}{3} left(frac{3}{8}right)^n frac{n+2}{2^{n+2}} binom{2n+4}{n+2}approx left(frac{3}{4}right)^{n-1}frac{sqrt{n+2}}{sqrt{pi}} quad (mbox{For large $n$})$.



I have used the recurrence relation $S_n=S_{n-1}+T_n$, along with the initial condition $S_1=frac{15}{16}$, in order to get a solution like this $$S_n=7.5+frac{T_n^2}{T_n-T_{n-1}}$$.



I am getting trouble in evaluating the limit of the second term as $n to infty$.



I haven't cross checked all the steps. Hope I would be pointed in case of any mistake.










share|cite|improve this question









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marked as duplicate by lab bhattacharjee sequences-and-series
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Jan 2 at 4:20


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    See math.stackexchange.com/questions/3054867/… and math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 4:20










  • $begingroup$
    @labbhattacharjee I do not think that the question is a duplicate of the cited one.
    $endgroup$
    – user
    Jan 7 at 15:07














3












3








3


0



$begingroup$



This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




Suppose $S=frac{15}{16}+frac{15}{16} times frac{21}{24}+frac{15}{16} times frac{21}{24} times frac{27}{32}+dots dots$

Does it converge? If so find the sum.



What I attempted:- On inspection of the successive terms, it easy to deduce that the $n^{th}$ term of the series is $t_{n}=left(frac{3}{4}right)^n quadfrac{5.7.9.dots dots (2n+3)}{4.6.8. dots dots (2n+2)}$



Thus $frac{t_{n+1}}{t_n}=frac{3}{4} times frac{2n+5}{2n+4}$. As $n to infty$ this ratio tends to $frac{3}{4}<1$. Hence by Ratio test it turns out to be convergent.



A similar type of question has already been asked here. One of the commenters has provided a nice method to evaluate the sum of such series using recurrence relation and finally using the asymptotic form of Catalan Number.



To proceed exactly in the similar way, I wrote $t_n$ as follows:-
$t_n=frac{1}{3} left(frac{3}{8}right)^n quadfrac{1.3.5.7.9.dots dots (2n+3)}{(n+1)!}=frac{1}{3} left(frac{3}{8}right)^n frac{n+2}{2^{n+2}} binom{2n+4}{n+2}approx left(frac{3}{4}right)^{n-1}frac{sqrt{n+2}}{sqrt{pi}} quad (mbox{For large $n$})$.



I have used the recurrence relation $S_n=S_{n-1}+T_n$, along with the initial condition $S_1=frac{15}{16}$, in order to get a solution like this $$S_n=7.5+frac{T_n^2}{T_n-T_{n-1}}$$.



I am getting trouble in evaluating the limit of the second term as $n to infty$.



I haven't cross checked all the steps. Hope I would be pointed in case of any mistake.










share|cite|improve this question









$endgroup$





This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




Suppose $S=frac{15}{16}+frac{15}{16} times frac{21}{24}+frac{15}{16} times frac{21}{24} times frac{27}{32}+dots dots$

Does it converge? If so find the sum.



What I attempted:- On inspection of the successive terms, it easy to deduce that the $n^{th}$ term of the series is $t_{n}=left(frac{3}{4}right)^n quadfrac{5.7.9.dots dots (2n+3)}{4.6.8. dots dots (2n+2)}$



Thus $frac{t_{n+1}}{t_n}=frac{3}{4} times frac{2n+5}{2n+4}$. As $n to infty$ this ratio tends to $frac{3}{4}<1$. Hence by Ratio test it turns out to be convergent.



A similar type of question has already been asked here. One of the commenters has provided a nice method to evaluate the sum of such series using recurrence relation and finally using the asymptotic form of Catalan Number.



To proceed exactly in the similar way, I wrote $t_n$ as follows:-
$t_n=frac{1}{3} left(frac{3}{8}right)^n quadfrac{1.3.5.7.9.dots dots (2n+3)}{(n+1)!}=frac{1}{3} left(frac{3}{8}right)^n frac{n+2}{2^{n+2}} binom{2n+4}{n+2}approx left(frac{3}{4}right)^{n-1}frac{sqrt{n+2}}{sqrt{pi}} quad (mbox{For large $n$})$.



I have used the recurrence relation $S_n=S_{n-1}+T_n$, along with the initial condition $S_1=frac{15}{16}$, in order to get a solution like this $$S_n=7.5+frac{T_n^2}{T_n-T_{n-1}}$$.



I am getting trouble in evaluating the limit of the second term as $n to infty$.



I haven't cross checked all the steps. Hope I would be pointed in case of any mistake.





