For $(z_n)$ complex, $limsuplimits_nn|z_n-1|$ finite iff $limsuplimits_nn|log(z_n)|$ finite












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Show that for $(z_n)_{ninmathbb N}$ in $mathbb C,$ $$limsuplimits_{n}n|z_n-1|<inftyifflimsuplimits_{n}|nlog(z_n)|<infty$$
And further that $$limlimits_{n}n(z_n-1)=limlimits_{n}nlog(z_n)$$ if one of the limits exists.




There is a hint that I should use the logarithmic series $sum_{n=1 } ^{infty } (-frac {1 } {k })^{k-1}(z-1)^k$ about the point $1 $. From which I can derive the inequality $|log(z)-(z-1)|le frac {1 } {2 } |z-1|^2 $for $|z-1|<1$. But here I get stuck.






sequences-and-series inequality complex-numbers







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    $begingroup$



    Show that for $(z_n)_{ninmathbb N}$ in $mathbb C,$ $$limsuplimits_{n}n|z_n-1|<inftyifflimsuplimits_{n}|nlog(z_n)|<infty$$
    And further that $$limlimits_{n}n(z_n-1)=limlimits_{n}nlog(z_n)$$ if one of the limits exists.




    There is a hint that I should use the logarithmic series $sum_{n=1 } ^{infty } (-frac {1 } {k })^{k-1}(z-1)^k$ about the point $1 $. From which I can derive the inequality $|log(z)-(z-1)|le frac {1 } {2 } |z-1|^2 $for $|z-1|<1$. But here I get stuck.






    sequences-and-series inequality complex-numbers







    share|cite|improve this question











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      0












      0








      0





      $begingroup$



      Show that for $(z_n)_{ninmathbb N}$ in $mathbb C,$ $$limsuplimits_{n}n|z_n-1|<inftyifflimsuplimits_{n}|nlog(z_n)|<infty$$
      And further that $$limlimits_{n}n(z_n-1)=limlimits_{n}nlog(z_n)$$ if one of the limits exists.




      There is a hint that I should use the logarithmic series $sum_{n=1 } ^{infty } (-frac {1 } {k })^{k-1}(z-1)^k$ about the point $1 $. From which I can derive the inequality $|log(z)-(z-1)|le frac {1 } {2 } |z-1|^2 $for $|z-1|<1$. But here I get stuck.






      sequences-and-series inequality complex-numbers







      share|cite|improve this question











      $endgroup$





      Show that for $(z_n)_{ninmathbb N}$ in $mathbb C,$ $$limsuplimits_{n}n|z_n-1|<inftyifflimsuplimits_{n}|nlog(z_n)|<infty$$
      And further that $$limlimits_{n}n(z_n-1)=limlimits_{n}nlog(z_n)$$ if one of the limits exists.




      There is a hint that I should use the logarithmic series $sum_{n=1 } ^{infty } (-frac {1 } {k })^{k-1}(z-1)^k$ about the point $1 $. From which I can derive the inequality $|log(z)-(z-1)|le frac {1 } {2 } |z-1|^2 $for $|z-1|<1$. But here I get stuck.






      sequences-and-series inequality complex-numbers




      sequences-and-series inequality complex-numbers






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 17 '18 at 8:29









      Did

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      asked Dec 17 '18 at 7:08









      MrFranzénMrFranzén

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          2 Answers
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          1












          $begingroup$

          I will give an answer using a slightly modified form of the hint. Here log denotes the principal branch of logarithm. We have $log (z-1)=(z-1)-frac {(z-1)^{2}} 2+frac {(z-1)^{3}} 2cdots$. Hence $|log(z-1)-(z-1)| leq |z-1|^{2}+ |z-1|^{3}+cdots =frac {|z-1|^{2}} {1-|z-1|}$. Note that if any of the limits in the question exist then $z_n to 1$. Hence, for $n$ sufficiently large we have $|z_n-1|<frac 1 2$. This gives $|log(z-1)-(z-1)| < 2|z_n-1|^{2}$ for such $n$. From this it should be fairly easy to draw the desired conclusions. [If $lim sup n|z_n-1|< infty$ there $|z_n-1|leq C/n$ for some constant $C$so $n|z_n-1|^{2} leq n(frac C n)^{2} to 0$ and this gives $|nlog(z_n)| leq n|z_n-1|+2n|z_n-1|^{2}$, etc].






