Decide about convergence of the series for $a>0$ and $qin mathbb R$












1














My series is: $$sum_{n=1}^{+ infty } frac{q^{n}n^{2}cos( frac{npi}{5} )}{ frac{n^{3}}{1444}+1001n^{2}+1444an }$$



for $q in mathbb{R}$ and $a > 1$.



I did this task for $|q|<1$ and $|q|>1$ using the Cauchy criterion and for $q=-1$ using the Leibnitz criterion. However I cannot do this task for $q=1$. My idea is to use Raabe test but there are big number and it is difficult to use it so I think exist easier way.






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    My series is: $$sum_{n=1}^{+ infty } frac{q^{n}n^{2}cos( frac{npi}{5} )}{ frac{n^{3}}{1444}+1001n^{2}+1444an }$$



    for $q in mathbb{R}$ and $a > 1$.



    I did this task for $|q|<1$ and $|q|>1$ using the Cauchy criterion and for $q=-1$ using the Leibnitz criterion. However I cannot do this task for $q=1$. My idea is to use Raabe test but there are big number and it is difficult to use it so I think exist easier way.






    real-analysis







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      1







      My series is: $$sum_{n=1}^{+ infty } frac{q^{n}n^{2}cos( frac{npi}{5} )}{ frac{n^{3}}{1444}+1001n^{2}+1444an }$$



      for $q in mathbb{R}$ and $a > 1$.



      I did this task for $|q|<1$ and $|q|>1$ using the Cauchy criterion and for $q=-1$ using the Leibnitz criterion. However I cannot do this task for $q=1$. My idea is to use Raabe test but there are big number and it is difficult to use it so I think exist easier way.






      real-analysis







      share|cite|improve this question















      My series is: $$sum_{n=1}^{+ infty } frac{q^{n}n^{2}cos( frac{npi}{5} )}{ frac{n^{3}}{1444}+1001n^{2}+1444an }$$



      for $q in mathbb{R}$ and $a > 1$.



      I did this task for $|q|<1$ and $|q|>1$ using the Cauchy criterion and for $q=-1$ using the Leibnitz criterion. However I cannot do this task for $q=1$. My idea is to use Raabe test but there are big number and it is difficult to use it so I think exist easier way.






      real-analysis




      real-analysis






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      edited Jan 4 at 23:16









      T. Ford

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      asked Jan 4 at 23:08









      MP3129MP3129

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          You can prove it converges using the Dirichlet Test. https://en.wikipedia.org/wiki/Dirichlet%27s_test



          When $q=1$ this sum will be less than
          $$1444sum_{n=1}^infty frac{cos{(frac{npi}{5})}}{n}$$



          $|sum_{n=1}^N cos{(frac{npi}{5})}| le 10$ for every integer $N$, this is easy to see because the sum repeats every $20$, and the sum at $20$ is $0$ (Note I choose $10$ randomly, you could get a way better bound if you wanted). And $frac1n$ is decreasing and goes to 0, so the sum converges.






          share|cite|improve this answer








          New contributor




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


















          • There are few unfortunate $a$s, which vanish the denominator.
            – user58697
            Jan 5 at 5:02












          • @user58697 The problem states that $a>1$ so the denominator will always be positive.
            – Erik Parkinson
            Jan 5 at 5:04










          • @ErikParkinson I checked this sum in Wolfram and I got that SumConvergence is for: "Abs[q] <= 1 && q != 1", so Wolfram think that for q=1, the series is divergent. Are you sure of your answer? Because I do not see an error in your reasoning, and yet Wolfram gives something else
            – MP3129
            Jan 5 at 10:54










          • @MP3129 I'm pretty sure, although I could be missing something. Do you know how Wolfram determines convergence? My guess is that it is doing something numerical and ends up guessing incorrectly that it diverges.
            – Erik Parkinson
            Jan 5 at 11:08










          • @ErikParkinson Unfortunately Wolfram must be right, because I am writing this series for q=1 and for several different $a$ - for example for 1,2 and really big $a$ to give Wolfram constant values and all the time I get that the series is divergent . Moreover I checked Sum for {n,1,100},{n,1,1000},{n,1,100000} and the sum is constantly growing so this series must be divergent
            – MP3129
            Jan 5 at 12:08











          Your Answer





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          active

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          1














          You can prove it converges using the Dirichlet Test. https://en.wikipedia.org/wiki/Dirichlet%27s_test



          When $q=1$ this sum will be less than
          $$1444sum_{n=1}^infty frac{cos{(frac{npi}{5})}}{n}$$



          $|sum_{n=1}^N cos{(frac{npi}{5})}| le 10$ for every integer $N$, this is easy to see because the sum repeats every $20$, and the sum at $20$ is $0$ (Note I choose $10$ randomly, you could get a way better bound if you wanted). And $frac1n$ is decreasing and goes to 0, so the sum converges.






