decide $sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}$ convergance












0












$begingroup$



$sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}$




how can one decide convergence / divergence of such sum ? i have seen context where they turn the integral and the sum why is it allowed ?




tried to substitute its tough integral to solve




$t = nx$



$dt = n~dx$



$int_0^{1} frac{dx}{1+n^4x^4} = frac{1}{n}int_0^{n} frac{dt}{1+t^4}$






integration sequences-and-series







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    $sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}$




    how can one decide convergence / divergence of such sum ? i have seen context where they turn the integral and the sum why is it allowed ?




    tried to substitute its tough integral to solve




    $t = nx$



    $dt = n~dx$



    $int_0^{1} frac{dx}{1+n^4x^4} = frac{1}{n}int_0^{n} frac{dt}{1+t^4}$






    integration sequences-and-series







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      $sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}$




      how can one decide convergence / divergence of such sum ? i have seen context where they turn the integral and the sum why is it allowed ?




      tried to substitute its tough integral to solve




      $t = nx$



      $dt = n~dx$



      $int_0^{1} frac{dx}{1+n^4x^4} = frac{1}{n}int_0^{n} frac{dt}{1+t^4}$






      integration sequences-and-series







      share|cite|improve this question











      $endgroup$





      $sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}$




      how can one decide convergence / divergence of such sum ? i have seen context where they turn the integral and the sum why is it allowed ?




      tried to substitute its tough integral to solve




      $t = nx$



      $dt = n~dx$



      $int_0^{1} frac{dx}{1+n^4x^4} = frac{1}{n}int_0^{n} frac{dt}{1+t^4}$






      integration sequences-and-series




      integration sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 9:39







      Mather

















      asked Jan 7 at 9:18









      Mather Mather

      3047




      3047






















          2 Answers
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          2












          $begingroup$

          Observe that
          $$begin{eqnarray}
          sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}&=&sum_{n=1}^{infty}frac{1}{n}int_0^{n} frac{du}{1+u^4}\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^4}du\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^2}du=frac{pi}{4}sum_{n=1}^infty frac{1}{n}=infty.
          end{eqnarray}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Really good solution did you use the fact that $ 0leq u leq 1 $ and so the function is greater in the second line ? so the integral is greater
            $endgroup$
            – Mather
            Jan 7 at 9:47










          • $begingroup$
            Sure I did :) I hope this helped you.
            $endgroup$
            – Song
            Jan 7 at 9:51



















          1












          $begingroup$

          Continuing with what you have done $ int_0^{n} frac 1 {1+t^{4}}dt$ tends to a finite positive limit as $n to infty$. Hence this the given sum is $geq delta sum frac 1 n$ for some $delta >0$ which makes the series divergent. In fact $ int_0^{n} frac 1 {1+t^{4}}dt geq int_0^{1} frac 1 2 dt =frac 1 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you sir could you explain the last inequality i cant mark that as a solution too it wont let me
            $endgroup$
            – Mather
            Jan 7 at 9:48






          • 1




            $begingroup$
            On $(0,1)$ we have $1+t^{4} <1+1=2$ so $frac 1 {1+t^{4}} >frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:50












          • $begingroup$
            @Mather You can only accept one answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:51










          • $begingroup$
            oh nice fast and clear
            $endgroup$
            – Mather
            Jan 7 at 9:54











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Observe that
          $$begin{eqnarray}
          sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}&=&sum_{n=1}^{infty}frac{1}{n}int_0^{n} frac{du}{1+u^4}\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^4}du\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^2}du=frac{pi}{4}sum_{n=1}^infty frac{1}{n}=infty.
          end{eqnarray}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Really good solution did you use the fact that $ 0leq u leq 1 $ and so the function is greater in the second line ? so the integral is greater
            $endgroup$
            – Mather
            Jan 7 at 9:47










          • $begingroup$
            Sure I did :) I hope this helped you.
            $endgroup$
            – Song
            Jan 7 at 9:51
















