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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.

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

- 2 months ago

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

- 2 months ago

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.

- 2 months ago

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...

- 2 months ago

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

Summoning @Mark Hennings + @Ishan Singh

- 2 months ago

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 $$A_{p,n}=(n-1)!C_{p,n}$$

- 2 months ago

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:

$$C_{n,n}=1$$

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

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

- 2 months ago