How can i put a log inside an infinite sum? [closed]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












iI would like to turn this



$$ e^z-1 = sum^infty_1 frac z^nn! $$



into this



$$ textsomething =sum^infty_1 frac z^nlog_e(n!) $$



Is this at all possible? thank you very much in advance!



Ps- i apologize of title not adequate , i did not know how to word it.




sequences-and-series complex-analysis complex-numbers infinity




share|cite|improve this question














closed as unclear what you're asking by Kavi Rama Murthy, Yves Daoust, Hans Lundmark, Lord Shark the Unknown, Jyrki Lahtonen Aug 24 at 14:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • First, the second series does not make sense since $log(0!) = log(1) = 0$ is in the denominator.
    – 4-ier
    Aug 24 at 5:55










  • i apologize , i just meant it as a general example , any value would work.
    – Victor Orta
    Aug 24 at 5:56






  • 2




    Are you asking to evaluate the modified sum ? What do you mean by "put a log inside" ???
    – Yves Daoust
    Aug 24 at 6:25















up vote
1
down vote

favorite












iI would like to turn this



$$ e^z-1 = sum^infty_1 frac z^nn! $$



into this



$$ textsomething =sum^infty_1 frac z^nlog_e(n!) $$



Is this at all possible? thank you very much in advance!



Ps- i apologize of title not adequate , i did not know how to word it.




sequences-and-series complex-analysis complex-numbers infinity




share|cite|improve this question














closed as unclear what you're asking by Kavi Rama Murthy, Yves Daoust, Hans Lundmark, Lord Shark the Unknown, Jyrki Lahtonen Aug 24 at 14:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • First, the second series does not make sense since $log(0!) = log(1) = 0$ is in the denominator.
    – 4-ier
    Aug 24 at 5:55










  • i apologize , i just meant it as a general example , any value would work.
    – Victor Orta
    Aug 24 at 5:56






  • 2




    Are you asking to evaluate the modified sum ? What do you mean by "put a log inside" ???
    – Yves Daoust
    Aug 24 at 6:25













up vote
1
down vote

favorite









up vote
1
down vote

favorite











iI would like to turn this



$$ e^z-1 = sum^infty_1 frac z^nn! $$



into this



$$ textsomething =sum^infty_1 frac z^nlog_e(n!) $$



Is this at all possible? thank you very much in advance!



Ps- i apologize of title not adequate , i did not know how to word it.




sequences-and-series complex-analysis complex-numbers infinity




share|cite|improve this question














iI would like to turn this



$$ e^z-1 = sum^infty_1 frac z^nn! $$



into this



$$ textsomething =sum^infty_1 frac z^nlog_e(n!) $$



Is this at all possible? thank you very much in advance!



Ps- i apologize of title not adequate , i did not know how to word it.





sequences-and-series complex-analysis complex-numbers infinity





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 6:20









Michael Hardy

205k23187463




205k23187463










asked Aug 24 at 5:41









Victor Orta

193




193




closed as unclear what you're asking by Kavi Rama Murthy, Yves Daoust, Hans Lundmark, Lord Shark the Unknown, Jyrki Lahtonen Aug 24 at 14:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as unclear what you're asking by Kavi Rama Murthy, Yves Daoust, Hans Lundmark, Lord Shark the Unknown, Jyrki Lahtonen Aug 24 at 14:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.













  • First, the second series does not make sense since $log(0!) = log(1) = 0$ is in the denominator.
    – 4-ier
    Aug 24 at 5:55










  • i apologize , i just meant it as a general example , any value would work.
    – Victor Orta
    Aug 24 at 5:56






  • 2




    Are you asking to evaluate the modified sum ? What do you mean by "put a log inside" ???
    – Yves Daoust
    Aug 24 at 6:25

















  • First, the second series does not make sense since $log(0!) = log(1) = 0$ is in the denominator.
    – 4-ier
    Aug 24 at 5:55










  • i apologize , i just meant it as a general example , any value would work.
    – Victor Orta
    Aug 24 at 5:56






  • 2




    Are you asking to evaluate the modified sum ? What do you mean by "put a log inside" ???
    – Yves Daoust
    Aug 24 at 6:25
















First, the second series does not make sense since $log(0!) = log(1) = 0$ is in the denominator.
– 4-ier
Aug 24 at 5:55