          2












          $begingroup$

          Observe that
          $$begin{eqnarray}
          sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}&=&sum_{n=1}^{infty}frac{1}{n}int_0^{n} frac{du}{1+u^4}\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^4}du\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^2}du=frac{pi}{4}sum_{n=1}^infty frac{1}{n}=infty.
          end{eqnarray}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Really good solution did you use the fact that $ 0leq u leq 1 $ and so the function is greater in the second line ? so the integral is greater
            $endgroup$
            – Mather
            Jan 7 at 9:47










          • $begingroup$
            Sure I did :) I hope this helped you.
            $endgroup$
            – Song
            Jan 7 at 9:51














          2












          2








          2





          $begingroup$

          Observe that
          $$begin{eqnarray}
          sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}&=&sum_{n=1}^{infty}frac{1}{n}int_0^{n} frac{du}{1+u^4}\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^4}du\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^2}du=frac{pi}{4}sum_{n=1}^infty frac{1}{n}=infty.
          end{eqnarray}$$






          share|cite|improve this answer









          $endgroup$



          Observe that
          $$begin{eqnarray}
          sum_{n=1}^{infty}int_0^{1} frac{dx}{1+n^4x^4}&=&sum_{n=1}^{infty}frac{1}{n}int_0^{n} frac{du}{1+u^4}\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^4}du\&ge&sum_{n=1}^{infty}frac{1}{n}int_0^{1} frac{1}{1+u^2}du=frac{pi}{4}sum_{n=1}^infty frac{1}{n}=infty.
          end{eqnarray}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 9:40









          SongSong

          10.8k628




          10.8k628












          • $begingroup$
            Really good solution did you use the fact that $ 0leq u leq 1 $ and so the function is greater in the second line ? so the integral is greater
            $endgroup$
            – Mather
            Jan 7 at 9:47










          • $begingroup$
            Sure I did :) I hope this helped you.
            $endgroup$
            – Song
            Jan 7 at 9:51


















          • $begingroup$
            Really good solution did you use the fact that $ 0leq u leq 1 $ and so the function is greater in the second line ? so the integral is greater
            $endgroup$
            – Mather
            Jan 7 at 9:47










          • $begingroup$
            Sure I did :) I hope this helped you.
            $endgroup$
            – Song
            Jan 7 at 9:51
















          $begingroup$
          Really good solution did you use the fact that $ 0leq u leq 1 $ and so the function is greater in the second line ? so the integral is greater
          $endgroup$
          – Mather
          Jan 7 at 9:47




          $begingroup$
          Really good solution did you use the fact that $ 0leq u leq 1 $ and so the function is greater in the second line ? so the integral is greater
          $endgroup$
          – Mather
          Jan 7 at 9:47












          $begingroup$
          Sure I did :) I hope this helped you.
          $endgroup$
          – Song
          Jan 7 at 9:51




          $begingroup$
          Sure I did :) I hope this helped you.
          $endgroup$
          – Song
          Jan 7 at 9:51











          1












          $begingroup$

          Continuing with what you have done $ int_0^{n} frac 1 {1+t^{4}}dt$ tends to a finite positive limit as $n to infty$. Hence this the given sum is $geq delta sum frac 1 n$ for some $delta >0$ which makes the series divergent. In fact $ int_0^{n} frac 1 {1+t^{4}}dt geq int_0^{1} frac 1 2 dt =frac 1 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you sir could you explain the last inequality i cant mark that as a solution too it wont let me
            $endgroup$
            – Mather
            Jan 7 at 9:48






          • 1




            $begingroup$
            On $(0,1)$ we have $1+t^{4} <1+1=2$ so $frac 1 {1+t^{4}} >frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:50












          • $begingroup$
            @Mather You can only accept one answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:51










          • $begingroup$
            oh nice fast and clear
            $endgroup$
            – Mather
            Jan 7 at 9:54
















          1












          $begingroup$

          Continuing with what you have done $ int_0^{n} frac 1 {1+t^{4}}dt$ tends to a finite positive limit as $n to infty$. Hence this the given sum is $geq delta sum frac 1 n$ for some $delta >0$ which makes the series divergent. In fact $ int_0^{n} frac 1 {1+t^{4}}dt geq int_0^{1} frac 1 2 dt =frac 1 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you sir could you explain the last inequality i cant mark that as a solution too it wont let me
            $endgroup$
            – Mather
            Jan 7 at 9:48






