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I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$



I do as follow




$sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$




I couldn't able to get the right hand



Any help will be appreciated! Thanks






calculus summation







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    0












    $begingroup$


    I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$



    I do as follow




    $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$




    I couldn't able to get the right hand



    Any help will be appreciated! Thanks






    calculus summation







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      0












      0








      0


      1



      $begingroup$


      I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$



      I do as follow




      $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$




      I couldn't able to get the right hand



      Any help will be appreciated! Thanks






      calculus summation







      share|cite|improve this question









      $endgroup$




      I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$



      I do as follow




      $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$




      I couldn't able to get the right hand



      Any help will be appreciated! Thanks






      calculus summation




      calculus summation






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      asked Dec 17 '18 at 6:53









      user62498user62498

      1,983714




      1,983714






















          3 Answers
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          0












          $begingroup$

          You might find it easier to start from the RHS and show that



          $$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$



          Actually, for me this is the only step. It's just how you multiply two series.



          Something more interesting would be to see what you can make of



          $$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$



          and the Binomial Theorem.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Hint



            Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
              $endgroup$
              – user62498
              Dec 17 '18 at 7:37





















            0












            $begingroup$

            By hint Claude Leibovici,




            $ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand



            $e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$




            Please let me know if this makes sense to you






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

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              active

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              active

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              0












              $begingroup$

              You might find it easier to start from the RHS and show that



              $$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$



              Actually, for me this is the only step. It's just how you multiply two series.



              Something more interesting would be to see what you can make of



              $$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$



              and the Binomial Theorem.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You might find it easier to start from the RHS and show that



                $$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$



                Actually, for me this is the only step. It's just how you multiply two series.



                Something more interesting would be to see what you can make of



                $$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$



                and the Binomial Theorem.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You might find it easier to start from the RHS and show that



                  $$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$



                  Actually, for me this is the only step. It's just how you multiply two series.



                  Something more interesting would be to see what you can make of



                  $$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$



                  and the Binomial Theorem.






                  share|cite|improve this answer









                  $endgroup$



                  You might find it easier to start from the RHS and show that



                  $$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$



                  Actually, for me this is the only step. It's just how you multiply two series.



                  Something more interesting would be to see what you can make of



                  $$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$



                  and the Binomial Theorem.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 7:12









                  Trevor GunnTrevor Gunn

                  14.9k32047




                  14.9k32047























                      1












                      $begingroup$

                      Hint



                      Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
                        $endgroup$
                        – user62498
                        Dec 17 '18 at 7:37


















                      1












                      $begingroup$

                      Hint



                      Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
                        $endgroup$
                        – user62498
                        Dec 17 '18 at 7:37
















                      1












                      1








                      1





                      $begingroup$

                      Hint



                      Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.






                      share|cite|improve this answer









                      $endgroup$



                      Hint



                      Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 17 '18 at 7:15









                      Claude LeiboviciClaude Leibovici

                      124k1157135




                      124k1157135












                      • $begingroup$
                        I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
                        $endgroup$
                        – user62498
                        Dec 17 '18 at 7:37




















                      • $begingroup$
                        I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
                        $endgroup$
                        – user62498
                        Dec 17 '18 at 7:37


















                      $begingroup$
                      I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
                      $endgroup$
                      – user62498
                      Dec 17 '18 at 7:37






                      $begingroup$
                      I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
                      $endgroup$
                      – user62498
                      Dec 17 '18 at 7:37













                      0












                      $begingroup$

                      By hint Claude Leibovici,




                      $ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand



                      $e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$




                      Please let me know if this makes sense to you






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        By hint Claude Leibovici,




                        $ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand



                        $e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$




                        Please let me know if this makes sense to you






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          By hint Claude Leibovici,




                          $ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand



                          $e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$




                          Please let me know if this makes sense to you






                          share|cite|improve this answer









                          $endgroup$



                          By hint Claude Leibovici,




                          $ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand



                          $e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$




                          Please let me know if this makes sense to you







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 17 '18 at 8:04









                          user62498user62498

                          1,983714




                          1,983714






























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