          share|cite|improve this answer








          New contributor




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


















          • There are few unfortunate $a$s, which vanish the denominator.
            – user58697
            Jan 5 at 5:02












          • @user58697 The problem states that $a>1$ so the denominator will always be positive.
            – Erik Parkinson
            Jan 5 at 5:04










          • @ErikParkinson I checked this sum in Wolfram and I got that SumConvergence is for: "Abs[q] <= 1 && q != 1", so Wolfram think that for q=1, the series is divergent. Are you sure of your answer? Because I do not see an error in your reasoning, and yet Wolfram gives something else
            – MP3129
            Jan 5 at 10:54










          • @MP3129 I'm pretty sure, although I could be missing something. Do you know how Wolfram determines convergence? My guess is that it is doing something numerical and ends up guessing incorrectly that it diverges.
            – Erik Parkinson
            Jan 5 at 11:08










          • @ErikParkinson Unfortunately Wolfram must be right, because I am writing this series for q=1 and for several different $a$ - for example for 1,2 and really big $a$ to give Wolfram constant values and all the time I get that the series is divergent . Moreover I checked Sum for {n,1,100},{n,1,1000},{n,1,100000} and the sum is constantly growing so this series must be divergent
            – MP3129
            Jan 5 at 12:08
















          1














          You can prove it converges using the Dirichlet Test. https://en.wikipedia.org/wiki/Dirichlet%27s_test



          When $q=1$ this sum will be less than
          $$1444sum_{n=1}^infty frac{cos{(frac{npi}{5})}}{n}$$



          $|sum_{n=1}^N cos{(frac{npi}{5})}| le 10$ for every integer $N$, this is easy to see because the sum repeats every $20$, and the sum at $20$ is $0$ (Note I choose $10$ randomly, you could get a way better bound if you wanted). And $frac1n$ is decreasing and goes to 0, so the sum converges.






          share|cite|improve this answer








          New contributor




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


















          • There are few unfortunate $a$s, which vanish the denominator.
            – user58697
            Jan 5 at 5:02












          • @user58697 The problem states that $a>1$ so the denominator will always be positive.
            – Erik Parkinson
            Jan 5 at 5:04










          • @ErikParkinson I checked this sum in Wolfram and I got that SumConvergence is for: "Abs[q] <= 1 && q != 1", so Wolfram think that for q=1, the series is divergent. Are you sure of your answer? Because I do not see an error in your reasoning, and yet Wolfram gives something else
            – MP3129
            Jan 5 at 10:54










          • @MP3129 I'm pretty sure, although I could be missing something. Do you know how Wolfram determines convergence? My guess is that it is doing something numerical and ends up guessing incorrectly that it diverges.
            – Erik Parkinson
            Jan 5 at 11:08










          • @ErikParkinson Unfortunately Wolfram must be right, because I am writing this series for q=1 and for several different $a$ - for example for 1,2 and really big $a$ to give Wolfram constant values and all the time I get that the series is divergent . Moreover I checked Sum for {n,1,100},{n,1,1000},{n,1,100000} and the sum is constantly growing so this series must be divergent
            – MP3129
            Jan 5 at 12:08














          1












          1








          1






          You can prove it converges using the Dirichlet Test. https://en.wikipedia.org/wiki/Dirichlet%27s_test



          When $q=1$ this sum will be less than
          $$1444sum_{n=1}^infty frac{cos{(frac{npi}{5})}}{n}$$



          $|sum_{n=1}^N cos{(frac{npi}{5})}| le 10$ for every integer $N$, this is easy to see because the sum repeats every $20$, and the sum at $20$ is $0$ (Note I choose $10$ randomly, you could get a way better bound if you wanted). And $frac1n$ is decreasing and goes to 0, so the sum converges.






          share|cite|improve this answer








          New contributor




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









          You can prove it converges using the Dirichlet Test. https://en.wikipedia.org/wiki/Dirichlet%27s_test



          When $q=1$ this sum will be less than
          $$1444sum_{n=1}^infty frac{cos{(frac{npi}{5})}}{n}$$



          $|sum_{n=1}^N cos{(frac{npi}{5})}| le 10$ for every integer $N$, this is easy to see because the sum repeats every $20$, and the sum at $20$ is $0$ (Note I choose $10$ randomly, you could get a way better bound if you wanted). And $frac1n$ is decreasing and goes to 0, so the sum converges.







          share|cite|improve this answer








          New contributor




          Erik Parkinson 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 answer



          share|cite|improve this answer






          New contributor




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









          answered Jan 5 at 0:45









          Erik ParkinsonErik Parkinson

          9159




          9159




          New contributor




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





          New contributor





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






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












          • There are few unfortunate $a$s, which vanish the denominator.
            – user58697
            Jan 5 at 5:02












