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Infinite sum of \(n^kx^n\)

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

\(S=1+x+x^2+x^3+\cdots=\sum\limits^{\infty}_{n=0}x^n\)

\(xS=x+x^2+x^3+\cdots=\sum\limits^{\infty}_{n=1}x^n=\left(\sum^{\infty}_{n=1}x^n\right)+x^0-x^0=\left(\sum\limits^{\infty}_{n=0}x^n\right)-1=S-1\)

\(xS=S-1\)

\(1=S-Sx\)

\(1=(1-x)S\)

\(S=\frac{1}{1-x}=\sum\limits^{\infty}_{n=0}x^n\)

This answers the question for \(k=0\), to solve this for other \(k\), 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):

\(\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{(1-x)^n}\right]=\frac{n}{(1-x)^{n+1}}\)

\(\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{1-x}\right]=\frac{\mathrm{d}}{\mathrm{d}x}\left[\sum\limits^{\infty}_{n=0}x^n\right]\)

\(\frac{1}{(1-x)^2}=\sum\limits^{\infty}_{n=0}nx^{n-1}\)

Now I will multiply everything by \(x\), to adjust the right side of the equality:

\(\frac{x}{(1-x)^2}=\sum\limits^{\infty}_{n=0}nx^{n}\)

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):

\(\frac{x}{(1-x)^n}=\frac{1}{(1-x)^n}-\frac{1}{(1-x)^{n-1}}\)

Then I get:

\(\frac{1}{(1-x)^2}-\frac{1}{1-x}=\sum\limits^{\infty}_{n=0}nx^{n}\)

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

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

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

\(\sum\limits^{p+1}_{k=1}\frac{A_{p,k}}{(1-x)^{k}}=\sum\limits^{\infty}_{k=0}k^px^k\)

I will do the derivative of both sides:

\(\sum\limits^{p+1}_{k=1}\frac{kA_{p,k}}{(1-x)^{k+1}}=\sum\limits^{\infty}_{k=0}k^{p+1}x^{k-1}\)

Then I will multiply both sides by \(x\):

\(\sum\limits^{p+1}_{k=1}\frac{kA_{p,k}x}{(1-x)^{k+1}}=\sum\limits^{\infty}_{k=0}k^{p+1}x^{k}\)

Applying partial fractions:

\(\sum\limits^{p+1}_{k=1}\frac{kA_{p,k}}{(1-x)^{k+1}}-\sum\limits^{p+1}_{k=1}\frac{kA_{p,k}}{(1-x)^{k}}=\sum\limits^{\infty}_{k=0}k^{p+1}x^{k}\)

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

\(\frac{kA_{p,p+1}}{(1-x)^{p+2}}+\sum\limits^{p+1}_{k=2}\left(\frac{(k-1)A_{p,k-1}}{(1-x)^{k}}-\frac{kA_{p,k}}{(1-x)^{k}}\right )-\frac{A_{p,1}}{1-x}=\sum\limits^{\infty}_{k=0}k^{p+1}x^{k}=\sum\limits^{p+2}_{k=1}\frac{A_{p+1,k}}{(1-x)^{k}}\)

With this, I can say some things:

\(A_{p+1,p+2}=(p+1)A_{p,p+1}\)

\(A_{p+1,1}=-A_{p,1}\)

\(A_{p+1,k}=(k-1)A_{p,k-1}-kA_{p,k}\)

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

\(A_{p,p+1}=p!\)

\(A_{p,1}=(-1)^n\)

\(A_{p,p}=-\frac{(p+1)!}{2}\)

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
2 months ago

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

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

Ishan Singh - 2 months ago

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

Jon Haussmann - 2 months ago

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

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

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

Jon Haussmann - 2 months ago

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

Mark Hennings - 2 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 - 2 months ago

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

\(\frac{1}{1-x}=\sum\limits^{\infty}_{n=0}x^n\)

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

\(\frac{(p-1)!}{(1-x)^p}=\sum\limits^{\infty}_{n=0}(n)_px^{n-p+1}\)

\(\frac{(p-1)!}{(1-x)^p}=\sum\limits^{\infty}_{n=0}n^{(p)}x^{n}\)

Well, consider the following question: how we can determine the coefficients \(C_{p,n}\) such that:

\(\sum\limits^{p}_{k=1}C_{p,k}x^{(k)}=x^p\)

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

Matheus Jahnke - 2 months ago

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

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

\((x+n)x^{(n)}=x^{(n+1)}\)

\(x^{(n)}x+nx^{(n)}=x^{(n+1)}\)

\(x^{(n)}x=x^{(n+1)}-nx^{(n)}\)

Now, trying to find the recurrence around \(C_{p,n}\):

\(\sum\limits^{p}_{k=1}C_{p,k}x^{(k)}=x^p\)

Multiplying everything by \(x\):

\(\sum\limits^{p}_{k=1}C_{p,k}x^{(k)}x=x^{p+1}\)

Using the recurrence of the Rising Factorial:

\(\sum\limits^{p}_{k=1}C_{p,k}(x^{(k+1)}-kx^{(k)})=x^{p+1}\)

\(\sum\limits^{p}_{k=1}C_{p,k}x^{(k+1)}-\sum\limits^{p}_{k=1}C_{p,k}kx^{(k)}=x^{p+1}\)

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

\(x^{(p+1)}C_{p,p}+\sum\limits^{p}_{k=2}x^{(k)}(C_{p,k-1}-kC_{p,k})x^{(k)}-C_{p,1}=x^{p+1}\)

With that I can say:

\(C_{p+1,p+1}=C_{p,p}\)

\(C_{p+1, k}=C_{p,k-1}-kC_{p,k}\)

\(C_{p+1,1}=-C_{p,1}\)

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

\(C_{n,n}=1\)

\(C_{n,n-1}=-\frac{n(n-1)}{2}\)

\(C_{n,1}=(-1)^{n+1}\)

Matheus Jahnke - 2 months ago

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