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What if $limsup n |log(z_n)| < infty $? As you note this leads to that $z_n to o $ which means $n |z_n-1| to n times infty $. How do I get a stronger bound her?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 11:01












          • $begingroup$
            @MrFranzén If $n, log, z_n $ is bounded then $n log ,z_n to 0$ and so $z_n to 1$
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 11:43












          • $begingroup$
            Yes I'm with you on that but how do $z_n to 1 $ imply $lim sup n|z_n-1| < infty$. As one term decreases but the other increases it does not seem obious that $n|z_n-1| $ converges. How do I show it is bounded?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 15:44










          • $begingroup$
            @MrFranzén Use the main inequality I have derived. Use the fact that $nfrac {|z_n-1|^{2}} {1-|z_n-1|} to 0$ if $|z_n-1| <frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 23:10





















          1












          $begingroup$

          Hint. First of all
          $$
          limsup n|z_n-1|<infty quadLongleftrightarrowquad {n|z_n-1|},,,text{bounded sequence}
          $$
          and
          $$
          {n|z_n-1|},,,text{bounded} quadLongleftrightarrowquad |z_n-1|<frac{c}{n}, ,,,text{for some $c>0$}.
          $$
          Hence, $|z_n-1|<1/2$, for large enough $n$, i.e., $nge n_0$, which means that $logbig(1+(z_n-1)big)$ is expressible as a power seires
          $$
          log z_n=logbig(1+(z_n-1)big)=(z_n-1)-frac{(z_n-1)^2}{2}+frac{(z_n-1)^3}{3}+cdots
          $$
          and thus
          $$
          nlog z_n=n(z_n-1)-frac{n(z_n-1)^2}{2}+frac{n(z_n-1)^3}{3}+cdots
          $$
          and hence
          $$
          |nlog z_n-n(z_n-1)|lefrac{n|z_n-1|^2}{2}+frac{n|z_n-1|^3}{3}+cdots
          \ =frac{(n|z_n-1|)^2}{2n}+frac{(n|z_n-1|)^3}{3n^2}+cdots \ le
          frac{1}{2ncdot 2^2}+frac{1}{3n^2cdot 2^3}+cdots<frac{1}{n}left(frac{1}{2^2}+frac{1}{2^3}+cdotsright)=frac{1}{2n}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The next to last inequality means $n|z_n-1|le 2^{-1} $, why is this so?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 10:13













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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          I will give an answer using a slightly modified form of the hint. Here log denotes the principal branch of logarithm. We have $log (z-1)=(z-1)-frac {(z-1)^{2}} 2+frac {(z-1)^{3}} 2cdots$. Hence $|log(z-1)-(z-1)| leq |z-1|^{2}+ |z-1|^{3}+cdots =frac {|z-1|^{2}} {1-|z-1|}$. Note that if any of the limits in the question exist then $z_n to 1$. Hence, for $n$ sufficiently large we have $|z_n-1|<frac 1 2$. This gives $|log(z-1)-(z-1)| < 2|z_n-1|^{2}$ for such $n$. From this it should be fairly easy to draw the desired conclusions. [If $lim sup n|z_n-1|< infty$ there $|z_n-1|leq C/n$ for some constant $C$so $n|z_n-1|^{2} leq n(frac C n)^{2} to 0$ and this gives $|nlog(z_n)| leq n|z_n-1|+2n|z_n-1|^{2}$, etc].