          • 1




            $begingroup$
            On $(0,1)$ we have $1+t^{4} <1+1=2$ so $frac 1 {1+t^{4}} >frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:50












          • $begingroup$
            @Mather You can only accept one answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:51










          • $begingroup$
            oh nice fast and clear
            $endgroup$
            – Mather
            Jan 7 at 9:54














          1












          1








          1





          $begingroup$

          Continuing with what you have done $ int_0^{n} frac 1 {1+t^{4}}dt$ tends to a finite positive limit as $n to infty$. Hence this the given sum is $geq delta sum frac 1 n$ for some $delta >0$ which makes the series divergent. In fact $ int_0^{n} frac 1 {1+t^{4}}dt geq int_0^{1} frac 1 2 dt =frac 1 2$.






          share|cite|improve this answer









          $endgroup$



          Continuing with what you have done $ int_0^{n} frac 1 {1+t^{4}}dt$ tends to a finite positive limit as $n to infty$. Hence this the given sum is $geq delta sum frac 1 n$ for some $delta >0$ which makes the series divergent. In fact $ int_0^{n} frac 1 {1+t^{4}}dt geq int_0^{1} frac 1 2 dt =frac 1 2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 9:39









          Kavi Rama MurthyKavi Rama Murthy

          56.1k42158




          56.1k42158












          • $begingroup$
            thank you sir could you explain the last inequality i cant mark that as a solution too it wont let me
            $endgroup$
            – Mather
            Jan 7 at 9:48






          • 1




            $begingroup$
            On $(0,1)$ we have $1+t^{4} <1+1=2$ so $frac 1 {1+t^{4}} >frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:50












          • $begingroup$
            @Mather You can only accept one answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:51










          • $begingroup$
            oh nice fast and clear
            $endgroup$
            – Mather
            Jan 7 at 9:54


















          • $begingroup$
            thank you sir could you explain the last inequality i cant mark that as a solution too it wont let me
            $endgroup$
            – Mather
            Jan 7 at 9:48






          • 1




            $begingroup$
            On $(0,1)$ we have $1+t^{4} <1+1=2$ so $frac 1 {1+t^{4}} >frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:50












          • $begingroup$
            @Mather You can only accept one answer.
            $endgroup$
            – Kavi Rama Murthy
            Jan 7 at 9:51










          • $begingroup$
            oh nice fast and clear
            $endgroup$
            – Mather
            Jan 7 at 9:54
















          $begingroup$
          thank you sir could you explain the last inequality i cant mark that as a solution too it wont let me
          $endgroup$
          – Mather
          Jan 7 at 9:48




          $begingroup$
          thank you sir could you explain the last inequality i cant mark that as a solution too it wont let me
          $endgroup$
          – Mather
          Jan 7 at 9:48




          1




          1




          $begingroup$
          On $(0,1)$ we have $1+t^{4} <1+1=2$ so $frac 1 {1+t^{4}} >frac 1 2$.
          $endgroup$
          – Kavi Rama Murthy
          Jan 7 at 9:50






          $begingroup$
          On $(0,1)$ we have $1+t^{4} <1+1=2$ so $frac 1 {1+t^{4}} >frac 1 2$.
          $endgroup$
          – Kavi Rama Murthy
          Jan 7 at 9:50














          $begingroup$
          @Mather You can only accept one answer.
          $endgroup$
          – Kavi Rama Murthy
          Jan 7 at 9:51




          $begingroup$
          @Mather You can only accept one answer.
          $endgroup$
          – Kavi Rama Murthy
          Jan 7 at 9:51












          $begingroup$
          oh nice fast and clear
          $endgroup$
          – Mather
          Jan 7 at 9:54




          $begingroup$
          oh nice fast and clear
          $endgroup$
          – Mather
          Jan 7 at 9:54


















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