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Today morning I had thought of a wonderful way to calculate the general sum of $$ \sum_{i=0}^{i=n} i^n$$

Using just things taught in elementary high school calculus. So, my method is as follows,

First consider the geometric series:

$$ S = \sum_{i=0}^{i=n} x^i$$

from the formula for geometric series,

$$ \frac{x^{n+1} - 1}{x-1} = \sum_{i=0}^{i=n} x^i$$

Rewriting this into the form in which leibniz product rule is more applicable,

$$ \sum_{i=0}^{i=n} x^i = (x^{n+1} - 1) (x-1)^{-1}$$

take a derivative on both sides,

$$ \sum_{i=0}^{i=n} i x^{i-1} = \frac{d}{dx} [ (x^{n+1} - 1) (x-1)^{-1}]$$

If I take limit as x goes to one on both sides I get formula for sum of first n integers,

now suppose I multiply both sides by x and take derivative,

$$ \sum_{i=0}^{i=n} i^2 x^{i-1} = \frac{d}{dx} [x\frac{d}{dx} [ (x^{n+1} - 1) (x-1)^{-1}]]$$

Taking limit as $ x \to 1$ gives me formula for first n squares,

now do same procedure again,

$$ \sum_{i=0}^{i=n} i^3 x^{i-1} = \frac{d}{dx} \left[x\frac{d}{dx} \left[x\frac{d}{dx} [ (x^{n+1} - 1) (x-1)^{-1}]\right]\right]$$

To generalize, let,

$$ P = x \frac{d}{dx}$$

And,

$$ P \cdot P =P^2 = x \frac{d}{dx} [ x \frac{d}{dx} ]$$

Then,

$$ \sum_{i=0}^{i=n} i^k = \lim_{x \to 1} P^k [ (x^{n+1} - 1) (x-1)^{-1}]$$

Is there a way to generally evaluate the limit? Thank you.

A-Level Student
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tryst with freedom
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    It seems OK now, but what's the question? – Vincenzo Tibullo Aug 25 '20 at 10:31
  • I want a way to generally evaluate the limit – tryst with freedom Aug 25 '20 at 10:33
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    This might be related to https://math.stackexchange.com/questions/2035188/methods-to-compute-sum-k-1nkp-without-faulhabers-formula – Arnaud D. Aug 25 '20 at 10:39
  • Would you mind justifying that the derivative of the summand with respect to $x$ is tantamount to the derivative of the formulae of the summation with respect to $n$? – A-Level Student Aug 25 '20 at 11:48
  • I have only seen verifications for it and ran with it. I haven't done a formal proof of it myself but from some searching I Found this stack exchange "https://math.stackexchange.com/questions/147869/interchanging-the-order-of-differentiation-and-summation#:~:text=While%20differentiation%20is%20linear%2C%20this%20does%20not%20extend%20to%20infinite%20sums.&text=Interchanging%20summation%20and%20differentiation%20is,for%20interchanging%20limits%20and%20differentials." – tryst with freedom Aug 25 '20 at 11:56
  • I find that the exponential series works better as a generating function, $$\frac{e^{(n+1)x}-e^x}{e^x-1}=\sum_{m=0}^\infty\frac{x^m}{m!}\sum_{i=1}^ni^m.$$ See also the Faulhaber formulas. – Lutz Lehmann Aug 25 '20 at 11:58
  • Looks good to me. If you wanted to prove it rigorously you would probably want to do a proof by induction. You might have to be a bit delicate when taking limits. Reminds me of the Rodriguez formula for Legendre polynomials. – Charles Hudgins Aug 25 '20 at 12:51
  • I'm not very well versed in exponential generating functions @LutzLehmann – tryst with freedom Aug 25 '20 at 15:10
  • Both sides are expressions for $\sum_{i=1}^ne^{ix}$. The left side can be treated as division of power series, then compare coefficients with the right side. This is best done using a computer algebra system, per hand you get perhaps to $m=5$ easily enough. – Lutz Lehmann Aug 25 '20 at 15:54
  • that did not answer my question..., why the close? – tryst with freedom Aug 25 '20 at 17:59
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    It certainly answers the titled question. I would think answers such as this one follow your line of logic, so it should answer anything more specific you are asking for. If by "generally evaluate" you are asking for a way to find a closed-form of the expression with the derivatives worked out, the end results are also given several times in the duplicate. – Simply Beautiful Art Aug 25 '20 at 18:02
  • Sidenote: The linked answer is the same as what @LutzLehmann was explaining, which is the same as your method but starting with $\sum e^{ix}$ instead of $\sum x^i$, which let's you avoid multiplying by $x$ between derivatives. The particular use of Taylor here is that you can avoid using derivative altogether by instead using algebraic manipulations of the expansion on the LHS, as they were suggesting. – Simply Beautiful Art Aug 25 '20 at 18:13
  • @LutzLehmann Now that I'm giving it another good look, it doesn't seem the method you've described is in the linked post. If you don't mind you could post an answer over there, otherwise I'll probably add such an answer sometime later. – Simply Beautiful Art Aug 25 '20 at 18:19
  • @SimplyBeautifulArt : I mean the same method as in your answer there, only instead of using differentiation to get to the coefficients, use Taylor arithmetic directly to get to the (truncated) power series. – Lutz Lehmann Aug 25 '20 at 18:40
  • @SimplyBeautifulArt I want the process of evaluation – tryst with freedom Aug 26 '20 at 08:08
  • As you have it, you simply apply your $P$ operator multiple times and take the limit. You'll find the limit may be easily reduced using algebraic means (long division) or other limit evaluation techniques (L'Hopital's rule for example). – Simply Beautiful Art Aug 27 '20 at 02:11

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