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What if $limsup n |log(z_n)| < infty $? As you note this leads to that $z_n to o $ which means $n |z_n-1| to n times infty $. How do I get a stronger bound her?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 11:01












          • $begingroup$
            @MrFranzén If $n, log, z_n $ is bounded then $n log ,z_n to 0$ and so $z_n to 1$
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 11:43












          • $begingroup$
            Yes I'm with you on that but how do $z_n to 1 $ imply $lim sup n|z_n-1| < infty$. As one term decreases but the other increases it does not seem obious that $n|z_n-1| $ converges. How do I show it is bounded?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 15:44










          • $begingroup$
            @MrFranzén Use the main inequality I have derived. Use the fact that $nfrac {|z_n-1|^{2}} {1-|z_n-1|} to 0$ if $|z_n-1| <frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 23:10


















          1












          $begingroup$

          I will give an answer using a slightly modified form of the hint. Here log denotes the principal branch of logarithm. We have $log (z-1)=(z-1)-frac {(z-1)^{2}} 2+frac {(z-1)^{3}} 2cdots$. Hence $|log(z-1)-(z-1)| leq |z-1|^{2}+ |z-1|^{3}+cdots =frac {|z-1|^{2}} {1-|z-1|}$. Note that if any of the limits in the question exist then $z_n to 1$. Hence, for $n$ sufficiently large we have $|z_n-1|<frac 1 2$. This gives $|log(z-1)-(z-1)| < 2|z_n-1|^{2}$ for such $n$. From this it should be fairly easy to draw the desired conclusions. [If $lim sup n|z_n-1|< infty$ there $|z_n-1|leq C/n$ for some constant $C$so $n|z_n-1|^{2} leq n(frac C n)^{2} to 0$ and this gives $|nlog(z_n)| leq n|z_n-1|+2n|z_n-1|^{2}$, etc].






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What if $limsup n |log(z_n)| < infty $? As you note this leads to that $z_n to o $ which means $n |z_n-1| to n times infty $. How do I get a stronger bound her?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 11:01












          • $begingroup$
            @MrFranzén If $n, log, z_n $ is bounded then $n log ,z_n to 0$ and so $z_n to 1$
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 11:43












          • $begingroup$
            Yes I'm with you on that but how do $z_n to 1 $ imply $lim sup n|z_n-1| < infty$. As one term decreases but the other increases it does not seem obious that $n|z_n-1| $ converges. How do I show it is bounded?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 15:44










          • $begingroup$
            @MrFranzén Use the main inequality I have derived. Use the fact that $nfrac {|z_n-1|^{2}} {1-|z_n-1|} to 0$ if $|z_n-1| <frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 23:10
















          1












          1








          1





          $begingroup$

          I will give an answer using a slightly modified form of the hint. Here log denotes the principal branch of logarithm. We have $log (z-1)=(z-1)-frac {(z-1)^{2}} 2+frac {(z-1)^{3}} 2cdots$. Hence $|log(z-1)-(z-1)| leq |z-1|^{2}+ |z-1|^{3}+cdots =frac {|z-1|^{2}} {1-|z-1|}$. Note that if any of the limits in the question exist then $z_n to 1$. Hence, for $n$ sufficiently large we have $|z_n-1|<frac 1 2$. This gives $|log(z-1)-(z-1)| < 2|z_n-1|^{2}$ for such $n$. From this it should be fairly easy to draw the desired conclusions. [If $lim sup n|z_n-1|< infty$ there $|z_n-1|leq C/n$ for some constant $C$so $n|z_n-1|^{2} leq n(frac C n)^{2} to 0$ and this gives $|nlog(z_n)| leq n|z_n-1|+2n|z_n-1|^{2}$, etc].






