Infinite sum of nkxnn^kx^n

To "solve" this problem, I will consider the following geometric series, of course, |x|<1:







This answers the question for k=0k=0, to solve this for other kk, I will do the derivative of both sides, before that, I will use the following "formula", to help get things done faster(I used the Power Rule and the Chain Rule to reach this result without too much work):




Now I will multiply everything by xx, to adjust the right side of the equality:


I will apply partial fractions on the left side of the equality, using the following formula(you can prove it by simply doing the operations):


Then I get:


Note to get the answer for k=2k=2, I can do the derivative of both sides, then multiply by xx, and then apply partial fractions, but then I can do for k=3k=3 and so on;

Does anyone know how to solve the recurrence relation I get?

Let's suppose that I know the answer for k=pk=p, there is a sequence of Ap,nA_{p,n} as the coefficients of 1(1x)n\frac{1}{(1-x)^n} such that:


I will do the derivative of both sides:


Then I will multiply both sides by xx:


Applying partial fractions:


This is the solution for k=p+1k=p+1, therefore I can say that this is equal to the sum of 1(1x)k\frac{1}{(1-x)^k}, with the appropriate coefficients Ap+1,kA_{p+1,k}, also, I will take out the last and first coefficients because they don't have pairs, with that I can say:

kAp,p+1(1x)p+2+k=2p+1((k1)Ap,k1(1x)kkAp,k(1x)k)Ap,11x=k=0kp+1xk=k=1p+2Ap+1,k(1x)k\frac{kA_{p,p+1}}{(1-x)^{p+2}}+\sum\limits^{p+1}_{k=2}\left(\frac{(k-1)A_{p,k-1}}{(1-x)^{k}}-\frac{kA_{p,k}}{(1-x)^{k}}\right )-\frac{A_{p,1}}{1-x}=\sum\limits^{\infty}_{k=0}k^{p+1}x^{k}=\sum\limits^{p+2}_{k=1}\frac{A_{p+1,k}}{(1-x)^{k}}

With this, I can say some things:




There are some patterns I can see(they can be proved by induction):




But I can't give a complete closed "simple" formula, like Faulhaber's Formula, does anyone know or has a suggestion to tackle this problem?

Note by Matheus Jahnke
3 years, 9 months ago

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The recurrence sequence can be expressed in terms of Stirling Numbers of the Second Kind.

k=0knxk=k=0nk!  S(n+1,k+1)(x1x)k+1 \sum _{k=0}^{\infty} k^n x^k = \sum_{k=0}^n k! \; S(n+1,k+1) \left( \dfrac x{1-x} \right)^{k+1}

Ishan Singh - 3 years, 9 months ago

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Absolutely, but I have not been able (yet) to obtain a closed expression the coefficients of 1(1x)k\tfrac{1}{(1-x)^k}. There is an obvious formula in terms of Stirling numbers and Binomial coefficients...

Mark Hennings - 3 years, 9 months ago

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Apparently k=1knxk=k=1n+1(1)n+k+1S(n+1,k)(k1)!(1x)k.\sum_{k = 1}^\infty k^n x^k = \sum_{k = 1}^{n + 1} \frac{(-1)^{n + k + 1} S(n + 1,k) (k - 1)!}{(1 - x)^k}.

Jon Haussmann - 3 years, 9 months ago

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And here is a proof by induction: It is easy to verify that the formula holds for n=1n = 1, so assume that it holds for some positive integer nn, i.e. k=1knxk=k=1n+1(1)n+k+1S(n+1,k)(k1)!(1x)k.\sum_{k = 1}^\infty k^n x^k = \sum_{k = 1}^{n + 1} \frac{(-1)^{n + k + 1} S(n + 1,k) (k - 1)!}{(1 - x)^k}.

Differentiating both sides, we get k=1kn+1xk1=k=1n+1(1)n+k+1S(n+1,k)k!(1x)k+1,\sum_{k = 1}^\infty k^{n + 1} x^{k - 1} = \sum_{k = 1}^{n + 1} \frac{(-1)^{n + k + 1} S(n + 1,k) k!}{(1 - x)^{k + 1}}, so k=1kn+1xk=k=1n+1(1)n+k+1S(n+1,k)k!x(1x)k+1.\sum_{k = 1}^\infty k^{n + 1} x^k = \sum_{k = 1}^{n + 1} \frac{(-1)^{n + k + 1} S(n + 1,k) k! x}{(1 - x)^{k + 1}}.

