Exponential series with $k$ as base [duplicate]












3












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This question already has an answer here:




  • Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$

    2 answers




I've tried to understand why $displaystylesum_{k=0}^{infty} frac{k^x}{k!}$ for lets say $x = 4$ equals $15e$.



It's clear why $displaystylesum_{k=0}^{infty} frac{x^k}{k!} = e^x$ and that $displaystylesum_{k=0}^{infty} frac{1^k}{k!}=e$



It's also unclear for me why $displaystylesum_{k=0}^{infty} frac{k}{k!}=e$



I've tried to argue that $displaystylesum_{k=0}^{infty} frac{e^k}{k!}= sum_{k=0}^{infty} frac{k}{ln(k!)}$ but that doesn't bring me further.



Hope someone here has got an idea for me



Thanks.






real-analysis exponential-function







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marked as duplicate by max_zorn, Lord Shark the Unknown, Claude Leibovici, Kemono Chen, Jyrki Lahtonen Jan 23 at 5:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    Maybe this question could help you understand how to prove the result math.stackexchange.com/questions/503451/…
    $endgroup$
    – Larry
    Jan 22 at 23:59










  • $begingroup$
    Thank you! It's worth noting that i have searched in the search bar for this series, no clue why you found it and i didn't
    $endgroup$
    – Quotenbanane
    Jan 23 at 0:01






  • 1




    $begingroup$
    It's quite a funny story. I have the same experience with you. I asked a question similar to yours, and people suggested me the link. I also didn't find this link when I tried to search it. Still, it is a good question.
    $endgroup$
    – Larry
    Jan 23 at 0:05












  • $begingroup$
    Seems like a broken search system. Glad to have people like you!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:33
















3












$begingroup$



This question already has an answer here:




  • Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$

    2 answers




I've tried to understand why $displaystylesum_{k=0}^{infty} frac{k^x}{k!}$ for lets say $x = 4$ equals $15e$.



It's clear why $displaystylesum_{k=0}^{infty} frac{x^k}{k!} = e^x$ and that $displaystylesum_{k=0}^{infty} frac{1^k}{k!}=e$



It's also unclear for me why $displaystylesum_{k=0}^{infty} frac{k}{k!}=e$



I've tried to argue that $displaystylesum_{k=0}^{infty} frac{e^k}{k!}= sum_{k=0}^{infty} frac{k}{ln(k!)}$ but that doesn't bring me further.



Hope someone here has got an idea for me



Thanks.






real-analysis exponential-function







share|cite|improve this question











$endgroup$



marked as duplicate by max_zorn, Lord Shark the Unknown, Claude Leibovici, Kemono Chen, Jyrki Lahtonen Jan 23 at 5:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    Maybe this question could help you understand how to prove the result math.stackexchange.com/questions/503451/…
    $endgroup$
    – Larry
    Jan 22 at 23:59










  • $begingroup$
    Thank you! It's worth noting that i have searched in the search bar for this series, no clue why you found it and i didn't
    $endgroup$
    – Quotenbanane
    Jan 23 at 0:01






  • 1




    $begingroup$
    It's quite a funny story. I have the same experience with you. I asked a question similar to yours, and people suggested me the link. I also didn't find this link when I tried to search it. Still, it is a good question.
    $endgroup$
    – Larry
    Jan 23 at 0:05












  • $begingroup$
    Seems like a broken search system. Glad to have people like you!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:33














3












3








3


1



$begingroup$



This question already has an answer here:




  • Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$

    2 answers




I've tried to understand why $displaystylesum_{k=0}^{infty} frac{k^x}{k!}$ for lets say $x = 4$ equals $15e$.



It's clear why $displaystylesum_{k=0}^{infty} frac{x^k}{k!} = e^x$ and that $displaystylesum_{k=0}^{infty} frac{1^k}{k!}=e$



It's also unclear for me why $displaystylesum_{k=0}^{infty} frac{k}{k!}=e$



I've tried to argue that $displaystylesum_{k=0}^{infty} frac{e^k}{k!}= sum_{k=0}^{infty} frac{k}{ln(k!)}$ but that doesn't bring me further.



Hope someone here has got an idea for me



Thanks.






real-analysis exponential-function







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$

    2 answers




I've tried to understand why $displaystylesum_{k=0}^{infty} frac{k^x}{k!}$ for lets say $x = 4$ equals $15e$.