          share|cite|improve this answer











          $endgroup$



          I will give an answer using a slightly modified form of the hint. Here log denotes the principal branch of logarithm. We have $log (z-1)=(z-1)-frac {(z-1)^{2}} 2+frac {(z-1)^{3}} 2cdots$. Hence $|log(z-1)-(z-1)| leq |z-1|^{2}+ |z-1|^{3}+cdots =frac {|z-1|^{2}} {1-|z-1|}$. Note that if any of the limits in the question exist then $z_n to 1$. Hence, for $n$ sufficiently large we have $|z_n-1|<frac 1 2$. This gives $|log(z-1)-(z-1)| < 2|z_n-1|^{2}$ for such $n$. From this it should be fairly easy to draw the desired conclusions. [If $lim sup n|z_n-1|< infty$ there $|z_n-1|leq C/n$ for some constant $C$so $n|z_n-1|^{2} leq n(frac C n)^{2} to 0$ and this gives $|nlog(z_n)| leq n|z_n-1|+2n|z_n-1|^{2}$, etc].







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 8:20

























          answered Dec 17 '18 at 8:11









          Kavi Rama MurthyKavi Rama Murthy

          66.6k53067




          66.6k53067












          • $begingroup$
            What if $limsup n |log(z_n)| < infty $? As you note this leads to that $z_n to o $ which means $n |z_n-1| to n times infty $. How do I get a stronger bound her?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 11:01












          • $begingroup$
            @MrFranzén If $n, log, z_n $ is bounded then $n log ,z_n to 0$ and so $z_n to 1$
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 11:43












          • $begingroup$
            Yes I'm with you on that but how do $z_n to 1 $ imply $lim sup n|z_n-1| < infty$. As one term decreases but the other increases it does not seem obious that $n|z_n-1| $ converges. How do I show it is bounded?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 15:44










          • $begingroup$
            @MrFranzén Use the main inequality I have derived. Use the fact that $nfrac {|z_n-1|^{2}} {1-|z_n-1|} to 0$ if $|z_n-1| <frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 23:10




















          • $begingroup$
            What if $limsup n |log(z_n)| < infty $? As you note this leads to that $z_n to o $ which means $n |z_n-1| to n times infty $. How do I get a stronger bound her?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 11:01












          • $begingroup$
            @MrFranzén If $n, log, z_n $ is bounded then $n log ,z_n to 0$ and so $z_n to 1$
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 11:43












          • $begingroup$
            Yes I'm with you on that but how do $z_n to 1 $ imply $lim sup n|z_n-1| < infty$. As one term decreases but the other increases it does not seem obious that $n|z_n-1| $ converges. How do I show it is bounded?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 15:44










          • $begingroup$
            @MrFranzén Use the main inequality I have derived. Use the fact that $nfrac {|z_n-1|^{2}} {1-|z_n-1|} to 0$ if $|z_n-1| <frac 1 2$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 17 '18 at 23:10


















          $begingroup$
          What if $limsup n |log(z_n)| < infty $? As you note this leads to that $z_n to o $ which means $n |z_n-1| to n times infty $. How do I get a stronger bound her?
          $endgroup$
          – MrFranzén
          Dec 17 '18 at 11:01






          $begingroup$
          What if $limsup n |log(z_n)| < infty $? As you note this leads to that $z_n to o $ which means $n |z_n-1| to n times infty $. How do I get a stronger bound her?
          $endgroup$
          – MrFranzén
          Dec 17 '18 at 11:01














          $begingroup$
          @MrFranzén If $n, log, z_n $ is bounded then $n log ,z_n to 0$ and so $z_n to 1$
          $endgroup$
          – Kavi Rama Murthy
          Dec 17 '18 at 11:43






          $begingroup$
          @MrFranzén If $n, log, z_n $ is bounded then $n log ,z_n to 0$ and so $z_n to 1$
          $endgroup$
          – Kavi Rama Murthy
          Dec 17 '18 at 11:43














          $begingroup$
          Yes I'm with you on that but how do $z_n to 1 $ imply $lim sup n|z_n-1| < infty$. As one term decreases but the other increases it does not seem obious that $n|z_n-1| $ converges. How do I show it is bounded?
          $endgroup$
          – MrFranzén
          Dec 17 '18 at 15:44




