Computation of a limit involving a series (related to Poisson distribution)












2












$begingroup$


Consider $lambda >0.$ I am reading a paper and the author states that



$$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$



I tried to compute such limit but I am getting anywhere. Someone could help me?



Thanks in advance!






analysis statistics







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$endgroup$

















    2












    $begingroup$


    Consider $lambda >0.$ I am reading a paper and the author states that



    $$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$



    I tried to compute such limit but I am getting anywhere. Someone could help me?



    Thanks in advance!






    analysis statistics







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Consider $lambda >0.$ I am reading a paper and the author states that



      $$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$



      I tried to compute such limit but I am getting anywhere. Someone could help me?



      Thanks in advance!






      analysis statistics







      share|cite|improve this question









      $endgroup$




      Consider $lambda >0.$ I am reading a paper and the author states that



      $$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$



      I tried to compute such limit but I am getting anywhere. Someone could help me?



      Thanks in advance!






      analysis statistics




      analysis statistics






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      asked 2 days ago









      math studentmath student

      2,39111018




      2,39111018






















          2 Answers
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          $begingroup$

          Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.






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            2












            $begingroup$

            The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).






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              2 Answers
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              active

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              active

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              6












              $begingroup$

              Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.






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                6












                $begingroup$

                Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.






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                  6












                  6








                  6





                  $begingroup$

                  Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.






                  share|cite|improve this answer









                  $endgroup$



                  Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.







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                  share|cite|improve this answer










                  answered 2 days ago









                  Mostafa AyazMostafa Ayaz

                  17k3939




                  17k3939























                      2












                      $begingroup$

                      The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).






                          share|cite|improve this answer









                          $endgroup$



                          The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 days ago









                          user66081user66081

                          3,2581126




                          3,2581126






























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