It's clear why $displaystylesum_{k=0}^{infty} frac{x^k}{k!} = e^x$ and that $displaystylesum_{k=0}^{infty} frac{1^k}{k!}=e$



It's also unclear for me why $displaystylesum_{k=0}^{infty} frac{k}{k!}=e$



I've tried to argue that $displaystylesum_{k=0}^{infty} frac{e^k}{k!}= sum_{k=0}^{infty} frac{k}{ln(k!)}$ but that doesn't bring me further.



Hope someone here has got an idea for me



Thanks.





This question already has an answer here:




  • Infinite Series $sumlimits_{k=1}^{infty}frac{k^n}{k!}$

    2 answers







real-analysis exponential-function




real-analysis exponential-function






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share|cite|improve this question













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share|cite|improve this question








edited Jan 23 at 6:11









El borito

666216




666216










asked Jan 22 at 23:54









QuotenbananeQuotenbanane

183




183




marked as duplicate by max_zorn, Lord Shark the Unknown, Claude Leibovici, Kemono Chen, Jyrki Lahtonen Jan 23 at 5:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by max_zorn, Lord Shark the Unknown, Claude Leibovici, Kemono Chen, Jyrki Lahtonen Jan 23 at 5:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Maybe this question could help you understand how to prove the result math.stackexchange.com/questions/503451/…
    $endgroup$
    – Larry
    Jan 22 at 23:59










  • $begingroup$
    Thank you! It's worth noting that i have searched in the search bar for this series, no clue why you found it and i didn't
    $endgroup$
    – Quotenbanane
    Jan 23 at 0:01






  • 1




    $begingroup$
    It's quite a funny story. I have the same experience with you. I asked a question similar to yours, and people suggested me the link. I also didn't find this link when I tried to search it. Still, it is a good question.
    $endgroup$
    – Larry
    Jan 23 at 0:05












  • $begingroup$
    Seems like a broken search system. Glad to have people like you!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:33














  • 1




    $begingroup$
    Maybe this question could help you understand how to prove the result math.stackexchange.com/questions/503451/…
    $endgroup$
    – Larry
    Jan 22 at 23:59










  • $begingroup$
    Thank you! It's worth noting that i have searched in the search bar for this series, no clue why you found it and i didn't
    $endgroup$
    – Quotenbanane
    Jan 23 at 0:01






  • 1




    $begingroup$
    It's quite a funny story. I have the same experience with you. I asked a question similar to yours, and people suggested me the link. I also didn't find this link when I tried to search it. Still, it is a good question.
    $endgroup$
    – Larry
    Jan 23 at 0:05












  • $begingroup$
    Seems like a broken search system. Glad to have people like you!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:33








1




1




$begingroup$
Maybe this question could help you understand how to prove the result math.stackexchange.com/questions/503451/…
$endgroup$
– Larry
Jan 22 at 23:59




$begingroup$
Maybe this question could help you understand how to prove the result math.stackexchange.com/questions/503451/…
$endgroup$
– Larry
Jan 22 at 23:59












$begingroup$
Thank you! It's worth noting that i have searched in the search bar for this series, no clue why you found it and i didn't
$endgroup$
– Quotenbanane
Jan 23 at 0:01




$begingroup$
Thank you! It's worth noting that i have searched in the search bar for this series, no clue why you found it and i didn't
$endgroup$
– Quotenbanane
Jan 23 at 0:01




1




1




$begingroup$
It's quite a funny story. I have the same experience with you. I asked a question similar to yours, and people suggested me the link. I also didn't find this link when I tried to search it. Still, it is a good question.
$endgroup$
– Larry
Jan 23 at 0:05






$begingroup$
It's quite a funny story. I have the same experience with you. I asked a question similar to yours, and people suggested me the link. I also didn't find this link when I tried to search it. Still, it is a good question.
$endgroup$
– Larry
Jan 23 at 0:05














$begingroup$
Seems like a broken search system. Glad to have people like you!
$endgroup$
– Quotenbanane
Jan 23 at 13:33




$begingroup$
Seems like a broken search system. Glad to have people like you!
$endgroup$
– Quotenbanane
Jan 23 at 13:33










1 Answer
1






active

oldest

votes


















2












$begingroup$

It is very more simple than you think, it is only a recursive propertie. When $x=1$



$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{k!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{1}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=0}^{infty}frac{1}{k!}=e
$$
When $x=2$:



