Finding the Laurent series (complex numbers)












4












$begingroup$


I have
$$
f(z)={frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=infty$.
I did
$$
{frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.






complex-analysis complex-numbers laurent-series







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    I have
    $$
    f(z)={frac{1}{z(1-z)}}
    $$
    Need to find the Laurent series around $z=0, z=1, z=infty$.
    I did
    $$
    {frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
    $$
    and found $A=1, B=1$. Therefore we get
    $$
    {frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
    $$
    But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.






    complex-analysis complex-numbers laurent-series







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      I have
      $$
      f(z)={frac{1}{z(1-z)}}
      $$
      Need to find the Laurent series around $z=0, z=1, z=infty$.
      I did
      $$
      {frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
      $$
      and found $A=1, B=1$. Therefore we get
      $$
      {frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
      $$
      But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.






      complex-analysis complex-numbers laurent-series







      share|cite|improve this question









      $endgroup$




      I have
      $$
      f(z)={frac{1}{z(1-z)}}
      $$
      Need to find the Laurent series around $z=0, z=1, z=infty$.
      I did
      $$
      {frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
      $$
      and found $A=1, B=1$. Therefore we get
      $$
      {frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
      $$
      But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.






      complex-analysis complex-numbers laurent-series




      complex-analysis complex-numbers laurent-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 17:56









      user3132457user3132457

      1336




      1336






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          Hints :



          For the $z=1$ case :



          You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :



          $$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
          $$=$$
          $$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$



          Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :



          $$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$



          For the $infty$ case :



          Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :



          $$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$



          Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
            $endgroup$
            – Mark Viola
            Dec 18 '18 at 18:13












          • $begingroup$
            @MarkViola I am working in an edit from earlier since I saw my error.
            $endgroup$
            – Rebellos
            Dec 18 '18 at 18:14










          • $begingroup$
            @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
            $endgroup$
            – Rebellos
            Dec 18 '18 at 18:21












          • $begingroup$
            I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
            $endgroup$
            – Mark Viola
            Dec 18 '18 at 18:24



















          3












          $begingroup$

          In the annulus $1<|z|<infty$, we have



          $$begin{align}
          frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
          &=frac{1}{1+(z-1)}+frac1{1-z}\\
          &=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
          &=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
          &=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
          end{align}$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            We have
            $$
            eqalign{
            & {1 over {zleft( {1 - z} right)}} = cr
            & = left{ matrix{
            - left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
            Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
            {1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
            - sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
            - left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
            Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
            $$






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Hints :



              For the $z=1$ case :



              You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :



              $$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
              $$=$$
              $$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$



              Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :



              $$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$



              For the $infty$ case :



              Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :



              $$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$



              Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:13












              • $begingroup$
                @MarkViola I am working in an edit from earlier since I saw my error.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:14










              • $begingroup$
                @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:21












              • $begingroup$
                I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:24
















              3












              $begingroup$

              Hints :



              For the $z=1$ case :



              You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :



              $$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
              $$=$$
              $$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$



              Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :



              $$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$



              For the $infty$ case :



              Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :



              $$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$



              Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:13












              • $begingroup$
                @MarkViola I am working in an edit from earlier since I saw my error.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:14










              • $begingroup$
                @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:21












              • $begingroup$
                I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:24














              3












              3








              3





              $begingroup$

              Hints :



              For the $z=1$ case :



              You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :



              $$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
              $$=$$
              $$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$



              Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :



              $$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$



              For the $infty$ case :



              Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :



              $$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$



              Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.






              share|cite|improve this answer











              $endgroup$



              Hints :



              For the $z=1$ case :



              You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :



              $$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
              $$=$$
              $$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$



              Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :



              $$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$



              For the $infty$ case :



              Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :



              $$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$



              Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 18 '18 at 18:18

























              answered Dec 18 '18 at 18:02









              RebellosRebellos

              14.5k31246




              14.5k31246












              • $begingroup$
                That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:13












              • $begingroup$
                @MarkViola I am working in an edit from earlier since I saw my error.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:14










              • $begingroup$
                @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:21












              • $begingroup$
                I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:24


















              • $begingroup$
                That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:13












              • $begingroup$
                @MarkViola I am working in an edit from earlier since I saw my error.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:14










              • $begingroup$
                @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
                $endgroup$
                – Rebellos
                Dec 18 '18 at 18:21












              • $begingroup$
                I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
                $endgroup$
                – Mark Viola
                Dec 18 '18 at 18:24
















              $begingroup$
              That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
              $endgroup$
              – Mark Viola
              Dec 18 '18 at 18:13






              $begingroup$
              That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
              $endgroup$
              – Mark Viola
              Dec 18 '18 at 18:13














              $begingroup$
              @MarkViola I am working in an edit from earlier since I saw my error.
              $endgroup$
              – Rebellos
              Dec 18 '18 at 18:14




              $begingroup$
              @MarkViola I am working in an edit from earlier since I saw my error.
              $endgroup$
              – Rebellos
              Dec 18 '18 at 18:14












              $begingroup$
              @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
              $endgroup$
              – Rebellos
              Dec 18 '18 at 18:21






              $begingroup$
              @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
              $endgroup$
              – Rebellos
              Dec 18 '18 at 18:21














              $begingroup$
              I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
              $endgroup$
              – Mark Viola
              Dec 18 '18 at 18:24




              $begingroup$
              I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
              $endgroup$
              – Mark Viola
              Dec 18 '18 at 18:24











              3












              $begingroup$

              In the annulus $1<|z|<infty$, we have



              $$begin{align}
              frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
              &=frac{1}{1+(z-1)}+frac1{1-z}\\
              &=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
              &=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
              &=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
              end{align}$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                In the annulus $1<|z|<infty$, we have



                $$begin{align}
                frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
                &=frac{1}{1+(z-1)}+frac1{1-z}\\
                &=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
                &=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
                &=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
                end{align}$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  In the annulus $1<|z|<infty$, we have



                  $$begin{align}
                  frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
                  &=frac{1}{1+(z-1)}+frac1{1-z}\\
                  &=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
                  &=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
                  &=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
                  end{align}$$






                  share|cite|improve this answer









                  $endgroup$



                  In the annulus $1<|z|<infty$, we have



                  $$begin{align}
                  frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
                  &=frac{1}{1+(z-1)}+frac1{1-z}\\
                  &=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
                  &=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
                  &=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
                  end{align}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 18:10









                  Mark ViolaMark Viola

                  131k1275171




                  131k1275171























                      0












                      $begingroup$

                      We have
                      $$
                      eqalign{
                      & {1 over {zleft( {1 - z} right)}} = cr
                      & = left{ matrix{
                      - left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
                      Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
                      {1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
                      - sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
                      - left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
                      Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        We have
                        $$
                        eqalign{
                        & {1 over {zleft( {1 - z} right)}} = cr
                        & = left{ matrix{
                        - left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
                        Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
                        {1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
                        - sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
                        - left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
                        Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
                        $$






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                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          We have
                          $$
                          eqalign{
                          & {1 over {zleft( {1 - z} right)}} = cr
                          & = left{ matrix{
                          - left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
                          Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
                          {1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
                          - sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
                          - left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
                          Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          We have
                          $$
                          eqalign{
                          & {1 over {zleft( {1 - z} right)}} = cr
                          & = left{ matrix{
                          - left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
                          Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
                          {1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
                          - sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
                          - left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
                          Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
                          $$







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                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 18 '18 at 18:28









                          G CabG Cab

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