How to check if the sum of infinite series is convergent?












4












$begingroup$


I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!






calculus sequences-and-series convergence trigonometric-series







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  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50


















4












$begingroup$


I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!






calculus sequences-and-series convergence trigonometric-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50
















4












4








4





$begingroup$


I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!






calculus sequences-and-series convergence trigonometric-series







share|cite|improve this question











$endgroup$




I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!






calculus sequences-and-series convergence trigonometric-series




calculus sequences-and-series convergence trigonometric-series






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 11 '18 at 22:43







Mark

















asked Dec 11 '18 at 21:37









MarkMark

233




233












  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50




















  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50


















$begingroup$
$0<sin^2x-sin x+1<1$ in that interval.
$endgroup$
– David Peterson
Dec 11 '18 at 21:43






$begingroup$
$0<sin^2x-sin x+1<1$ in that interval.
$endgroup$
– David Peterson
Dec 11 '18 at 21:43














$begingroup$
why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
$endgroup$
– Mark
Dec 11 '18 at 21:45






$begingroup$
why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
$endgroup$
– Mark
Dec 11 '18 at 21:45














$begingroup$
For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
$endgroup$
– David Peterson
Dec 11 '18 at 21:50






$begingroup$
For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
$endgroup$
– David Peterson
Dec 11 '18 at 21:50












4 Answers
4






active

oldest

votes


















2












$begingroup$

hint



$$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



The limit of the ratio is



$$L=sin^2(x)-sin(x)+1$$



but



$$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



thus



$$0<L<1$$
the series is convergent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:49










  • $begingroup$
    @Mark The ratio of $ln$ goes to one.
    $endgroup$
    – hamam_Abdallah
    Dec 11 '18 at 21:50



















2












$begingroup$

Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



$$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
    $endgroup$
    – Mark
    Dec 11 '18 at 22:48












  • $begingroup$
    Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
    $endgroup$
    – zhw.
    Dec 11 '18 at 22:57












  • $begingroup$
    (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
    $endgroup$
    – Mark Viola
    Dec 11 '18 at 23:09












  • $begingroup$
    @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
    $endgroup$
    – zhw.
    Dec 11 '18 at 23:16










  • $begingroup$
    @zhw. Happy Holiday to you Z-man!
    $endgroup$
    – Mark Viola
    Dec 12 '18 at 2:27



















1












$begingroup$

As an alternative by root test



$$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



and on the interval



$$0<sin^2(x) - sin (x) +1<1$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    PARTIAL PROOF



    Find the radius of convergence of the series
    $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



    As n tends to ∞,
    Using L'Hospital's Rule,



    $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



    =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



    This must be <1 for the series to converge



    Therefore,



    $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



    Prove that the interval lies in these values.






    share|cite|improve this answer









    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50
















      2












      $begingroup$

      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50














      2












      2








      2





      $begingroup$

      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.






      share|cite|improve this answer









      $endgroup$



      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 11 '18 at 21:45









      hamam_Abdallahhamam_Abdallah

      38.1k21634




      38.1k21634












      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50


















      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50
















      $begingroup$
      Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
      $endgroup$
      – Mark
      Dec 11 '18 at 21:49




      $begingroup$
      Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
      $endgroup$
      – Mark
      Dec 11 '18 at 21:49












      $begingroup$
      @Mark The ratio of $ln$ goes to one.
      $endgroup$
      – hamam_Abdallah
      Dec 11 '18 at 21:50




      $begingroup$
      @Mark The ratio of $ln$ goes to one.
      $endgroup$
      – hamam_Abdallah
      Dec 11 '18 at 21:50











      2












      $begingroup$

      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27
















      2












      $begingroup$

      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27














      2












      2








      2





      $begingroup$

      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






      share|cite|improve this answer









      $endgroup$



      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 11 '18 at 22:08









      zhw.zhw.

      73.6k43175




      73.6k43175












      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27


















      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27
















      $begingroup$
      I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
      $endgroup$
      – Mark
      Dec 11 '18 at 22:48






      $begingroup$
      I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
      $endgroup$
      – Mark
      Dec 11 '18 at 22:48














      $begingroup$
      Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
      $endgroup$
      – zhw.
      Dec 11 '18 at 22:57






      $begingroup$
      Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
      $endgroup$
      – zhw.
      Dec 11 '18 at 22:57














      $begingroup$
      (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
      $endgroup$
      – Mark Viola
      Dec 11 '18 at 23:09






      $begingroup$
      (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
      $endgroup$
      – Mark Viola
      Dec 11 '18 at 23:09














      $begingroup$
      @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
      $endgroup$
      – zhw.
      Dec 11 '18 at 23:16




      $begingroup$
      @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
      $endgroup$
      – zhw.
      Dec 11 '18 at 23:16












      $begingroup$
      @zhw. Happy Holiday to you Z-man!
      $endgroup$
      – Mark Viola
      Dec 12 '18 at 2:27




      $begingroup$
      @zhw. Happy Holiday to you Z-man!
      $endgroup$
      – Mark Viola
      Dec 12 '18 at 2:27











      1












      $begingroup$

      As an alternative by root test



      $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



      and on the interval



      $$0<sin^2(x) - sin (x) +1<1$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        As an alternative by root test



        $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



        and on the interval



        $$0<sin^2(x) - sin (x) +1<1$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          As an alternative by root test



          $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



          and on the interval



          $$0<sin^2(x) - sin (x) +1<1$$






          share|cite|improve this answer









          $endgroup$



          As an alternative by root test



          $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



          and on the interval



          $$0<sin^2(x) - sin (x) +1<1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 21:50









          gimusigimusi

          92.9k84494




          92.9k84494























              0












              $begingroup$

              PARTIAL PROOF



              Find the radius of convergence of the series
              $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



              As n tends to ∞,
              Using L'Hospital's Rule,



              $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



              =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



              This must be <1 for the series to converge



              Therefore,



              $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



              Prove that the interval lies in these values.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                PARTIAL PROOF



                Find the radius of convergence of the series
                $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



                As n tends to ∞,
                Using L'Hospital's Rule,



                $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



                =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



                This must be <1 for the series to converge



                Therefore,



                $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



                Prove that the interval lies in these values.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  PARTIAL PROOF



                  Find the radius of convergence of the series
                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



                  As n tends to ∞,
                  Using L'Hospital's Rule,



                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



                  =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



                  This must be <1 for the series to converge



                  Therefore,



                  $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



                  Prove that the interval lies in these values.






                  share|cite|improve this answer









                  $endgroup$



                  PARTIAL PROOF



                  Find the radius of convergence of the series
                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



                  As n tends to ∞,
                  Using L'Hospital's Rule,



                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



                  =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



                  This must be <1 for the series to converge



                  Therefore,



                  $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



                  Prove that the interval lies in these values.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 22:07









                  Pratik ApshingePratik Apshinge

                  305




                  305






























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