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How would I come to a general form of this type of summation, similar to

$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ and $$\sum_{i=1}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ ? I found this post about calculating higher powers of i, but am unsure if this would apply for negative powers as well: Feyman's trick for Discrete calculus?, as his formula: $$\sum_{n=0}^N n^{m+1} = \frac{N^{m+2}}{m+2}+N^{m+1}$$ results in dividing by 0 (zero) for m = -2. Also, How could I extrapolate for the more general form $$\sum_{i=1}^n \frac{1}{i^k}$$ where k is some constant? This is my first StackExchange post, so feel free to give me tips on how to improve my posts. Thank you.

Edit: I have now learned that there is no rational closed form expression of the Harmonic numbers, But can someone explain why, or simplify another proof explaining why?

AlienRem
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Jonah
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    These are the so-called Harmonic Numbers and no simple closed form is known. – lulu Jan 01 '18 at 18:44
  • Note: This sum does not converge! – For the love of maths Jan 01 '18 at 18:44
  • And you are asking us to find $\zeta (1)$ except the limit is not $\infty$. – For the love of maths Jan 01 '18 at 18:46
  • Thank you lulu. Should I close this question or leave it open? – Jonah Jan 01 '18 at 18:47
  • @Mohammad Zuhair Khan: All the sums in Jonah's question (first version, in case later versions show up) are finite sums. – Dave L. Renfro Jan 01 '18 at 18:47
  • Why does convergence matter? These are finite sums. – Randall Jan 01 '18 at 18:48
  • @DaveL.Renfro I know that but I just stated these facts so that he can verify his answers by using $\lim_{n \to \infty}$ for sums with $k > 1$. – For the love of maths Jan 01 '18 at 18:49
  • @Randall I was just notifying him so that he does not treat it like a converging geometric series. – For the love of maths Jan 01 '18 at 18:50
  • Oops my bad! I forgot that it does not have any common ratio or difference. My apologies. – For the love of maths Jan 01 '18 at 18:53
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    "should I close or leave open?" Assuming the link to wikipedia that lulu provided fully answers your question, I would close it or answer it yourself, if you feel guilty about points then maybe as a community wiki. As for how you might improve your posts in the future, this is a rather well formed question including some of your background knowledge like it should. The only thing it suffers from is that it has a well known answer that a short phrase and a wikipedia link ought to be enough to satisfy all of your curiosity on the subject. Keep asking questions and welcome to the site! – JMoravitz Jan 01 '18 at 18:54
  • @JMoravitz well said. That piece of advice will also help me. Thanks! – For the love of maths Jan 01 '18 at 18:57
  • Thank you JMoravitz. I may leave it open in case of someone having an even better answer. – Jonah Jan 01 '18 at 18:58
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    Jonah, if you want to read about the divergence behavior of this series (as $n \rightarrow \infty),$ see my 1999 handout and Havil's book Gamma: Exploring Euler's Constant. Incidentally, I wrote that handout several years before Havil's book came out -- at the time there wasn't anything easily available I could share with my students about how slowly the series diverged. I had thought of expanding it, but then Havil's book appeared (which I then cited). – Dave L. Renfro Jan 01 '18 at 18:58
  • So, upon expanding something I did last summer over the sum of a series of fractions with consecutive denominators, I think I came up with an expression that may work. $\frac {\sum_{i=1}^N \frac{N!}{i}}{N!} What is wrong with this? It works for k =1. I did this right now. I will open a second post inquiring about this. – Jonah Jan 01 '18 at 19:09
  • The expression may work but it doesn't give you a closed form, you just rewrote the original series. – kingW3 Jan 01 '18 at 21:02
  • I realized that after posting. – Jonah Jan 01 '18 at 21:04
  • Sorry to have to vote to close this as a dupe. Your extra question whether this sum can be written as a rational funcion of $n$ is easy to handle. Take a look! – Jyrki Lahtonen Jan 01 '18 at 22:59
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    Welcome to Math.SE! Thanks for sharing your thoughts (+1) - you did fine. Better luck next time not to hit a duplicate :-) – Jyrki Lahtonen Jan 01 '18 at 23:02

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