This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers








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asked Jan 2 at 3:33







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marked as duplicate by lab bhattacharjee sequences-and-series
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Jan 2 at 4:20


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by lab bhattacharjee sequences-and-series
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Jan 2 at 4:20


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    See math.stackexchange.com/questions/3054867/… and math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 4:20










  • $begingroup$
    @labbhattacharjee I do not think that the question is a duplicate of the cited one.
    $endgroup$
    – user
    Jan 7 at 15:07


















  • $begingroup$
    See math.stackexchange.com/questions/3054867/… and math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 4:20










  • $begingroup$
    @labbhattacharjee I do not think that the question is a duplicate of the cited one.
    $endgroup$
    – user
    Jan 7 at 15:07
















$begingroup$
See math.stackexchange.com/questions/3054867/… and math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 2 at 4:20




$begingroup$
See math.stackexchange.com/questions/3054867/… and math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 2 at 4:20












$begingroup$
@labbhattacharjee I do not think that the question is a duplicate of the cited one.
$endgroup$
– user
Jan 7 at 15:07




$begingroup$
@labbhattacharjee I do not think that the question is a duplicate of the cited one.
$endgroup$
– user
Jan 7 at 15:07










1 Answer
1






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4












$begingroup$

Let
$$f(x)=sum_{n=1}^infty t_nx^n=sum_{n=1}^inftyfrac{5times7timescdots
times(2n+3)}{4times6timescdots
times(2n+2)}x^n.$$

Then
$$t_n=frac23frac1{(n+1)!}left(frac32right)left(frac52right)cdots
left(frac{2n+3}2right)=frac23u_{n+1}$$

where
$$u_n=frac{(3/2)(5/2)cdots((2n+1)/2)}{n!}.$$
Then, for $|x|<1$,
$$sum_{n=0}^infty u_nx^n=frac1{(1-x)^{3/2}}$$
by the binomial theorem.
Then
$$f(x)=frac23sum_{n=1}^infty u_{n+1}x^n=frac23sum_{n=2}^infty
u_nx^{n-1}=frac2{3x}left(frac1{(1-x)^{3/2}}-1-frac{3x}2right)$$

Now insert $x=3/4$.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Let
    $$f(x)=sum_{n=1}^infty t_nx^n=sum_{n=1}^inftyfrac{5times7timescdots
    times(2n+3)}{4times6timescdots
    times(2n+2)}x^n.$$

    Then
    $$t_n=frac23frac1{(n+1)!}left(frac32right)left(frac52right)cdots
    left(frac{2n+3}2right)=frac23u_{n+1}$$

    where
    $$u_n=frac{(3/2)(5/2)cdots((2n+1)/2)}{n!}.$$
    Then, for $|x|<1$,
    $$sum_{n=0}^infty u_nx^n=frac1{(1-x)^{3/2}}$$
    by the binomial theorem.
    Then
    $$f(x)=frac23sum_{n=1}^infty u_{n+1}x^n=frac23sum_{n=2}^infty
    u_nx^{n-1}=frac2{3x}left(frac1{(1-x)^{3/2}}-1-frac{3x}2right)$$

    Now insert $x=3/4$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Let
      $$f(x)=sum_{n=1}^infty t_nx^n=sum_{n=1}^inftyfrac{5times7timescdots
      times(2n+3)}{4times6timescdots
      times(2n+2)}x^n.$$

      Then
      $$t_n=frac23frac1{(n+1)!}left(frac32right)left(frac52right)cdots
      left(frac{2n+3}2right)=frac23u_{n+1}$$

      where
      $$u_n=frac{(3/2)(5/2)cdots((2n+1)/2)}{n!}.$$
      Then, for $|x|<1$,
      $$sum_{n=0}^infty u_nx^n=frac1{(1-x)^{3/2}}$$
      by the binomial theorem.
      Then
      $$f(x)=frac23sum_{n=1}^infty u_{n+1}x^n=frac23sum_{n=2}^infty
      u_nx^{n-1}=frac2{3x}left(frac1{(1-x)^{3/2}}-1-frac{3x}2right)$$

      Now insert $x=3/4$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Let
        $$f(x)=sum_{n=1}^infty t_nx^n=sum_{n=1}^inftyfrac{5times7timescdots
        times(2n+3)}{4times6timescdots
        times(2n+2)}x^n.$$

        Then
        $$t_n=frac23frac1{(n+1)!}left(frac32right)left(frac52right)cdots
        left(frac{2n+3}2right)=frac23u_{n+1}$$

        where
        $$u_n=frac{(3/2)(5/2)cdots((2n+1)/2)}{n!}.$$
        Then, for $|x|<1$,
        $$sum_{n=0}^infty u_nx^n=frac1{(1-x)^{3/2}}$$
        by the binomial theorem.
        Then
        $$f(x)=frac23sum_{n=1}^infty u_{n+1}x^n=frac23sum_{n=2}^infty
        u_nx^{n-1}=frac2{3x}left(frac1{(1-x)^{3/2}}-1-frac{3x}2right)$$

        Now insert $x=3/4$.






        share|cite|improve this answer









        $endgroup$



        Let
        $$f(x)=sum_{n=1}^infty t_nx^n=sum_{n=1}^inftyfrac{5times7timescdots
        times(2n+3)}{4times6timescdots
        times(2n+2)}x^n.$$

        Then
        $$t_n=frac23frac1{(n+1)!}left(frac32right)left(frac52right)cdots
        left(frac{2n+3}2right)=frac23u_{n+1}$$

        where
        $$u_n=frac{(3/2)(5/2)cdots((2n+1)/2)}{n!}.$$
        Then, for $|x|<1$,
        $$sum_{n=0}^infty u_nx^n=frac1{(1-x)^{3/2}}$$
        by the binomial theorem.
        Then
        $$f(x)=frac23sum_{n=1}^infty u_{n+1}x^n=frac23sum_{n=2}^infty
        u_nx^{n-1}=frac2{3x}left(frac1{(1-x)^{3/2}}-1-frac{3x}2right)$$

        Now insert $x=3/4$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 3:43









        Lord Shark the UnknownLord Shark the Unknown

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