          $begingroup$
          Yes I'm with you on that but how do $z_n to 1 $ imply $lim sup n|z_n-1| < infty$. As one term decreases but the other increases it does not seem obious that $n|z_n-1| $ converges. How do I show it is bounded?
          $endgroup$
          – MrFranzén
          Dec 17 '18 at 15:44












          $begingroup$
          @MrFranzén Use the main inequality I have derived. Use the fact that $nfrac {|z_n-1|^{2}} {1-|z_n-1|} to 0$ if $|z_n-1| <frac 1 2$.
          $endgroup$
          – Kavi Rama Murthy
          Dec 17 '18 at 23:10






          $begingroup$
          @MrFranzén Use the main inequality I have derived. Use the fact that $nfrac {|z_n-1|^{2}} {1-|z_n-1|} to 0$ if $|z_n-1| <frac 1 2$.
          $endgroup$
          – Kavi Rama Murthy
          Dec 17 '18 at 23:10













          1












          $begingroup$

          Hint. First of all
          $$
          limsup n|z_n-1|<infty quadLongleftrightarrowquad {n|z_n-1|},,,text{bounded sequence}
          $$
          and
          $$
          {n|z_n-1|},,,text{bounded} quadLongleftrightarrowquad |z_n-1|<frac{c}{n}, ,,,text{for some $c>0$}.
          $$
          Hence, $|z_n-1|<1/2$, for large enough $n$, i.e., $nge n_0$, which means that $logbig(1+(z_n-1)big)$ is expressible as a power seires
          $$
          log z_n=logbig(1+(z_n-1)big)=(z_n-1)-frac{(z_n-1)^2}{2}+frac{(z_n-1)^3}{3}+cdots
          $$
          and thus
          $$
          nlog z_n=n(z_n-1)-frac{n(z_n-1)^2}{2}+frac{n(z_n-1)^3}{3}+cdots
          $$
          and hence
          $$
          |nlog z_n-n(z_n-1)|lefrac{n|z_n-1|^2}{2}+frac{n|z_n-1|^3}{3}+cdots
          \ =frac{(n|z_n-1|)^2}{2n}+frac{(n|z_n-1|)^3}{3n^2}+cdots \ le
          frac{1}{2ncdot 2^2}+frac{1}{3n^2cdot 2^3}+cdots<frac{1}{n}left(frac{1}{2^2}+frac{1}{2^3}+cdotsright)=frac{1}{2n}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The next to last inequality means $n|z_n-1|le 2^{-1} $, why is this so?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 10:13


















          1












          $begingroup$

          Hint. First of all
          $$
          limsup n|z_n-1|<infty quadLongleftrightarrowquad {n|z_n-1|},,,text{bounded sequence}
          $$
          and
          $$
          {n|z_n-1|},,,text{bounded} quadLongleftrightarrowquad |z_n-1|<frac{c}{n}, ,,,text{for some $c>0$}.
          $$
          Hence, $|z_n-1|<1/2$, for large enough $n$, i.e., $nge n_0$, which means that $logbig(1+(z_n-1)big)$ is expressible as a power seires
          $$
          log z_n=logbig(1+(z_n-1)big)=(z_n-1)-frac{(z_n-1)^2}{2}+frac{(z_n-1)^3}{3}+cdots
          $$
          and thus
          $$
          nlog z_n=n(z_n-1)-frac{n(z_n-1)^2}{2}+frac{n(z_n-1)^3}{3}+cdots
          $$
          and hence
          $$
          |nlog z_n-n(z_n-1)|lefrac{n|z_n-1|^2}{2}+frac{n|z_n-1|^3}{3}+cdots
          \ =frac{(n|z_n-1|)^2}{2n}+frac{(n|z_n-1|)^3}{3n^2}+cdots \ le
          frac{1}{2ncdot 2^2}+frac{1}{3n^2cdot 2^3}+cdots<frac{1}{n}left(frac{1}{2^2}+frac{1}{2^3}+cdotsright)=frac{1}{2n}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The next to last inequality means $n|z_n-1|le 2^{-1} $, why is this so?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 10:13
















