Proof that for infinite sum: $sum_{mgeq n}^infty frac{(lambda (1-p)t)^m}{(m-n)!}=((1-p)lambda t)^n...











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How can I prove that:



$$sum_{mgeq n}^infty frac{(lambda (1-p)t)^m}{(m-n)!}=((1-p)lambda t)^n e^{(1-p)lambda t}$$






sequences-and-series







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  • The left side not even defined.
    – Kavi Rama Murthy
    2 days ago










  • @KaviRamaMurthy Better now?
    – Dole
    2 days ago










  • Perhaps the denominator is $(m - n)!$.
    – Awe Kumar Jha
    2 days ago










  • Is $n$ a free choice, or should you be considering $sum_n sum_{m geq n} ldots$?
    – Kevin
    2 days ago

















up vote
0
down vote

favorite












How can I prove that:



$$sum_{mgeq n}^infty frac{(lambda (1-p)t)^m}{(m-n)!}=((1-p)lambda t)^n e^{(1-p)lambda t}$$






sequences-and-series







share|cite|improve this question
























  • The left side not even defined.
    – Kavi Rama Murthy
    2 days ago










  • @KaviRamaMurthy Better now?
    – Dole
    2 days ago










  • Perhaps the denominator is $(m - n)!$.
    – Awe Kumar Jha
    2 days ago










  • Is $n$ a free choice, or should you be considering $sum_n sum_{m geq n} ldots$?
    – Kevin
    2 days ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











How can I prove that:



$$sum_{mgeq n}^infty frac{(lambda (1-p)t)^m}{(m-n)!}=((1-p)lambda t)^n e^{(1-p)lambda t}$$






sequences-and-series







share|cite|improve this question















How can I prove that:



$$sum_{mgeq n}^infty frac{(lambda (1-p)t)^m}{(m-n)!}=((1-p)lambda t)^n e^{(1-p)lambda t}$$






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited 2 days ago

























asked 2 days ago









Dole

865514




865514












  • The left side not even defined.
    – Kavi Rama Murthy
    2 days ago










  • @KaviRamaMurthy Better now?
    – Dole
    2 days ago










  • Perhaps the denominator is $(m - n)!$.
    – Awe Kumar Jha
    2 days ago










  • Is $n$ a free choice, or should you be considering $sum_n sum_{m geq n} ldots$?
    – Kevin
    2 days ago




















  • The left side not even defined.
    – Kavi Rama Murthy
    2 days ago










  • @KaviRamaMurthy Better now?
    – Dole
    2 days ago










  • Perhaps the denominator is $(m - n)!$.
    – Awe Kumar Jha
    2 days ago










  • Is $n$ a free choice, or should you be considering $sum_n sum_{m geq n} ldots$?
    – Kevin
    2 days ago


















The left side not even defined.
– Kavi Rama Murthy
2 days ago




The left side not even defined.
– Kavi Rama Murthy
2 days ago












@KaviRamaMurthy Better now?
– Dole
2 days ago




@KaviRamaMurthy Better now?
– Dole
2 days ago












Perhaps the denominator is $(m - n)!$.
– Awe Kumar Jha
2 days ago




Perhaps the denominator is $(m - n)!$.
– Awe Kumar Jha
2 days ago












Is $n$ a free choice, or should you be considering $sum_n sum_{m geq n} ldots$?
– Kevin
2 days ago






Is $n$ a free choice, or should you be considering $sum_n sum_{m geq n} ldots$?
– Kevin
2 days ago












1 Answer
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I am interpreting $m-n!$ as $(m-n)!$. Just change the variable from $m$ to $j=m-n$. Let $x=lambda (1-p)t$. You will get $sum_{j=1}^{infty} frac {x^{j+n}} {j!}$. Pull out $x^{n}$ and use the series for $e^{x}$.






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    1 Answer
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    active

    oldest

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    1 Answer
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    active

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    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    I am interpreting $m-n!$ as $(m-n)!$. Just change the variable from $m$ to $j=m-n$. Let $x=lambda (1-p)t$. You will get $sum_{j=1}^{infty} frac {x^{j+n}} {j!}$. Pull out $x^{n}$ and use the series for $e^{x}$.






    share|cite|improve this answer

























      up vote
      5
      down vote



      accepted










      I am interpreting $m-n!$ as $(m-n)!$. Just change the variable from $m$ to $j=m-n$. Let $x=lambda (1-p)t$. You will get $sum_{j=1}^{infty} frac {x^{j+n}} {j!}$. Pull out $x^{n}$ and use the series for $e^{x}$.






      share|cite|improve this answer























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        I am interpreting $m-n!$ as $(m-n)!$. Just change the variable from $m$ to $j=m-n$. Let $x=lambda (1-p)t$. You will get $sum_{j=1}^{infty} frac {x^{j+n}} {j!}$. Pull out $x^{n}$ and use the series for $e^{x}$.






        share|cite|improve this answer












        I am interpreting $m-n!$ as $(m-n)!$. Just change the variable from $m$ to $j=m-n$. Let $x=lambda (1-p)t$. You will get $sum_{j=1}^{infty} frac {x^{j+n}} {j!}$. Pull out $x^{n}$ and use the series for $e^{x}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Kavi Rama Murthy

        39.1k31748




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