First, the second series does not make sense since $log(0!) = log(1) = 0$ is in the denominator.
– 4-ier
Aug 24 at 5:55












i apologize , i just meant it as a general example , any value would work.
– Victor Orta
Aug 24 at 5:56




i apologize , i just meant it as a general example , any value would work.
– Victor Orta
Aug 24 at 5:56




2




2




Are you asking to evaluate the modified sum ? What do you mean by "put a log inside" ???
– Yves Daoust
Aug 24 at 6:25





Are you asking to evaluate the modified sum ? What do you mean by "put a log inside" ???
– Yves Daoust
Aug 24 at 6:25











2 Answers
2






active

oldest

votes

















up vote
2
down vote













This is exactly same as your first sum as you said anything would work



$$something =sum^infty_1 frac z^nlog e^n!$$






share|cite|improve this answer



























    up vote
    0
    down vote













    I'm not even sure what you're asking, but $displaystyle sum_2^inftyfracz^nln(n!)>sum_2^inftyfracz^nln(n^n)=sum_2^inftyfracz^nnln(n)>sum_2^inftyfracz^nn^2$, the last of which does not converge (ratio between consecutive terms > $1$).



    Therefore, $displaystyle sum_2^inftyfracz^nln(n!)$ does not exist, but can a series have a log in the denominator? sure.






    share|cite|improve this answer



























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote













      This is exactly same as your first sum as you said anything would work



      $$something =sum^infty_1 frac z^nlog e^n!$$






      share|cite|improve this answer
























        up vote
        2
        down vote













        This is exactly same as your first sum as you said anything would work



        $$something =sum^infty_1 frac z^nlog e^n!$$






        share|cite|improve this answer






















          up vote
          2
          down vote










          up vote
          2
          down vote









          This is exactly same as your first sum as you said anything would work



          $$something =sum^infty_1 frac z^nlog e^n!$$






          share|cite|improve this answer












          This is exactly same as your first sum as you said anything would work



          $$something =sum^infty_1 frac z^nlog e^n!$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 24 at 6:03









          Deepesh Meena

          2,688721




          2,688721




















              up vote
              0
              down vote













              I'm not even sure what you're asking, but $displaystyle sum_2^inftyfracz^nln(n!)>sum_2^inftyfracz^nln(n^n)=sum_2^inftyfracz^nnln(n)>sum_2^inftyfracz^nn^2$, the last of which does not converge (ratio between consecutive terms > $1$).



              Therefore, $displaystyle sum_2^inftyfracz^nln(n!)$ does not exist, but can a series have a log in the denominator? sure.






              share|cite|improve this answer
























                up vote
                0
                down vote













                I'm not even sure what you're asking, but $displaystyle sum_2^inftyfracz^nln(n!)>sum_2^inftyfracz^nln(n^n)=sum_2^inftyfracz^nnln(n)>sum_2^inftyfracz^nn^2$, the last of which does not converge (ratio between consecutive terms > $1$).



                Therefore, $displaystyle sum_2^inftyfracz^nln(n!)$ does not exist, but can a series have a log in the denominator? sure.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  I'm not even sure what you're asking, but $displaystyle sum_2^inftyfracz^nln(n!)>sum_2^inftyfracz^nln(n^n)=sum_2^inftyfracz^nnln(n)>sum_2^inftyfracz^nn^2$, the last of which does not converge (ratio between consecutive terms > $1$).



                  Therefore, $displaystyle sum_2^inftyfracz^nln(n!)$ does not exist, but can a series have a log in the denominator? sure.






                  share|cite|improve this answer












                  I'm not even sure what you're asking, but $displaystyle sum_2^inftyfracz^nln(n!)>sum_2^inftyfracz^nln(n^n)=sum_2^inftyfracz^nnln(n)>sum_2^inftyfracz^nn^2$, the last of which does not converge (ratio between consecutive terms > $1$).



                  Therefore, $displaystyle sum_2^inftyfracz^nln(n!)$ does not exist, but can a series have a log in the denominator? sure.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 24 at 6:49









                  Saketh Malyala

                  7,1301532




                  7,1301532












                      -

                      這個網誌中的熱門文章

                      How to combine Bézier curves to a surface?

                      圖片
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
                      閱讀完整內容

                      Mutual Information Always Non-negative

                      圖片
                      Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
                      閱讀完整內容

                      Why am i infinitely getting the same tweet with the Twitter Search API?

                      圖片
                      up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
                      閱讀完整內容