We can write x(1x)k+1=(x1)+1(1x)k+1=1(1x)k+11(1x)k,\frac{x}{(1 - x)^{k + 1}} = \frac{(x - 1) + 1}{(1 - x)^{k + 1}} = \frac{1}{(1 - x)^{k + 1}} - \frac{1}{(1 - x)^k}, so k=1n+1(1)n+k+1S(n+1,k)k!x(1x)k+1=k=1n+1((1)n+k+1S(n+1,k)k!(1x)k+1(1)n+k+1S(n+1,k)k!(1x)k)=(1)n+11x+k=2n+1((1)n+kS(n+1,k1)(k1)!(1x)k(1)n+k+1S(n+1,k)k!(1x)k)+(n+1)!(1x)n+2=(1)n+11x+k=2n+1(1)n+k(k1)!(1x)k[S(n+1,k1)+kS(n+1,k)]+(n+1)!(1x)n+2=(1)n+11x+k=2n+1(1)n+kS(n+2,k)(k1)!(1x)k+(n+1)!(1x)n+2=k=1n+2(1)n+kS(n+2,k)(k1)!(1x)k. \begin{aligned} &\sum_{k = 1}^{n + 1} \frac{(-1)^{n + k + 1} S(n + 1,k) k! x}{(1 - x)^{k + 1}} \\ &= \sum_{k = 1}^{n + 1} \left( \frac{(-1)^{n + k + 1} S(n + 1,k) k!}{(1 - x)^{k + 1}} - \frac{(-1)^{n + k + 1} S(n + 1,k) k!}{(1 - x)^k} \right) \\ &= \frac{(-1)^{n + 1}}{1 - x} + \sum_{k = 2}^{n + 1} \left( \frac{(-1)^{n + k} S(n + 1,k - 1) (k - 1)!}{(1 - x)^k} - \frac{(-1)^{n + k + 1} S(n + 1,k) k!}{(1 - x)^k} \right) + \frac{(n + 1)!}{(1 - x)^{n + 2}} \\ &= \frac{(-1)^{n + 1}}{1 - x} + \sum_{k = 2}^{n + 1} \frac{(-1)^{n + k} (k - 1)!}{(1 - x)^k} [S(n + 1,k - 1) + kS(n + 1,k)] + \frac{(n + 1)!}{(1 - x)^{n + 2}} \\ &= \frac{(-1)^{n + 1}}{1 - x} + \sum_{k = 2}^{n + 1} \frac{(-1)^{n + k} S(n + 2,k) (k - 1)!}{(1 - x)^k} + \frac{(n + 1)!}{(1 - x)^{n + 2}} \\ &= \sum_{k = 1}^{n + 2} \frac{(-1)^{n + k} S(n + 2,k) (k - 1)!}{(1 - x)^k}. \end{aligned} This completes the induction step.

Jon Haussmann - 3 years, 9 months ago

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This is an interesting question. My first thought is Falling factorials, but I don't think that's helpful.

Summoning @Mark Hennings + @Ishan Singh

Pi Han Goh - 3 years, 9 months ago

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There is a chance of this(or the Rising Factorial) being helpful:


dp1dp1x[11x]=dp1dp1x[n=0xn]\frac{\mathrm{d}^{p-1}}{\mathrm{d}^{p-1}x}\left[\frac{1}{1-x}\right ]=\frac{\mathrm{d}^{p-1}}{\mathrm{d}^{p-1}x}\left[\sum\limits^{\infty}_{n=0}x^n \right ]



Well, consider the following question: how we can determine the coefficients Cp,nC_{p,n} such that:


This question is closely related to the problem I discussed earlier, because then Ap,n=(n1)!Cp,nA_{p,n}=(n-1)!C_{p,n}

Matheus Jahnke - 3 years, 9 months ago

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This problem may be easier to tackle because it has an easier recurrence relation:

x(n)=x(x+1)(x+2)x(x+n1)x^{(n)}=x(x+1)(x+2)\cdots x(x+n-1)




Now, trying to find the recurrence around Cp,nC_{p,n}:


Multiplying everything by xx:


Using the recurrence of the Rising Factorial:



I will take the first and last terms because they don't have pairs:


With that I can say:


Cp+1,k=Cp,k1kCp,kC_{p+1, k}=C_{p,k-1}-kC_{p,k}


Again, I can find some patterns that can be proved by induction:




Matheus Jahnke - 3 years, 9 months ago

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