$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k+1}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1$:
$$
sum_{k=0}^{infty}frac{k^2}{k!}=e+e=2e
$$
When $x=3$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{(k+1)^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{k^2}{k!}+2sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=2$ and $x=1$:
$$
sum_{k=0}^{infty}frac{k^3}{k!}=2e+2e+e=5e
$$
When $x=4$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^4}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{(k+1)^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{k^3}{k!}+3sum_{k=0}^{infty}frac{k^2}{k!}+3sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1,2,3$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=5e+6e+3e+e=15e
$$
And that's all.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:31


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

It is very more simple than you think, it is only a recursive propertie. When $x=1$



$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{k!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{1}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=0}^{infty}frac{1}{k!}=e
$$
When $x=2$:



$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k+1}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1$:
$$
sum_{k=0}^{infty}frac{k^2}{k!}=e+e=2e
$$
When $x=3$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{(k+1)^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{k^2}{k!}+2sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=2$ and $x=1$:
$$
sum_{k=0}^{infty}frac{k^3}{k!}=2e+2e+e=5e
$$
When $x=4$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^4}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{(k+1)^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{k^3}{k!}+3sum_{k=0}^{infty}frac{k^2}{k!}+3sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1,2,3$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=5e+6e+3e+e=15e
$$
And that's all.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:31
















2












$begingroup$

It is very more simple than you think, it is only a recursive propertie. When $x=1$



$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{k!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{1}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=0}^{infty}frac{1}{k!}=e
$$
When $x=2$:



$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k+1}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1$:
$$
sum_{k=0}^{infty}frac{k^2}{k!}=e+e=2e
$$
When $x=3$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{(k+1)^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{k^2}{k!}+2sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=2$ and $x=1$:
$$
sum_{k=0}^{infty}frac{k^3}{k!}=2e+2e+e=5e
$$
When $x=4$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^4}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{(k+1)^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{k^3}{k!}+3sum_{k=0}^{infty}frac{k^2}{k!}+3sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1,2,3$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=5e+6e+3e+e=15e
$$
And that's all.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:31














2












2








2





$begingroup$

It is very more simple than you think, it is only a recursive propertie. When $x=1$



$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{k!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{1}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=0}^{infty}frac{1}{k!}=e
$$
When $x=2$:



$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k+1}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1$:
$$
sum_{k=0}^{infty}frac{k^2}{k!}=e+e=2e
$$
When $x=3$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{(k+1)^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{k^2}{k!}+2sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=2$ and $x=1$:
$$
sum_{k=0}^{infty}frac{k^3}{k!}=2e+2e+e=5e
$$
When $x=4$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^4}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{(k+1)^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{k^3}{k!}+3sum_{k=0}^{infty}frac{k^2}{k!}+3sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1,2,3$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=5e+6e+3e+e=15e
$$
And that's all.






share|cite|improve this answer











$endgroup$



It is very more simple than you think, it is only a recursive propertie. When $x=1$



$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{k!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=1}^{infty}frac{1}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k}{k!}=sum_{k=0}^{infty}frac{1}{k!}=e
$$
When $x=2$:



$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=1}^{infty}frac{k}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k+1}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^2}{k!}=sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1$:
$$
sum_{k=0}^{infty}frac{k^2}{k!}=e+e=2e
$$
When $x=3$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!,k}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=1}^{infty}frac{k^2}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{(k+1)^2}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^3}{k!}=sum_{k=0}^{infty}frac{k^2}{k!}+2sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=2$ and $x=1$:
$$
sum_{k=0}^{infty}frac{k^3}{k!}=2e+2e+e=5e
$$
When $x=4$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^4}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=1}^{infty}frac{k^3}{(k-1)!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{(k+1)^3}{k!}
$$
$$
sum_{k=0}^{infty}frac{k^4}{k!}=sum_{k=0}^{infty}frac{k^3}{k!}+3sum_{k=0}^{infty}frac{k^2}{k!}+3sum_{k=0}^{infty}frac{k}{k!}+sum_{k=0}^{infty}frac{1}{k!}
$$
From $x=1,2,3$:
$$
sum_{k=0}^{infty}frac{k^4}{k!}=5e+6e+3e+e=15e
$$
And that's all.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 23 at 11:26

























answered Jan 23 at 4:41









El boritoEl borito

666216




666216








  • 1




    $begingroup$
    My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:31














  • 1




    $begingroup$
    My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks!
    $endgroup$
    – Quotenbanane
    Jan 23 at 13:31








1




1




$begingroup$
My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks!
$endgroup$
– Quotenbanane
Jan 23 at 13:31




$begingroup$
My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks!
$endgroup$
– Quotenbanane
Jan 23 at 13:31



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