Proving a recursive sequence of polynomials is equal to a series

Question

The sequence of polynomials $f_k(x)$ is defined recursively as:

$$f_1(x)=x$$ and $$f_k(x)=x(1-x)f'_{k-1}(x) +kxf_{k-1}(x). $$

I also know that $f_k(1)=k!$ and $f'_k(0)=1$.

The identity I want to prove is:

$$\frac{f_k(x)}{(1-x)^{k+1}}=\sum_{n=1}^{\infty} n^kx^n $$

assuming that $|x|<1.$

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So far I've done this:

Starting with the LHS:

$$\frac{f_k(x)}{(1-x)^{k+1}}=\frac{x(1-x)f'_{k-1}(x) +kxf_{k-1}(x)}{(1-x)^{k+1}}=\frac{xf'_{k-1}(x)}{(1-x)^{k}}+\frac{kxf_{k-1}(x)}{(1-x)^{k+1}}$$

And I know the RHS expands to:

$$\sum_{n=1}^{\infty} n^kx^n=1^kx^1+2^kx^2+3^kx^3+...$$

After this I'm not sure how to link the two sides.

Can anyone give me any hints on how to manipulate the LHS further?

Answer 1

Perhaps assuming this inductively will help, since then you know by the inductive hypothesis, $$ \frac{f_{k-1}(x)}{(1-x)^k} = \sum_{n=1}^\infty n^{k-1}x^n, $$ or in other words, $$ f_{k-1}(x) = (1-x)^k \sum_{n=1}^\infty n^{k-1}x^n, $$ and you can find $f_{k-1}'(x)$ explicitly. Use it and prove for $f_k(x)$, and validate for $k=0$ or $k=1$ and you are done for all integer $k$ after your initial validation...

Answer 2

You already have $$\frac{f_k(x)}{(1-x)^{k+1}}=\frac{x\,f'_{k-1}(x)}{(1-x)^{k}}+\frac{k\,x\,f_{k-1}(x)}{(1-x)^{k+1}}=x\,\frac{\text{d}}{\text{d}x}\,\frac{f_{k-1}(x)}{(1-x)^k}\,.$$ Then, use induction on $k$.