          1












          1








          1





          $begingroup$

          Hint. First of all
          $$
          limsup n|z_n-1|<infty quadLongleftrightarrowquad {n|z_n-1|},,,text{bounded sequence}
          $$
          and
          $$
          {n|z_n-1|},,,text{bounded} quadLongleftrightarrowquad |z_n-1|<frac{c}{n}, ,,,text{for some $c>0$}.
          $$
          Hence, $|z_n-1|<1/2$, for large enough $n$, i.e., $nge n_0$, which means that $logbig(1+(z_n-1)big)$ is expressible as a power seires
          $$
          log z_n=logbig(1+(z_n-1)big)=(z_n-1)-frac{(z_n-1)^2}{2}+frac{(z_n-1)^3}{3}+cdots
          $$
          and thus
          $$
          nlog z_n=n(z_n-1)-frac{n(z_n-1)^2}{2}+frac{n(z_n-1)^3}{3}+cdots
          $$
          and hence
          $$
          |nlog z_n-n(z_n-1)|lefrac{n|z_n-1|^2}{2}+frac{n|z_n-1|^3}{3}+cdots
          \ =frac{(n|z_n-1|)^2}{2n}+frac{(n|z_n-1|)^3}{3n^2}+cdots \ le
          frac{1}{2ncdot 2^2}+frac{1}{3n^2cdot 2^3}+cdots<frac{1}{n}left(frac{1}{2^2}+frac{1}{2^3}+cdotsright)=frac{1}{2n}
          $$






          share|cite|improve this answer









          $endgroup$



          Hint. First of all
          $$
          limsup n|z_n-1|<infty quadLongleftrightarrowquad {n|z_n-1|},,,text{bounded sequence}
          $$
          and
          $$
          {n|z_n-1|},,,text{bounded} quadLongleftrightarrowquad |z_n-1|<frac{c}{n}, ,,,text{for some $c>0$}.
          $$
          Hence, $|z_n-1|<1/2$, for large enough $n$, i.e., $nge n_0$, which means that $logbig(1+(z_n-1)big)$ is expressible as a power seires
          $$
          log z_n=logbig(1+(z_n-1)big)=(z_n-1)-frac{(z_n-1)^2}{2}+frac{(z_n-1)^3}{3}+cdots
          $$
          and thus
          $$
          nlog z_n=n(z_n-1)-frac{n(z_n-1)^2}{2}+frac{n(z_n-1)^3}{3}+cdots
          $$
          and hence
          $$
          |nlog z_n-n(z_n-1)|lefrac{n|z_n-1|^2}{2}+frac{n|z_n-1|^3}{3}+cdots
          \ =frac{(n|z_n-1|)^2}{2n}+frac{(n|z_n-1|)^3}{3n^2}+cdots \ le
          frac{1}{2ncdot 2^2}+frac{1}{3n^2cdot 2^3}+cdots<frac{1}{n}left(frac{1}{2^2}+frac{1}{2^3}+cdotsright)=frac{1}{2n}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 8:49









          Yiorgos S. SmyrlisYiorgos S. Smyrlis

          63.4k1385164




          63.4k1385164












          • $begingroup$
            The next to last inequality means $n|z_n-1|le 2^{-1} $, why is this so?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 10:13




















          • $begingroup$
            The next to last inequality means $n|z_n-1|le 2^{-1} $, why is this so?
            $endgroup$
            – MrFranzén
            Dec 17 '18 at 10:13


















          $begingroup$
          The next to last inequality means $n|z_n-1|le 2^{-1} $, why is this so?
          $endgroup$
          – MrFranzén
          Dec 17 '18 at 10:13






          $begingroup$
          The next to last inequality means $n|z_n-1|le 2^{-1} $, why is this so?
          $endgroup$
          – MrFranzén
          Dec 17 '18 at 10:13




















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