          • @user58697 The problem states that $a>1$ so the denominator will always be positive.
            – Erik Parkinson
            Jan 5 at 5:04










          • @ErikParkinson I checked this sum in Wolfram and I got that SumConvergence is for: "Abs[q] <= 1 && q != 1", so Wolfram think that for q=1, the series is divergent. Are you sure of your answer? Because I do not see an error in your reasoning, and yet Wolfram gives something else
            – MP3129
            Jan 5 at 10:54










          • @MP3129 I'm pretty sure, although I could be missing something. Do you know how Wolfram determines convergence? My guess is that it is doing something numerical and ends up guessing incorrectly that it diverges.
            – Erik Parkinson
            Jan 5 at 11:08










          • @ErikParkinson Unfortunately Wolfram must be right, because I am writing this series for q=1 and for several different $a$ - for example for 1,2 and really big $a$ to give Wolfram constant values and all the time I get that the series is divergent . Moreover I checked Sum for {n,1,100},{n,1,1000},{n,1,100000} and the sum is constantly growing so this series must be divergent
            – MP3129
            Jan 5 at 12:08


















          • There are few unfortunate $a$s, which vanish the denominator.
            – user58697
            Jan 5 at 5:02












          • @user58697 The problem states that $a>1$ so the denominator will always be positive.
            – Erik Parkinson
            Jan 5 at 5:04










          • @ErikParkinson I checked this sum in Wolfram and I got that SumConvergence is for: "Abs[q] <= 1 && q != 1", so Wolfram think that for q=1, the series is divergent. Are you sure of your answer? Because I do not see an error in your reasoning, and yet Wolfram gives something else
            – MP3129
            Jan 5 at 10:54










          • @MP3129 I'm pretty sure, although I could be missing something. Do you know how Wolfram determines convergence? My guess is that it is doing something numerical and ends up guessing incorrectly that it diverges.
            – Erik Parkinson
            Jan 5 at 11:08










          • @ErikParkinson Unfortunately Wolfram must be right, because I am writing this series for q=1 and for several different $a$ - for example for 1,2 and really big $a$ to give Wolfram constant values and all the time I get that the series is divergent . Moreover I checked Sum for {n,1,100},{n,1,1000},{n,1,100000} and the sum is constantly growing so this series must be divergent
            – MP3129
            Jan 5 at 12:08
















          There are few unfortunate $a$s, which vanish the denominator.
          – user58697
          Jan 5 at 5:02






          There are few unfortunate $a$s, which vanish the denominator.
          – user58697
          Jan 5 at 5:02














          @user58697 The problem states that $a>1$ so the denominator will always be positive.
          – Erik Parkinson
          Jan 5 at 5:04




          @user58697 The problem states that $a>1$ so the denominator will always be positive.
          – Erik Parkinson
          Jan 5 at 5:04












          @ErikParkinson I checked this sum in Wolfram and I got that SumConvergence is for: "Abs[q] <= 1 && q != 1", so Wolfram think that for q=1, the series is divergent. Are you sure of your answer? Because I do not see an error in your reasoning, and yet Wolfram gives something else
          – MP3129
          Jan 5 at 10:54




          @ErikParkinson I checked this sum in Wolfram and I got that SumConvergence is for: "Abs[q] <= 1 && q != 1", so Wolfram think that for q=1, the series is divergent. Are you sure of your answer? Because I do not see an error in your reasoning, and yet Wolfram gives something else
          – MP3129
          Jan 5 at 10:54












          @MP3129 I'm pretty sure, although I could be missing something. Do you know how Wolfram determines convergence? My guess is that it is doing something numerical and ends up guessing incorrectly that it diverges.
          – Erik Parkinson
          Jan 5 at 11:08




          @MP3129 I'm pretty sure, although I could be missing something. Do you know how Wolfram determines convergence? My guess is that it is doing something numerical and ends up guessing incorrectly that it diverges.
          – Erik Parkinson
          Jan 5 at 11:08












          @ErikParkinson Unfortunately Wolfram must be right, because I am writing this series for q=1 and for several different $a$ - for example for 1,2 and really big $a$ to give Wolfram constant values and all the time I get that the series is divergent . Moreover I checked Sum for {n,1,100},{n,1,1000},{n,1,100000} and the sum is constantly growing so this series must be divergent
          – MP3129
          Jan 5 at 12:08




          @ErikParkinson Unfortunately Wolfram must be right, because I am writing this series for q=1 and for several different $a$ - for example for 1,2 and really big $a$ to give Wolfram constant values and all the time I get that the series is divergent . Moreover I checked Sum for {n,1,100},{n,1,1000},{n,1,100000} and the sum is constantly growing so this series must be divergent
          – MP3129
          Jan 5 at 12:08


















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