Prove $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$












1














I was wondering if this proof and my later assertions were correct.



$f_n(x) = x/(1+x)^n$



Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.



$sum_{n=1}^{infty} M_n$ converges by the ratio test.



Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.



Also, since it is uniformly convergent, it is convergent.



Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.



Thanks ahead of time.










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    1














    I was wondering if this proof and my later assertions were correct.



    $f_n(x) = x/(1+x)^n$



    Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.



    $sum_{n=1}^{infty} M_n$ converges by the ratio test.



    Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.



    Also, since it is uniformly convergent, it is convergent.



    Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.



    Thanks ahead of time.










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      1












      1








      1


      1





      I was wondering if this proof and my later assertions were correct.



      $f_n(x) = x/(1+x)^n$



      Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.



      $sum_{n=1}^{infty} M_n$ converges by the ratio test.



      Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.



      Also, since it is uniformly convergent, it is convergent.



      Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.



      Thanks ahead of time.










      share|cite|improve this question













      I was wondering if this proof and my later assertions were correct.



      $f_n(x) = x/(1+x)^n$



      Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.



      $sum_{n=1}^{infty} M_n$ converges by the ratio test.



      Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.



      Also, since it is uniformly convergent, it is convergent.



      Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.



      Thanks ahead of time.







      real-analysis proof-verification uniform-convergence






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      asked Dec 1 at 3:56









      user591271

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          2 Answers
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          Your application of $M_n$ test is wrong



          Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test



          $f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$



          $f_n'(x)<0$ for $n>1$ as $xin [1,2]$



          SO $f_n(x)=x/(1+x)^nleq f_n(1)$



          i.e $M_n =1/2^n$



          $sum_1^{infty}1/2^n =1$



          SO given series is uniformly convergent by Weierstrass $M_n$ Test






          share|cite|improve this answer





















          • Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
            – user591271
            Dec 1 at 17:28






          • 1




            Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
            – Shubham
            Dec 1 at 17:57



















          0














          No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.

          Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Your application of $M_n$ test is wrong



            Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test



            $f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$



            $f_n'(x)<0$ for $n>1$ as $xin [1,2]$



            SO $f_n(x)=x/(1+x)^nleq f_n(1)$



            i.e $M_n =1/2^n$



            $sum_1^{infty}1/2^n =1$



            SO given series is uniformly convergent by Weierstrass $M_n$ Test






            share|cite|improve this answer





















            • Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
              – user591271
              Dec 1 at 17:28






            • 1




              Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
              – Shubham
              Dec 1 at 17:57
















            1














            Your application of $M_n$ test is wrong



            Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test



            $f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$



            $f_n'(x)<0$ for $n>1$ as $xin [1,2]$



            SO $f_n(x)=x/(1+x)^nleq f_n(1)$



            i.e $M_n =1/2^n$



            $sum_1^{infty}1/2^n =1$



            SO given series is uniformly convergent by Weierstrass $M_n$ Test






            share|cite|improve this answer





















            • Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
              – user591271
              Dec 1 at 17:28






            • 1




              Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
              – Shubham
              Dec 1 at 17:57














            1












            1








            1






            Your application of $M_n$ test is wrong



            Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test



            $f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$



            $f_n'(x)<0$ for $n>1$ as $xin [1,2]$



            SO $f_n(x)=x/(1+x)^nleq f_n(1)$



            i.e $M_n =1/2^n$



            $sum_1^{infty}1/2^n =1$



            SO given series is uniformly convergent by Weierstrass $M_n$ Test






            share|cite|improve this answer












            Your application of $M_n$ test is wrong



            Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test



            $f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$



            $f_n'(x)<0$ for $n>1$ as $xin [1,2]$



            SO $f_n(x)=x/(1+x)^nleq f_n(1)$



            i.e $M_n =1/2^n$



            $sum_1^{infty}1/2^n =1$



            SO given series is uniformly convergent by Weierstrass $M_n$ Test







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 4:21









            Shubham

            1,5851519




            1,5851519












            • Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
              – user591271
              Dec 1 at 17:28






            • 1




              Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
              – Shubham
              Dec 1 at 17:57


















            • Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
              – user591271
              Dec 1 at 17:28






            • 1




              Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
              – Shubham
              Dec 1 at 17:57
















            Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
            – user591271
            Dec 1 at 17:28




            Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
            – user591271
            Dec 1 at 17:28




            1




            1




            Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
            – Shubham
            Dec 1 at 17:57




            Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
            – Shubham
            Dec 1 at 17:57











            0














            No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.

            Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.






            share|cite|improve this answer


























              0














              No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.

              Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.






              share|cite|improve this answer
























                0












                0








                0






                No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.

                Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.






                share|cite|improve this answer












                No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.

                Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 at 4:13









                jmerry

                1,48817




                1,48817






























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