Radius of convergence for three series












0












$begingroup$


I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



1) $sum_{n=0}^{infty}x^{n!}$

2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$

3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$




1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



Thus the radius of convergence is 1.

On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.




Are my results correct ? If not, what did I do wrong or forgot ?

Thanks for your help !






real-analysis sequences-and-series convergence power-series divergent-series







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



    1) $sum_{n=0}^{infty}x^{n!}$

    2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$

    3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$




    1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



    Thus the radius of convergence is 1.

    On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



    2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

    We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



    3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


    On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.




    Are my results correct ? If not, what did I do wrong or forgot ?

    Thanks for your help !






    real-analysis sequences-and-series convergence power-series divergent-series







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



      1) $sum_{n=0}^{infty}x^{n!}$

      2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$

      3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$




      1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



      Thus the radius of convergence is 1.

      On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



      2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

      We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



      3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


      On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.




      Are my results correct ? If not, what did I do wrong or forgot ?

      Thanks for your help !






      real-analysis sequences-and-series convergence power-series divergent-series







      share|cite|improve this question











      $endgroup$




      I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



      1) $sum_{n=0}^{infty}x^{n!}$

      2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$

      3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$




      1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



      Thus the radius of convergence is 1.

      On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



      2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

      We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



      3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


      On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.




      Are my results correct ? If not, what did I do wrong or forgot ?

      Thanks for your help !






      real-analysis sequences-and-series convergence power-series divergent-series




      real-analysis sequences-and-series convergence power-series divergent-series






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      share|cite|improve this question













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      edited Dec 20 '18 at 13:04









      David G. Stork

      10.9k31432




      10.9k31432










      asked Dec 20 '18 at 12:58









      PoujhPoujh

      612516




      612516






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42



















          1












          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22













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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42
















          1












          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42














          1












          1








          1





          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$



          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 17:30









          user10354138

          7,3872925




          7,3872925










          answered Dec 20 '18 at 13:09









          hamam_Abdallahhamam_Abdallah

          38k21634




          38k21634












          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42


















          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42
















          $begingroup$
          Something is wrong with the total rep.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:13




          $begingroup$
          Something is wrong with the total rep.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:13












          $begingroup$
          What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:15




          $begingroup$
          What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:15












          $begingroup$
          @Poujh This post if for managers.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:16




          $begingroup$
          @Poujh This post if for managers.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:16




          1




          1




          $begingroup$
          @Poujh Your answerS are correct.but no need for the border.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:19




          $begingroup$
          @Poujh Your answerS are correct.but no need for the border.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:19




          1




          1




          $begingroup$
          I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
          $endgroup$
          – R.Jackson
          Dec 20 '18 at 13:42




          $begingroup$
          I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
          $endgroup$
          – R.Jackson
          Dec 20 '18 at 13:42











          1












          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22


















          1












          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22
















          1












          1








          1





          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$



          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 13:30

























          answered Dec 20 '18 at 13:10









          Yadati KiranYadati Kiran

          1,763619




          1,763619












          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22




















          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22


















          $begingroup$
          Just for 1), do you think my method would also be correct (even if yours is easier) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:19






          $begingroup$
          Just for 1), do you think my method would also be correct (even if yours is easier) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:19






          1




          1




          $begingroup$
          @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
          $endgroup$
          – Yadati Kiran
          Dec 20 '18 at 13:22






          $begingroup$
          @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
          $endgroup$
          – Yadati Kiran
          Dec 20 '18 at 13:22




















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